Tailieu.vncty.com c++ for beginners......masters 2007
4 likes2,542 views
This document discusses the evolution of computers and programming languages. It describes how Charles Babbage invented the difference engine in 1822, which executed tasks by changing gears. The ENIAC computer used in 1942 operated using electrical signals instead of physical motion. John von Neumann developed the concepts of shared program technique and conditional control transfer in 1945, enabling computers to be reprogrammed faster using subroutines. The first computing language was developed in 1949 called Short Code, requiring programmers to manually convert statements into binary. This led to the development of machine language using 0s and 1s.
![145Part I: Chapter 5
Arrays
void main()
{
clrscr();
int num1, num2, num3, num4, num5;
int sum=0;
cout<<” Enter the five values”;
sum=num1+ num2+ num3+ num4+ num5;
cout<<” The sum of the five values “<<sum;
}
The above piece of code declares five different locations in the computer memory for
processing five integer values. By using arrays, we can allocate these five memory
locations by a single variable name called array with a subscript value ‘5’. A subscript is
a number, within brackets, that differentiates one element of an array from another.
Elements are the individual variables in an array. To declare an array the following
syntax is used:
Syntax:
Where ‘data_type’ is any C++ data type, ‘array_name’ is the name of the array and
‘subscripted range’ is the length of the array.
For example,
int employees[5];
This statement declares an array ‘employees’ of type integer and it can store ‘5’ integer
values. Fig 5.2 shows how this array ‘employees’ is represented in the computer’s
memory.
data_type array_name[subscripted range];
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-160-320.jpg)
![146
Figure 5.2: Array 'employees' in the computer’s memory
The indexing of an array always starts from ‘zero till ‘size of the array -1’. The array
elements are always stored in contagious memory locations (‘100’,’102’...). Data entries
in an array are accessed by using square brackets with an index value. For example, to
access the first index of array ‘employees’ we would write ‘employees [0];’. Similarly to
access the second index of the array we would write ‘employees [1];’ and so on.
5.2.1 Solved Examples
Example 5.2.1.1
Write a program to enter five integer values from the user and print their sum on the
screen.
Solution of 5.2.1.1
// Header Files
# include<iostream.h>
# include<conio.h>
// Main Function
void main()
{
// Declaring an array 'num'
int num[5];
int i, sum=0;
// Inputting the values in the array
for ( i=0; i<5;i++ )
{
cout<<" Enter value "<<i<<"->";
cin>>num[i];
}
// Calculating the sum of the values
for ( i=0; i<5;i++)
{
sum = sum + num[i];
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-161-320.jpg)
![147Part I: Chapter 5
Arrays
cout<<" The sum of all the five elements is"<<sum;
}
Output Screen
Let us try to understand how this program actually works. We begin by declaring an array
by the statement: ‘int num[5];’. This tells the compiler to set aside the memory space for
5 integer variables grouped together under the name ‘num’. As one might notice in the
figure given below these ‘5’ integer variables which are grouped together are stored in
contagious memory locations and can be accessed by changing the subscripts assigned
with the array.
Figure 5.3: Array declaration
Once this is done next comes entering elements in our array ‘num’. In example 5.2.1.1
this is done using the following piece of code.
// Inputting the values in the array
for ( i=0; i<5;i++ )
{
cout<<” Enter value “<<i<<”->”;
Enter value 0 -> 100
Enter value 1 -> 50
Enter value 2 -> 400
Enter value 3 -> 300
Enter value 4 -> 250
The sum of all the five elements is 1100
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-162-320.jpg)
![148
cin>>num[i];
}
The index variable ‘i’ takes values from ‘0’ to ‘4’ within the ‘for’ loop which are used to
enter the array elements of array ‘num’. Let us assume that the user enters the values
‘100’, ‘50’, ‘400’, ‘300’ and ‘250’. These values are then assigned to num[0], num[1], …
num[4], i.e., in the order they were entered by the user (see fig. 5.4) which are finally
used to calculate the sum of the array elements displayed on the screen.
Figure 5.4: Array 'num’
Example 5.2.1.2
Suppose that in your computer science class, each student secures a certain percentage of
marks and the teacher decides to give grace of 10 % to each student. If the class contains
20 students, write a C++ program to do the needful?
Solution of 5.2.1.2
// Header Files
# include <iostream.h>
# include <conio.h>
// Main Function
void main()
{
// Variable Declrations
float per[20];
int i;
cout<<" Enter the students percentages for the 20 students";](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-163-320.jpg)
![149Part I: Chapter 5
Arrays
// Entering student information from the student
for( i=0;i<20;i++)
{
cin>>per[i];
}
// Display the increased percentage of the 20 students
for (i=0;i<20;i++)
{
per[i] = per[i] + 10;
cout<<per[i]<<endl;
}
getch();
}
5.3 Passing arrays as parameters to functions
Arrays are passed to functions as reference parameters. To pass an array argument to a
function the array name is specified without any brackets. The syntax for passing arrays
to functions is given below:
The subscripts of all the arrays should always begin with 0. For
example in the array int a [5], the subscripts should begin with 0
as explained above.
When using loops to initialize values into the array one should
make sure that the index variable always contains a positive
value.
If the number of values assigned exceeds the size of the array or
the program tries to access an index outside the bounds of the
array such a statement is an example of a logical error. As the
concurrent C++ compilers are very user friendly the compiler
may not give an error but it will although produce unexpected
results.
For example,
int A[3]={9,5,2,3} is an incorrect statement as we
are assigning an integer value ‘3’ to unallocated set of
memory
Common Programming Errors
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-164-320.jpg)
![150
Syntax:
return_type function_name( data_type array_name)
For example, if array ‘weeklyupdate’ is defined as:
int weeklyupdate[10];
then the function call for a function ‘modifyarray’ which takes in an array as a parameter
would look like ‘modifyarray(weeklyupdate);’. Let us take up a program to further
understand the concept of passing arrays to a function.
Program 5.1
// Header Files
# include<iostream.h>
# include<conio.h>
void calculate( int arr[5])
{
for (int i=0 ; i<5 ;i++)
{
arr[i] = arr[i] + 10;
}
}
void main()
{
clrscr();
int array_example[5];
cout<<" Enter the elements of the array";
for (int a=0; a<5; a++)
{
cin>>array_example[a];
}
// Modifying array
calculate(array_example);
cout<<" The modified array is as followsn";](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-165-320.jpg)
![151Part I: Chapter 5
Arrays
for (a=0; a<5;a++)
{
cout<<array_example[a]<<"n";
}
getch();
}
Now in this program, we have passed the array ‘array_example’ as a parameter to
function ‘calculate’ which increments each array entry by 10 using the concept that
arrays are passed to functions as reference parameters. That is the array ‘arr’ defined in
the function call for functions ‘calculate’ acts as a formal parameter to ‘array_example’.
Hence any changes made to this formal parameter are also reflected back to the actual
parameter ‘array_example’. The output of this program is given below (Fig. 5.5):
Output Screen
Figure 5.5: Output Screen
5.4 Traversing of Arrays
It is an operation in which each element of an array (i.e. from first element of an array to
the last element of the array) is visited. Processing of the visited element can be done
according to the requirements of the problem. Program 5.2 shows how traversal of arrays
takes place.
Program 5.2: Traversal in an array
// Preprocessor Directives
# include <iostream.h>
# include <conio.h>
void main()
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-166-320.jpg)
![152
{
clrscr();
int a[15],i,n;
do
{
cout<<” Enter the number of elements in array less than 16”;
cin>>n;
}
while(n>15);
// Traversing
// Reading the array
for(i=0;i<n;i++)
{
cout<<” Enter “<<i+1<<” element : “;
cin>>a[i];
// Displaying original array
cout<<” n Original Array “;
for (i=0;i<n;i++)
cout<<a[i]<<” “;
// Traversing (a[i]=a[i]*10)
for(i=0;i<n;i++)
a[i]=a[i]*10;
// Displaying the transformed array
cout<<” n The transformed array :- n“;
for (i=0;i<n;i++)
cout<<a[i]<<” “;
getch();
}
5.5 Searching element in an array
The process of finding a particular element in an array is called as searching. This process
can be carried out in two ways.
5.5.1 Linear Search
In linear search, each element of the array is compared with a search ‘key’. The process
of carrying out linear search is given below in program 5.3.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-167-320.jpg)
![153Part I: Chapter 5
Arrays
Program 5.3: Linear Search
// Header Files
# include<iostream.h>
# include<conio.h>
// Function Prototypes
void display (int [], int);
void input (int [], int &);
int lsearch (int a[],int n, int data);
// Main Function
void main()
{
// Variable and array declrations
int a[10], n, data;
input (a, n);
cout<<" n The Array ->"<<endl;
display (a,n);
cout<<" n Enter element to be searched for: ";
cin>>data;
int x= lsearch(a,n,data);
if (x!=-1)
{
cout<<data<<" Present at location "<<x<<endl;
}
else
{
cout<<" Not present";
}
getch();
}
/*****************
Function:lsearch
Parameters: a[], n, data
Purpose: To search for the desired data element
entered by the user in the array
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-168-320.jpg)
![154
************************************************/
int lsearch (int a[],int n,int data)
{
int c=0,found=0;
while ((c<n) && !(found))
{
if (a[c]==data)
{
found++;
}
else
c++;
}
if (found)
{
return (c+1);
}
else
{
return (-1);
}
}
/*****************
Function:input
Parameters: a[], n
Purpose: To enter the elements of the array
*********************************************/
void input (int a[],int &n)
{
int i;
do
{
cout<<" Enter the number of elements <= 10:";
cin>>n;
}
for(i=0;i<n;i++)
{
cin>>a[i];
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-169-320.jpg)
![155Part I: Chapter 5
Arrays
/*****************
Function:display
Parameters: a[], n
Purpose: To display the elements of the array
**********************************************/
void display (int a[], int n)
{
int i;
cout<<" The elements of the array are: n";
for (i=0;i<n;i++)
{
cout<<a[i]<<endl;
}
}
5.5.2 Binary search
To understand the concept of binary search lets take up the following example. Let’s
say one is looking for the name ‘Paul’ in a phone book. Open up the phone book
approximately half way and look for the name on the opened page. What does it say?
Probably one sees a name that begins from ‘M’. One would then think does ‘Paul’
come before or after ‘M’ in the phone book and realize that it comes in the latter
portion of the phone book As a result of this we can ignore the entire first half of the
phone book (Part before ‘M’). Now open the remaining portion of the book half way.
One is probably some where near the data entries starting with the alphabet ‘T’ in the
phone book.
We can then again ask our selves the same question does ‘Paul’ come before or after
data entries starting from alphabet ‘T’. As the data entry ‘Paul’ comes before data
entries starting from alphabet ‘T’, we can ignore the latter half of the phone book and
then keep searching in the volume between letters ‘M’ and ‘T’.
Repeating this process, we would eventually find the data entry ‘Paul’ we are looking
for. This is how most of us search for data in huge volumes like a dictionary or a
phone book.
This process of searching data through a sorted list is also known as binary search. In
brief the process is to search a sorted array by repeatedly dividing the search interval
in half. Begin with an interval covering the whole array. If the value of the search key
is less than the item in the middle of the interval, narrow the interval to the lower half.
Otherwise narrow it to the upper half. Repeatedly check until the value is found or the
interval is empty.
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-170-320.jpg)
![156
The algorithm for binary search is given below:
Algorithm:
o The element to be searched for is taken from the user say the ‘Key’.
o The given data array is taken. Any arbitrary lower limit or the lower bound
for the data values is taken. Similarly one defines the upper limit or the upper
bound. Then we obtain the middle value using.
Middle = (lower bound + upper bound)/2;
o The search ‘key’ is compared to the Middle value if ‘Key=Middle’ then
the search is complete. If Key<Middle, the program continues and the search
is confined to the first half only. On the other hand if Key>Middle then also
the program continues and the search is confined to the latter half. This
process is continued till the time ‘Key = Middle’.
Using binary search in a phone book containing one million entries, one can locate
any specific data entry in at most 20 comparisons. The program to carry out binary
search is given below:
Problem 5.4: Binary Search
/*************
Name:bsearch
Parameters: ‘a’, ‘n’, ‘data’
Purpose: To perform binary search to sorted array ‘a’
*******************************************/
int bsearch ( int a[], int n, int data)
{
int mid,lbound=0,ubound=n-1,found=0;
while ((lbound<=ubound) && !(found))
{
mid = (lbound + ubound)/2;
if (data > a[mid] )
lbound = mid +1;
if (data <a [mid] )
ubound = mid -1;
else
found++;
}
if (found)
return (mid + 1); // Returning location, if present
else
return (-1); // Returning -1, if not present](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-171-320.jpg)
![157Part I: Chapter 5
Arrays
}
Common Programming Errors
Binary search only works when the array to which it is applied is sorted.
5.6 Merging of arrays
This operation merges two sorted arrays into a single sorted array. Suppose ‘a’ is a sorted
array with ‘r’ elements and ‘b’ is the sorted array with ‘s’ elements, then the operation
that combines the elements of ‘a’ and ‘b’ into a single array ‘c’ is called merging.
Program 5.4 provides you with the code to merge two sorted arrays ‘a’ and ‘b’.
Program 5.4: To merge two arrays (sorted in ascending order)?
// Header Files
# include<iostream.h>
# include<conio.h>
// Main Function
void main()
{
const int r=5; // size of array A
const int c=5; // size of array B
const int m=10; // size of the merged array A and array B
int a[r],b[c],res[m],i,j,k;
// Enter Details of the first sorted array
cout<<" Enter the first sorted array ==> "<<endl;
for(i=0;i<r;i++)
{
cout<<" Element "<<i+1<<":";
cin>>a[i];
}
// Enter the details of the second array
cout<<" Enter the second sorted array ===>"<<endl;
for(i=0;i<c;i++)
{
cout<<" Element "<<i+1<<":";
cin>>b[i];
}
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-172-320.jpg)
![158
// Initializing all index values to zero
i=0, j=0, k=0;
// Merging Arrays 'A' and 'B'
while ((i<r) && (j<c))
{
if(a[i]<b[j])
{
res[k++]=a[i++];
}
else if (a[i]>b[j])
{
res[k++]=b[j++];
}
else
{
res[k++]=a[i++];
j++;
}
}
int t;
for (t=i;t<r;t++)
{
res[k++]=a[t];
}
for (t=j;t<c;t++)
{
res[k++]=b[t];
}
// Displaying Merged Array
cout<<" Merged array ==>";
for (i=0;i<k;i++)
{
cout<<endl<<res[i];
}
getch();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-173-320.jpg)
![159Part I: Chapter 5
Arrays
5.7 Sorting with arrays
Sorting is one of the most important operations performed by computers. The history of
sorting goes back to the days when magnetic tapes were used for data storage .Before the
emergence of modern databases, sorting was certainly the most common operation
performed by computers as most database updating was done by sorting transactions.
Sorting still plays an important part for presentation of data extracted from databases as
most people prefer to get reports sorted into some relevant order before wading through
series of data. We will discuss two of the simplest method of sorting ‘bubble’ and
‘selection’ which will give you an idea of how sorting takes place.
Bubble Sort
Bubble sort is the simplest but unfortunately also the slowest method of sorting arrays.
The basic idea when using bubble sort is to compare two neighboring objects and to swap
them if they are present in the incorrect order. Given an integer array ‘X’, of size ‘N’ the
piece of code to employ bubble sort is given below:
// Bubble Sort in Ascending Order
for (i=0; i<N; i++)
{
for (j=0; j<N-1-i;j++)
{
if (X[j+1]<X[j]) /* Compare two neighboring elements */
{
temp = X[j]; /* Swap X[j] and X[j+1] */
X[j] = X[j+1];
X[j+1]=temp;
}
}
}
As one might observe the piece of code given above consists of two nested for loops. The
inner loop having ‘j’ as the index variable is used to compare adjacent entries in the array
( ‘j’ and ‘j+1’). While the outer loop causes the inner loop to make repeated passes
through the array in the process sorting the array. If the data elements are being sorted in
the ascending order then after the first pass the largest element is shifted to the last
position in the array. Similarly, after the second pass the second largest data element is
placed next to the largest element. Which is the reason that in the inner loop the upper
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-174-320.jpg)
![161Part I: Chapter 5
Arrays
the smallest data element in the unsorted list and then exchanging it with the data element
in first position of the array. Once this is done, it then selects the second smallest element
and exchanges it with the data element at the second position of the array. This process is
carried on till the entire array has been sorted. Given an integer array ‘X’, of size ‘N’ the
piece of code to employ selection sort is given below:
for (i = 0; i < N-1; i++)
{
min = i;
for (j = i+1; j < N; j++)
{
if (X[j] < X[min])
min = j;
}
temp = X[i];
X[i] = X[min];
X[min] = temp;
}
Figure 5.6 explains the working of selection sort with the help of an example.
Array X
Pass 1
Pass 2
Pass 3
Pass 4
Figure 5.6: Selection Sort
5.8 Arrays as strings
C++ does not have a string data type rather it implements strings as single-dimension
character arrays. A string is defined as a character array that is terminated by a null character
‘/0’. For this reason, the character arrays are declared one character longer than the largest
string they can hold. This makes room for the null character at the end of the string.
Individual characters of a string can be easily accessed as they make the elements of the
character array. The index – refers to the first character, the index 1 to the second, 2 to the
third, and so forth. The end of a string is determined by checking for a null character. For
example,
21 60 29 52 17
17 60 29 52 21
17 21 29 52 60
17 21 29 52 60
17 21 29 52 60
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-176-320.jpg)
![162
A character array can be initialized using a string literal. For example,
char myfirststring[] = “Hello”;
This statement declares a character string ‘myfirststring’. The size of the string is
automatically determined by the compiler based upon the length of the string. In this
example, the compiler would determine the size of the string as ‘6’. As this string consists of
‘5’ characters and a special null character (‘0’) included at the end of each string.
Strings can also be declared using the following syntax:
char myfirststring[] = {‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘0’};
When assigning the null character ‘0’ you might think that we are assigning two characters
’’ and ‘0’ but this is not true the ‘’ symbol just tells the compiler that you are dealing with a
special character which happens to be the null character ‘0’. As strings also are array of
characters, they can also be accessed as any integer array. For example, myfirststring[0] is
the character ‘H’ and myfirststring[1] is the character ‘e’. Figure 5.7 given below shows how
strings are stored in the computer’s memory.
Figure 5.7: 'myfirststring'
The functions to display and print the string on the screen are defined under the <stdio.h>
header file and are given in table 5.1.
Table 5.1: The <stdio.h> header file
gets(string_name);
The function assigns the string
inputted by the user to the parameter
passed to it.
For example,
char string_name[15];
Function Usage](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-177-320.jpg)
![5.8.1 String functions
The <string.h> header file enlists a number of useful functions for dealing with null
terminated strings. A brief description of the functions is given in table 5.2.
Table 5.2: The <string.h> header file
gets(string_name);
Returns the string inputted by the user to the character
array ‘string_name’. It works exactly like the ‘cin’
function defined under the <conio.h> header file which
is used to enter integers, floats and characters.
puts(string_name);
The function prints the character array passed to it as a
parameter on the screen. It works exactly like the ‘cout’
function which is used to display integers, floating
points and characters on the screen.
For example,
char string_name[]=“This is my first string”;
puts(string_name);
Output Screen
Function Description
strlen(string_name);
‘string_name’ is a string
The function calculates the length of the string passed to it as a
parameter (‘string_name’) by counting the number of
characters from the 0th
index of the array to the index
containing the null character marking the end of the string.
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-178-320.jpg)
![164
For example,
char string_name[20];
cout<<" Please enter the string: ";
gets(string_name);
cout<<" The length of the string
is:"<<strlen(string_name);
Output Screen
strcpy(string1,string2)
Where ‘string1’ is the destination
string and
‘string2’ is the source
string.
C++ does not allow the programmer to directly assign one
string to the other. For example the statement ‘string1 =
string2’ will result in a syntax error. Hence for copying or
assigning strings to a character array the ‘strcpy’ function is
used. The function copies the string stored in ‘string2’ to
the first string ‘string1’ which are passed to it as
parameters. Let us look at a few examples to understand
how this function works:
Examples,
Output Screen
char string1[]="My first string";
char string2[]="My last string";
cout<<strcmp(string1,string2);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-179-320.jpg)
![165Part I: Chapter 5
Arrays
Output Screen
Strcmp(string1,string2 The function is used to compare two strings passed to it as
parameters by the programmer. The function starts comparing
from the first character in each string and continues with
subsequent characters until the corresponding characters differ
or until the end of the strings is reached. Based upon this
comparison the function returns the following values.
Return Value Comparison
‘0’
It returns a value ‘0’ if both the strings
passed to it as parameters (‘string1’,
‘string2) are equivalent.
Example,
‘>0’
It returns a value ‘>0’ if ‘string1’ is greater than
‘strings2’. The magnitude of the value returned
is the difference between the ASCII values of
the first dissimilar characters observed during the
char string1[20], string2[20];
cout<<" Enter a string:";
gets(string2);
strcpy(string1,string2);
cout<<" The entered string is:";
puts(string1);
char string1[]="My first string";
char string2[]="My last string";
cout<<strcmp(string1,string2);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-180-320.jpg)
![166
Strcat(string1,strings2);
Where ‘string1’ and
‘string2’ are both strings.
String comparison
For example,
The function will return the value ‘6’ as a
result of the comparison between the strings
passed to it. This is so because the first
difference between the two strings occurs at
the 4th
index. As ‘1’ and ‘f’ are not equivalent
the difference in their ASCII values comes out
to be ‘+6’.
The functions returns a value (<0) if ‘string2’
is greater than ‘string1’.
For example,
When this piece of code is executed then a
value of ‘-6 is observed on the screen instead
of ‘+6’ as in the previous example when
‘string1’ was greater than ‘string2’.
The functions strcat appends a copy of ‘string2’ at the end of
‘string1’. As a result of which the length of ‘string1’ becomes
(strlen(string1) + strlen (string2)).
For example,
char string1[]="My first string";
char string2[]="My last string";
cout<<strcmp(string1,string2);
char string1[]="My first string";
char string2[]="My last string";
cout<<strcmp(string1,string2);
Char string1 []=”My first”;
char string1[20];
strcpy(string1,"My first string");
puts(string1);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-181-320.jpg)
![167Part I: Chapter 5
Arrays
5.8.2 Solved Examples
Example 5.8.2.1
Write a program to check whether the string entered by the user is a palindrome or not?
Solution 5.8.2.1
// Preprocessor Directives
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<string.h>
// Main Function
void main()
{
clrscr();
char string[40],i;
int j=0,flag=0;
cout<<" Enter string";
gets(string);
j=strlen(string)-1;
for (i=0;i<strlen(string)-1;i++)
{
if (string[i] != string[j])
{
flag++;
}
j--;
}
if (flag==0)
Output Screen
ARRAYS
My first string](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-182-320.jpg)
![168
{
cout<<" n |Entered string is palindrome|";
}
else
cout<<" Entered string is not a palindrome";
getch();
}
OUTPUT SCREEN
Example 5.8.2.2
Write a program to enter a string from the user and then display the inputted string with
all the spaces converted into ‘#’s?
Solution of 5.8.2.2
// Preprocessor Directives
# include<iostream.h>
# include<conio.h>
# include<string.h>
# include<stdio.h>
// Main Function
void main()
{
clrscr();
char string[15];
cout<<" Enter the string";
gets(string);
for (int i=0;i<strlen(string);i++)
{
if (string[i]==' ')
{
string[i]='#';
}
}
cout<<" n Displaying the converted string n";
puts(string);
Enter the string MaM
|Entered string is a palindrome|](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-183-320.jpg)
![169Part I: Chapter 5
Arrays
getch();
}
Output Screen
Example 5.8.2.3
Write a program to count the no. of vowels in the given string?
Solution of 5.8.2.3
// Header Files
# include<iostream.h>
# include<conio.h>
# include<string.h>
# include<stdio.h>
void main()
{
clrscr();
char string[15];
int vowels=0;
cout<<" Enter the string";
gets(string);
for (int i=0;i<strlen(string);i++)
{
if (string[i]=='a' || string[i]=='e' || string[i]=='o' ||string[i]=='i' || string[i]=='u')
{
vowels++;
}
}
cout<<" n The no. of vowels are"<<vowels;
getch();
}
Enter the string My sisters name is Deepti
Displaying the converted string
My#sisters#name#is#Deepti
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-184-320.jpg)
![170
OUTPUT SCREEN
5.9 Two-dimensional arrays
A two-dimensional array is an array in which each element is itself an array. For instance,
an array A[M][N] is an M by N table with M rows and N columns containing M * N
elements figure 5.8.
Figure 5.8: Two Dimensional Arrays
The number of elements in a 2-D array can be determined by multiplying number of rows
with number of columns. For example, the number of elements in an array A[5][3] is
calculated as 15.
This two-dimensional array may also be visualized as a one-dimensional array of arrays.
An alternative view of this array is given in the figure 5.9.
Enter the string Ankit
The no. of vowels are 2](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-185-320.jpg)
![171Part I: Chapter 5
Arrays
Figure 5.9: Two dimensional Arrays
The simplest form of a multi-dimensional array, the two-dimensional array, is an array
having single-dimension arrays as its elements. The general form of a two-dimensional
array declaration in C++ as follows:
Type array-name[rows][columns]
Where type is the base data type of the array having name array-name; rows, the first
index, refers to the number of rows in the array and columns, the second index, refers to
the number of columns in the array. The following declaration declares a character array
‘name’ having ‘5’ rows and ‘25’ columns.
Just like one dimensional arrays two dimensional arrays can also be initialized at the time
of declaration. The following declaration assigns values for three grades for each of the
first two students in the array.
int grades[Student_Number][Grades_achieved] = { {55, 60, 65},{95, 90, 85}};
Note: Data entries for each row are enclosed in individual curly braces.
5.9.1 Accessing Two Dimensional Arrays
Two dimensional arrays are accessed/ initialized with the help of two square brackets.
The following piece of code given below is used to access an array ‘A’ of size ‘M*N’
(‘M rows, N columns) and initialize all its members to zero.
char name[5][25];
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-186-320.jpg)
![172
…
{
…
// Accessing two dimensional arrays
int i, j;
for(i = 0; i < M; i++)
{
for(j = 0; j < N; j++)
{
A[i][j] = 0;
}
}
}
When accessing two dimensional arrays it is important to put each subscript in its own,
correct pair of brackets. Trying to initialize or access elements of a two dimensional array
would produce syntax errors. For example,
/* Incorrect Statements */
int A[M, N];
A[i, j] = 0;
A[j][i] = 0;
5.9.2 Passing Two Dimensional Arrays as a Parameter to Functions
As one dimensional array, two dimension arrays are also passed to functions by reference
automatically. Program 5.5 declares a two dimensional array ‘actual_2darray’ which is
initialized by the user and then passed to the function ‘display’ where it is displayed in
matrix form.
Program 5.5: Two dimensional arrays as function arguments
// Header Files
# include<iostream.h>
# include<conio.h>
// Function prototype
void display(int formal_2darray[3][3]);
// Main Function
void main()
{
clrscr();
int actual_2darray[3][3];
// Enter 2d Array](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-187-320.jpg)
![173Part I: Chapter 5
Arrays
cout<<" Enter the elements of the 2 dimensional array:";
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
cin>>actual_2darray[i][j];
}
}
//Passing 2d array to function
display(actual_2darray);
}
/******************************
Name: display
Purpose: To display the 2d array in Matrix form
**************************************/
void display(int formal_2darray[3][3])
{
cout<<endl<<" Displaying the 2d array in Matrix Form... n"<<endl;
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
cout<<"t"<<formal_2darray[i][j]<<" ";
}
cout<<endl;
}
}
The following output is observed when the program 5.5 is executed:
OUTPUT SCREEN
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-188-320.jpg)
![174
5.9.3 Mapping of Two Dimensional Arrays
Although, we visualize two dimensional arrays by rectangular tables this is not the way
two/multi dimension arrays are stored in the computers memory. The computer’s
memory only supports a linear placement of data items akin to a one-dimensional array as
a result of which the memory simulates the two/multi dimensional array as a one
dimensional array. This process is called as Mapping of an array. For example, for a two
dimensional array ‘Two_dimension_array[M][N];’ having ‘M*N’ as the maximum
number of elements that can be stored into it. The computer maps the two dimensional
array into one dimension using the following formula:
Two_dimension_array[ i * N + j ]
Where ‘i’ represents the corresponding row, ‘j’ represents the corresponding column and
‘N’ represents the maximum number of columns allowed in the two dimensional array
‘Two_dimension_array’.
Assuming ‘N’ to be ‘10’ the data entry stored at ‘Two_dimension_array[2][3]’ is mapped
into a one dimensional array say, Single_dimension_array[2*10 +3]. Hence, we can
also use this for ‘Single_dimension_array[i * N+j]’ in place of
‘Two_dimension_array[i][j]’ as both point or reflect to the same memory block. The
entire process of mapping is explained in the part III of the book.
5.9.4 Solved Examples
Example 5.9.4.1
Write a program to accept a sequence of strings, storing them in a two dimensional array,
and then print them on the screen?
Solution of 5.9.4.1
// Header Files
# include<iostream.h>
# include<string.h>
# include<stdio.h>
// Main Function
void main()
{
clrscr();
char name[5][25];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-189-320.jpg)
![175Part I: Chapter 5
Arrays
cout<<"Enter the names with a maximum of 24 characters";
for(int i=0;i<5;i++)
gets(name[i]);
cout<<" n The names stored are";
for (i=0;i<5;i++)
puts(name[i]);
getch();
}
OUTPUT SCREEN
5.10 Solved Examples
Example 5.10.1
Write a program to create a matrix of order 4*4 containing integer elements using a
function swap which swaps the value of two integers print the transpose of this matrix
(without using a 2nd
matrix)?
Solution of 5.10.1
// Header Files
# include <iostream.h>
# include <conio.h>
# include <iomanip.h>
// Function Prototype
void swap(int e, int f);
Enter the names with the maximum of 24 characters
Ankit Asthana
Kartik Gupta
Deepti Asthana
Karan Chopra
Neha Maheshwari
The names stored are
Ankit Asthana
Kartik Gupta
Deepti Asthana
Karan Chopra
Neha Maheshwari
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-190-320.jpg)
![176
// Main Function
void main()
{
int a,b,c,d,e,f;
int trans[5][5];
cout<<"Enter the size of the 2-dimensional array A(Rows(<5) and Columns(<5))";
cin>>a>>b;
swap(a,b);
cout<<"Thank you for using the program ";
getch();
}
/********************************
Name: swap
Parameters: int e, int f
Purpose: Transpose the matrix
********************************/
void swap( int e, int f)
{
int arr1[5][5];
int arr2[5][5];
int a,b,c,d;
// Transposing the array
cout<<" Enter array A";
for (int p=0;p<e;p++)
for (int q=0;q<f;q++)
cin>>arr1[p][q];
for (int i=0;i<e;i++)
for (int j=0;j<f;j++)
arr2[j][i]=arr1[i][j];
cout<<" Displaying";
for (int t=0;t<10;t++)
{
cout<<".";
}
cout<<" Elements of A"<<endl;
for (i=0;i<e;i++)
{
for (int j=0;j<f;j++)
cout<<setw(8)<<arr1[i][j];
cout<<endl;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-191-320.jpg)
![177Part I: Chapter 5
Arrays
cout<<" Transposed array A as follows nn";
for (int j=0;j<f;j++)
{
for (i=0;i<e;i++)
cout<<setw(8)<<arr2[j][i];
cout<<endl;
}
}
OUTPUT SCREEN
Example 5.10.2
Write a program to create two matrices A and B and to store the product of these two
matrices into another matrix C. The program should use a separate function to calculate
the product of the two matrices.
Solution of 5.10.2
// Header Files
# include <iostream.h>
# include <conio.h>
# include <dos.h>
// Function Prototype
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-192-320.jpg)
![178
void multiplication (float arr1[5][5], float arr2[5][5], int s, int t, int u, int v);
// Main Function
void main()
{
float A[5][5],B[5][5];
int a,b,c,d;
cout<<"n Enter the size of array 1 ";
cin>>a>>b;
cout<<" Enter the elements of array 1";
for (int e=0;e<a;e++)
{
for (int f=0;f<b;f++)
{
cin>>A[e][f];
}
}
cout<<"n Enter the size of the array 2";
cin>>c>>d;
cout<<" Enter the elements of the array 2";
for (e=0;e<c;e++)
{
for (int f=0 ; f<d;f++)
{
cin>>B[e][f];
}
}
multiplication (A,B,a,b,c,d);
}
/*********************
Name: multiplication
Purpose: To calculate the product of two matrices passed to it as parameters
**************************************************************/
void multiplication (float arr1[5][5], float arr2[5][5], int s, int t, int u, int v)
{
int C[5][5];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-193-320.jpg)
![179Part I: Chapter 5
Arrays
if (t==u)
{
for (int i=0;i<s;i++)
{
for (int j=0;j<v;j++)
{ C[i][j]=0;
for (int k=0;k<t;k++)
C[i][j]= C[i][j] + arr1[i][k] * arr2[k][j];
}
}
// Displaying array 1
cout<<" nArray 1 in matrix form is as follows nn";
for (i=0;i<s;i++)
{
for (int j=0;j<t;j++)
{
cout<<"t"<<arr1[i][j]<<" ";
}
cout<<endl;
}
//Displaying array 2
cout<< "nArray 2 in matrix form is as follows nn";
for (i=0;i<u;i++)
{
for (int j=0;j<v;j++)
{
cout<<"t"<<arr2[i][j]<<" ";
}
cout<<endl;
}
// Displaying array 3
cout<<"nArray C = A * B in matrix form is as follows n"<<endl;
for (i=0;i<s;i++)
{
for (int j=0;j<v;j++)
{
cout<<"t"<<C[i][j]<<" ";
}
cout<<endl;
}
}
else
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-194-320.jpg)
![180
{
cout<<"n Matrices not Compatiable";
}
}
Example 5.10.3
Write a program to create a matrix of the order of ‘5 * 5’ containing integer elements.
Write user defined functions to do the following:
(1) Find the sum of the elements lying on the left diagonal
(2) Find the sum of the elements lying on the right diagonal
(3) Find the sum of the elements lying on both the diagonals
Solution of 5.10.3
// Header Files
# include <iostream.h>
# include <conio.h>
// Function Prototypes
void suml ( int arr1[ ][5], int c, int d) ;
void sumr ( int arr1[ ][5], int c, int d) ;
void sumb ( int arr1[ ][5], int c, int d) ;
void display (int arr1[ ][5],int c ,int d);
// Main Function
void main()
{
int a,b,e,f;
int arr[5][5];
do
{
cout<<"nEnter the dimensions of the square matrix(rows and columns):";
cin>>a>>b;
if (a!=b)
{
cout<<"nInvalid Dimensions.n";
}
}while (a!=b);
cout<<"nEnter the elements of the array:";
for (e=0;e<a;e++)
for (f=0;f<b;f++)
cin>>arr[e][f];
display(arr,a,b);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-195-320.jpg)
![181Part I: Chapter 5
Arrays
suml(arr,a,b);
sumr(arr,a,b);
sumb(arr,a,b);
getch();
}
/*********
Name:suml
Parameters: arr1[][], c, d
Purpose: To calculate the sum of the left diagonal
***************************************************/
void suml( int arr1[5][5], int c,int d )
{
int lsum=0;
int i;
int j;
// sum of left diagonal
for (i=0; i<c; i++)
{
for (j=0; j<d; j++)
if (i==j)
{
lsum+= arr1[i][j];
}
}
cout<<"nSum of the elements on the left diagonal "<<lsum<<endl;
}
/*********
Name:sumr
Parameters: arr1[][], c, d
Purpose: To calculate the sum of the right diagonal
***************************************************/
void sumr(int arr1[5][5],int c,int d)
{
int rsum=0;
int j=d-1;
// Sum of the right diagonal
for (int i=0;i<c;i++)
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-196-320.jpg)
![182
{
rsum+=arr1[i][j];
j--;
}
cout<<endl<<"The sum of the elements on the right diagonal
"<<rsum<<endl;
}
/*********
Name:sumb
Parameters: arr1[][], c, d
Purpose: To calculate the sum of both diagonal
***************************************************/
void sumb(int arr1[5][5], int c,int d)
{
int lsum=0;
// Sum of left diagonal
for (int i=0;i<c;i++)
{
for (int j=0;j<d;j++)
if (i==j)
{
lsum+=arr1[i][j];
}
}
int lrsum=0;
int rsum=0;
int j=d-1;
// Sum of the right diagonal
for (i=0;i<c;i++)
{
rsum+=arr1[i][j];
j--;
}
// Sum of both the diagonals
if (c%2==0)
{
lrsum= lsum+rsum ;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-197-320.jpg)
![183Part I: Chapter 5
Arrays
else
{
int b=c/2;
lrsum= lsum + rsum - arr1[b][b];
}
cout<<endl<<"The sum of both the diagonals "<<lrsum<<endl;
}
/*********
Name:display
Parameters: arr1[][], m, n
Purpose: To display the matrix
***************************************************/
void display (int arr1[5][5], int m,int n )
{
// To clear the screen
cout<<"nMatrix -> n"<<endl;
for (int i=0;i<m;i++)
{
for (int j=0;j<n;j++)
{
cout<<arr1[i][j]<<" t ";
}
cout<<endl<<endl;
}
}
5.11 Review Exercise
1/2 mark questions
Q1) Is the following declaration valid?
‘unsigned int arr(7);’
Q2) State whether the following statements are true or false:
(i) An array element is accessed using a dot operator.
(ii) An array can store many different types of values.
(iii) This is an syntactically correct statement int arr[2.3];
(iv)An individual array element is passed to a function and not the entire array.
Q3) Arrays are passed to functions as:
(i) reference parameters.
(ii) formal parameters.
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-198-320.jpg)
![184
(iii) actual parameters
(iv) (i) and (iii) are correct
(v) None of the above
Q4) The number of bytes occupied by the two dimensional array A[10][10] are:
a) 100 bytes
b) 10 bytes
c) 20 bytes
d) 40 bytes
e) None of the above
Q5) State whether the following are true or false:
a) Arrays can have more than two dimensions
b) There is no difference between a character array and a string.
c) Arrays are a collection of elements belonging to the same data type.
d) An array element is accessed using the dot operator.
Q6) Write C++ statements for the following questions:
a) Array birds having 50 elements of type int.
b) A float array sum of index value 23.
c) A statement that checks for the occurrence of string “Dog” in the string s1.
Q7) Are the following statements syntactically correct:
a) int x[2][2] = { {0,1},{1,2}};
x[1,1]=5;
b) int name1[15];
char name2[15];
strcat(name1,name2);
c) int x[4][0] = { (0,0)};
Q8) Write statements to accomplish the following:
i) Display the value of the fifth element of array ‘A’.
ii) Print a value into element 5 of single-subscripted long int array A.
iii) Initialize each of the eight members of single subscripted array A to 0.
iv) Print the grand total of integer array ‘num’ containing 100 elements.
v) Copy all the elements of integer array ‘A’ into floating-point array ‘B’.
vi) Determine the smallest and the largest integer values from a character array ‘A’
containing 15 elements. Hint: Use type casting (ASCII Values)
Q9) Consider a 5 by 5 integer array ‘A’.
a) Write C++ statement to define this array?
b) How much space (in bytes) does this declaration involve using the declaration?
c) How many columns does it have?
d) How many rows does it have?
e) How many elements does it have?
f) Write the names of all the elements in the second row in array A.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-199-320.jpg)
![185Part I: Chapter 5
Arrays
g) Using loops initialize each member of array A to a value of 0.
h) Using loops input values for each member of Array A.
i) Determine the sum of elements on the main diagonal of this array A inputted in the
above statements.
j) Print this array A in matrix (tabular) form.
Q10) Using binary search how many comparisons will the array of following size take
i) 1023 elements
ii) 1048576 elements
(Hint: Binary search eliminates one half of the elements in a sorted array after each
comparison)
4/5 mark Questions
Q11) Write a program to initialize an integer array numbers containing 100 elements.
Initialize these elements values from 1 to 100 in ascending order. Create another
array even which stores those elements of array numbers that are even. Also
display all the elements of array even?
Q12) Write a program to create two matrices a of the order m*n and matrix b of the order
p *q. Write a user defined functions to do the following:
a. Create another matrix ‘c’ where each element of ‘c’ is the sum of the
corresponding elements of the array ‘a’ and ‘b’.
b. Create another matrix‘d’ where each element‘d’ is the difference of the
corresponding elements of the arrays ‘a’ and ‘b’.
c. Write a function display to display the following matrices ‘a’, ‘b’, ‘c’ and ‘d’.
d. Find the transpose of matrix a and display it
Q13) Write user defined functions to do the following
a. Create 2 matrices ‘x’ & ‘y’ of the order ‘a*b’, ‘c*d’ respectively containing
integer values
b. Create a 3rd
matrices which has the sum of the array X and array Y
5.12 Programming Project
In section 5.8.4, we studied how the computer memory stores two dimension arrays by
simulating them in a single dimension. The programming project for this chapter is to
state how C++ maps a three dimensional array (i.e. array_name[row][col][depth]) to one
dimensional memory. Write a fully commented and structured ANSI C++ program that
uses a single looping structure (of your choice) and your 3-D to 1-D mapping to print out
the following array: int map_3d[2][2][2] = {5,10,15,20,25,30,35,40}.
ARRAYS](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-200-320.jpg)
![190
190
student s1;
This statement would declare a variable ‘s1’ of type student. We could easily understand
this by comparing it to a normal declaration that is declaring a variable ‘s1’ of say, type
integer.
int s1;
The only difference being in the two declarations is that in the 1st
case s1 is a variable of
type ‘student’ and in the second case s1 is a variable of type int.
Now let me ask you a very interesting question what is the size of this structure variable
‘s1’ defined above?
The answer to this question is very simple, as we know that a structure is nothing but a
heterogeneous collection of data hence the size of a structure variable can be calculated
easily by adding up individually the size of all the structure members present in the
structure. Using this logic we can easily calculate the size of the structure variable ‘s1’
which comes out to be 21 bytes. Fig (6.2) gives an idea of how memory is allocated by
the computer to the structure variable ‘s1’.
Figure 6.2: Memory allocation for structure variables
(s1 occupies memory locations say, from 99 to 122, i.e., 21 bytes)
In addition to this one can also define multiple data structures, also all the information
stored in one structure can also be copied into another structure of the same type which
we will study in more detail in the next section.
6.3.1 Solved Examples
Example 6.3.1.1
(a) Declare a structure ‘date’ having the following members:
day -- A string of size[3]
Represents the day part of the date
month -- A string of size[20]
Represents month name of a date](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-205-320.jpg)
![191
Part I: Chapter 6
Structures
year -- A string of size [4]
Represents the year part of the date
(b) What is size of a variable of type date hence created?
Solution of 6.3.1.1
a)
struct date
{
char day[3];
char month[20];
char year[4];
};
b) The size of a structure variable of type ‘date’ will occupy 27 bytes of computers
memory.
Note: The size of structure might differ from one machine to another also the size of
the structure is also determined by a concept known as ‘word alignment’ although
this concept is out of the scope of the book.
Example 6.3.1.2
(a) Declare a structure ‘Employee’ having the following members:
Emp_Name -- A string of size[15]
Represents the name of the employee
Emp_IdNo. -- A Long int variable
Represents the employees Id number
Emp_Designation -- A string of size[20]
Represents the designation of the employee
Emp_salary -- A float variable
Represents the salary of the employee
(b) Declare C++ statement to declare two structure variables ‘Emp1’, ‘Emp2’ of type
‘Employee’
Solution of 6.3.1.2
(a)
struct Employee
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-206-320.jpg)
![192
192
char Emp_Name[15];
long int Emp_IdNo;
char Emp_Designation[20];
float Emp_salary;
};
(b) Employee Emp1,Emp2;
We can also define the structure variables after the structure, in such a case the tag name
becomes optional but then one cannot declare any more structure variables in the main
function.
For example, consider the structure ‘Employee’, for this structure we can also declare the
structure variables ‘Emp1’, ‘Emp2‘without giving a name to the structure as shown
below.
struct <optional tag/name>
{
char Emp_Name[15];
long int Emp_IdNo;
char Emp_Designation;
float Emp_salary;
}Emp1,Emp2;
The only disadvantage using this feature is that now we cannot create any more structure
variables of this type.
6.4 Accessing members of a structure
We can access members of a structure with the help of two operators in C++. The
operator (.) and the structure pointer operator (->) also known as the arrow
operator. The structure member is accessed with the help of the structure name.
Lets take an example, to access the structure member ‘Emp_salary’ of structure
variable Emp1 of type Employee defined in the example above one would write
the following C++ statement.
Emp1.Emp_salary = 2000.20 ;
The important thing to note here is that the part of the statement before the (.) operator is
the structure variable name ‘Emp1’ and the part after the dot operator ‘Emp_salary’ is the](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-207-320.jpg)
![195
Part I: Chapter 6
Structures
Structure Members and Type: roll_no: int
student_name[15]: char
avg: float
********************************************/
struct student
{
//Data_type1 Variable1
int roll_no;
//Data_type2 Variable2
char student_name[15];
//Data_type3 Variable3
float avg;
};
//Main Program
void main()
{
// Declaring structure variable s1 of type student
student s1;
cout<<" ENTER DETAILS";
// Entering Data for student
cout<<endl<<" Enter Roll No:";
cin>>s1.roll_no; // Accessing the Structure member 'roll_no'](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-210-320.jpg)
![197
Part I: Chapter 6
Structures
6.4.1 Solved Examples
Example 6.4.1.1
Using the structure definition for structure ‘Library’ having the following members write
a c++ program which declares a structure variable L1:
Book_Name -- A string of size[15]
Represents the name of the book
Book_IdNo -- A int variable
Represents the books Id number
Book_Shelf -- A string of size[5]
Represents the Book shelf number where
the book is stored.
The user should then be asked to enter the details for the structure variable. The program
should also check whether the book name is “Cleopatra” if so, the program should print a
suitable message on the screen.
Solution of 6.4.1.1
// Header Files
# include <iostream.h>
# include <conio.h>
# include <string.h>
# include <stdio.h>
// Structure Library
struct Library
{
char Book_Name[15];
int Book_IdNo;
char Book_Shelf[5];
};
// Main Function
void main()
{
Library L1;
char str[]="Cleopatra";
cout<<"Enter Respective Details:";
cout<<endl<<"Book Name";](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-212-320.jpg)
![198
198
gets(L1.Book_Name);
cout<<endl<<"Book_IdNo";
cin>>L1.Book_IdNo;
cout<<endl<<"Book Shelf";
gets(L1.Book_Shelf);
if (strcmp(str,L1.Book_Name)==0)
{
cout<<endl<<"Condition Satisfied";
}
else
{
cout<<endl<<"Condition not Satisfied";
}
}
COMMON PROGRAMMING ERRORS
• Not ending the structure definition with a (;) semicolon is a compilation error.
• Not using the dot operator (.) to reference to structure members is a compilation
error.
6.5 Structure Arrays
Before we learn about arrays of structures lets compare arrays and structures. Both arrays
and structures are collection of data types. The major difference in them being that an
array is a collection of data of the same type, where as a structure helps to group together
data that may vary in type. Let us look at the following comparison between array and
structures.
Structures Arrays
struct Employee char Emp_Name [15];
{
char Emp_Name[15]; char Emp_Designation[15];
int Emp_IdNo.;
char Emp_Designation[15];
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-213-320.jpg)
![199
Part I: Chapter 6
Structures
Now as we can see the structure ‘employee’ groups data of three different data types:
char, int and float where as the array ‘name[15]’ is a collection only of character type of
data. Structures allow us to build data bases, especially if one considers the possibility of
building arrays of them but is it possible to have arrays of structures?
The answer to this question is ‘Yes’. Arrays of structure are declared by declaring the
structure variables as arrays. Let us look at program 6.2 given below:
Program 6.2:
// Header Files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Structure Employee
struct Employee
{
char Emp_Name[15];
int Emp_IdNo;
char Emp_Designation[15];
};
// Main Program
void main()
{
Employee E1[10]; // Array of structures
for (int i = 0; i<10;i++)
{
cout<<" Enter Respective Details for Employee"<<i+1;
cout<<endl<<"1. Name";
gets(E1[i].Emp_Name);
cout<<endl<<"2. Employee Id No.";
cin>>E1[i].Emp_IdNo;
cout<<endl<<"3. Employee Designation";
cin>>E1[i].Emp_Designation;
}
cout<<endl<< " Printing Employee database";](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-214-320.jpg)
![200
200
for (i=0;i<10;i++)
{
cout<<endl<<"1. Name"<<E1[i].Emp_Name;
cout<<endl<<"2. Employee Id No."<<E1[i].Emp_IdNo;
cout<<endl<<"3. Employee Designation"<<E1[i].Emp_Designation;
}
}
Arrays of structure have the properties same as a normal array the only difference being
is that this time they are collection of a ‘struct’ data type. All operations valid on ordinary
Arrays are also valid with arrays of structure. For example, in the program above we have
declared an array ‘E1’ with index value 10 of type ‘Employee’. This array of structure
helps us to create a database for a set of 10 employees with the name, designation and Id
No. for each employee. Each index value in this array of structure is accessed in a similar
manner with the help of the (.) operator. Let us look at the following C++ statements
from program 6.2 above.
One uses for loop to enter the details for employee name into each index of array ‘E1’.
Each index entry of this array is accessed using ‘E1[i].Emp_Name’. The variable ‘i’
refers to each index of the array ‘E1’ and the part after the (.) operator refers to the
‘Emp_Name’ of the structure variable ‘E1’. Similarly, the other for loop is used to print
the details of this database on the screen.
An array of structure is very similar to a two-dimensional array, it can be thought as if it
consists of columns, each of which may consist of distinct data types. Fig 6.3 given
below represents the array ‘E1’ in such a manner.
for (int i = 0; i<10;i++)
{
cout<<" Enter Respective Details for Employee"<<i+1;
cout<<endl<<"1. Name";
cin>>E1[i].Emp_Name;
. . .
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-215-320.jpg)
![201
Part I: Chapter 6
Structures
Figure 6.3: Memory Allocation for Structure variable E1
6.5.1 Solved Examples
Example 6.5.1.1
(a) Declare a structure ‘Team’ having the following members:
Player_Name -- A string of size[15]
Represents the name of the employee
Player_No. -- An integer variable
Represents the employees Id number
Player_Type -- A string of size[20]
Represents the designation of the employee
(b) Write C++ statements to declare an array of structure variables ‘Players’ for a team
of 11 players and create a menu driven program which takes in details from the user
for creating a team, displays the created team and also searches for a specific player
in the created team using the Player_No. as the search key?
Solution of 6.5.1.1
// Header Files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Structure Definition
struct team
{
char Player_Name[15];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-216-320.jpg)
![202
202
int Player_No;
char Player_Type[20];
};
// Main Function
void main()
{
team Players[11];
int flag=0;
int search=0;
int ch;
do
{
cout<<endl<<"1. Create Team";
cout<<endl<<"2. View Team";
cout<<endl<<"3. Search for a player";
cout<<endl<<"4. Exit";
cin>>ch;
switch(ch)
{
case 1:
{
for (int i=0; i<11;i++)
{
cout<<" Enter Details for player "<<i+1;
cout<<endl<<" Enter Name:";
gets(Players[i].Player_Name);
cout<<endl<<" Enter Player No.";
cin>>Players[i].Player_No;
cout<<endl<<" Enter Player Type";
gets(Players[i].Player_Type);
flag++;
}
break;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-217-320.jpg)
![203
Part I: Chapter 6
Structures
}
case 2:
if (flag!=0)
{
cout<<"=== TEAM ===";
for (int i=0;i<11;i++)
{
cout<<endl<<" Player No:"<<Players[i].Player_No;
cout<<endl<<" Player Name:"<<Players[i].Player_Name;
cout<<endl<<" Player Type:"<<Players[i].Player_Type;
}
}
break;
case 3:
cout<<" Enter the Player No. to be searched";
cin>>search;
for (int i=0;i<11;i++)
{
if (search == Players[i].Player_No)
{
cout<<" Player Found";
cout<<endl<<" Name"<<Players[i].Player_Name;
cout<<endl<<" Type"<<Players[i].Player_Type;
cout<<endl<<" Player No."<<Players[i].Player_No;
break;
}
}
}
} while (ch!=4);
}
6.6 Structures with functions
As variables of any other data type, structure variables can also be passed to functions.
Structure variables can be passed to functions in the following ways:
Passing the entire structure variable as a parameter
Passing individual structure members to the function
Passing a pointer to a specific structure*
* Topic of pointers is introduced and explained in Chapter 7.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-218-320.jpg)
![209
Part I: Chapter 6
Structures
Now let me ask you a very interesting question can a structure contain an instance of its
own type?
The answer to this question is ‘no’ we cannot. We can only declare structures as structure
members to a structure provided the structure whose variable is a member to the nested
structure is already defined.
6.7.1 Solved Examples
Example 6.7.1.1
Write a structure definition for structure ‘Race’ having the following attributes. A
character array ‘Player_Name’ which represents the name of the player playing in the
race, an integer variable ‘Player_No’ representing the player number and a structure
variable ‘Player_Time’ of structure ‘Time’ which represents the time taken by each
player in the race. The definition for structure ‘Time’ is given below:
struct Time
{
int hrs;
int mins;
int secs;
};
Solution of 6.7.1.1
struct Time
{
int hrs;
int mins;
int secs;
};
struct Race
{
char Player_name[20];
int Player_No;
struct Time Player_Time;
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-224-320.jpg)
![210
210
6.8 Solved Examples
Example 6.8.1
Write a program to create an array of the given structure ‘Emp’
struct Emp
{
int Emp_No;
char Emp_Name[15];
float Emp_Basic;
};
Create user defined functions to
(a) Sort the array of structures declared in ascending order of Emp_Basic
(b) Calculate the net salary of each employee using :
Net Salary = Basic + H.R.A.+ D.A. – P.F.
H.R.A. = 25 % of Basic
D.A = 10 % of Basic
P.F. = 10% of Basic
Solution of 6.8.1
/* Header Files */
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Structure 'Employee'
// Structure Members: Employee_No: Represents Employee Number
// : Employee_Name: Represents Employee Name
// : Employee_Basic: Represents Employee Basic
struct Employee
{
int Employee_No;
char Employee_Name[15];
float Employee_Basic;
};
// Function Headers
void accept (Employee E1[], int index);
void display (Employee E1[], int index);
void sort (Employee E1[], int index);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-225-320.jpg)
![211
Part I: Chapter 6
Structures
void Netsalary (Employee E1[], int index);
// Main Function
void main()
{
clrscr();
Employee E[50];
int ch;
int flag=0;
int Array_index;
do
{
clrscr();
cout<<"n =========== DATABASE: EMPLOYEE =============="<<endl;
cout<<" 1. Create Employee database"<<endl;
cout<<" 2. Print Enteries form database"<<endl;
cout<<" 3. Sort the Employees in Order of Basic Salaries"<<endl;
cout<<" 4. Calculate the Net salary of each Employee"<<endl;
cout<<" 5. Exit Database"<<endl;
cout<<" !. Press Choice";
cin>>ch;
switch (ch)
{
case 1:
if (flag==0)
{
clrscr();
cout<<" For how many employees should the database be created";
cin>>Array_index;
accept(E,Array_index);
flag++;
}
else
{
cout<<" Option Not Available";
getch();
}
break;
case 2:
clrscr();
display(E,Array_index);
break;
case 3:](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-226-320.jpg)
![212
212
sort(E,Array_index);
break;
case 4:
Netsalary(E,Array_index);
break;
}
}
while (ch!=5);
getch();
}
/*******************************************
Name: accept
Formal Parameters: E1[], index
Purpose: Enter values into structure array
*******************************************/
void accept( Employee E1[], int index)
{
cout<<" Enter the following Details";
for (int i=0; i<index;i++)
{
cout<<endl<<" Employee No:";
cin>>E1[i].Employee_No;
cout<<" Employee Name:";
gets(E1[i].Employee_Name);
cout<<" Employee Basic:";
cin>>E1[i].Employee_Basic;
cout<<"--------------------";
}
}
/*******************************************
Name: display
Formal Parameters: E1[], index
Purpose: Display the structure array
*******************************************/
void display( Employee E1[], int index)
{
for (int i=0; i<index;i++)
{
cout<<endl<<" Employee No:"<<E1[i].Employee_No;
cout<<endl<<" Employee Name:";
puts(E1[i].Employee_Name);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-227-320.jpg)
![213
Part I: Chapter 6
Structures
cout<<" Employee Basic:"<<E1[i].Employee_Basic;
cout<<endl<<"--------------------";
}
getch();
}
/**********************************************************************
Name: sort
Formal Parameters: E1[], index
Purpose: Sort the structure array in ascending order of basic salary
**********************************************************************/
void sort (Employee E1[], int index)
{
clrscr();
struct Employee temp;
for (int i=0; i<index;i++)
for (int j=0; j<index-i-1;j++)
if (E1[j].Employee_Basic>E1[j+1].Employee_Basic)
{
temp = E1[j];
E1[j]=E1[j+1];
E1[j+1]=temp;
}
cout<<"Sorting Complete";
getch();
}
/*******************************************
Name: Netsalary
Formal Parameters: E1[], index
Purpose: Calculate the Net salary of the employees
using Net Salary = Basic + 25 % Basic(H.R.A) + 10%Basic(D.A.) -10%
Basic(P.F.)
*******************************************/
void Netsalary (Employee E1[], int index)
{
clrscr();
cout<<"==================================="<<endl;
cout<<"Net Salaries for all employees"<<endl;
float Salary[50];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-228-320.jpg)
![214
214
for ( int i=0; i<index;i++)
{
Salary[i]= (1.25) * E1[i].Employee_Basic;
cout<<"Employee No: "<<E1[i].Employee_No<<" Net Salary: "<<Salary[i]<<endl;
}
getch();
}
Example 6.8.2
Write a program to define a structure called ‘Award’ with the following data members
(i) Name of the winner: String of length 30 characters
(ii) Prize Money of Award: Long integer variable
Declare an array of the structure with n elements. Write user defined functions for:
(a) Input data values into structure members;
(b) Display the contents in tabular form;
(c) Display the details of students who have achieved awards having prize money >
$50000; and
(d) Display details of the student achieving the highest prize money.
Solution of 6.8.2
// Header Files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Structure definition for'Award'
struct Award
{
char Name[30];
char type[15];
long prize;
};
// Function Headers
void input (Award A[], int n);
void display (Award A[],int n);
void details (Award A[],int n);
void largest (Award A[],int n);
// Main Function
void main()](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-229-320.jpg)
![215
Part I: Chapter 6
Structures
{
Award A[50];
int n;
int flag=0;
int ch;
do
{
clrscr();
cout<<" ================== Award Database =====================";
cout<<endl<<" 1. Input Award Information ";
cout<<endl<<" 2. Display Award Information";
cout<<endl<<" 3. Details: Awards awarded for prize money > 50000$";
cout<<endl<<" 4. largest Prize money award";
cout<<endl<<" 5. Exit";
cout<<endl<<" Enter Choice :?: ";
cin>>ch;
switch(ch)
{
case 1:
if (flag==0)
{
clrscr();
cout<<" For How Many Awards";
cin>>n;
input (A,n);
flag++;
}
else
cout<<" Option not available";
getch();
break;
case 2:
display (A,n);
getch();
break;
case 3:
details (A,n);
break;
case 4:
largest (A,n);
break;
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-230-320.jpg)
![216
216
while (ch!=5);
getch();
}
/****************************************************
Name: input
Parameters: A[], n
Purpose: Input data in structure array 'A[]'
****************************************************/
void input ( Award A[], int n)
{
clrscr();
for (int i=0; i<n;i++)
{
cout<<" Enter Name of person";
gets(A[i].Name);
cout<<" Enter type";
cin>>A[i].type;
cout<<" Enter Prize Money $";
cin>> A[i].prize;
cout<<"======================"<<endl;
}
}
/****************************************************
Name: Details
Parameters: A[], n
Purpose: Print all members of array structure 'A[]'
****************************************************/
void display (Award A[], int n)
{
clrscr();
for (int i=0; i<n;i++)
{
cout<<"Name :";
puts(A[i].Name);
cout<<"Type :"<<A[i].type<<endl;
cout<<"Prize Money :"<<A[i].prize;
cout<<endl<<"====================="<<endl;
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-231-320.jpg)
![217
Part I: Chapter 6
Structures
/****************************************************
Name: Details
Parameters: A[], n
Purpose: Print details of structure members of Array 'A[]'
where prize money>50000$
****************************************************/
void details (Award A[], int n)
{
clrscr();
for (int i=0;i<n;i++)
{
if (A[i].prize>50000)
{
cout<<endl<<"Name:"<< A[i].Name;
cout<<endl<<"Type:"<< A[i].type;
cout<<endl<<"Prize Money:"<<A[i].prize;
}
}
getch();
}
/****************************************************
Name: largest
Parameters: A[], n
Purpose: Print details of structure array 'A[]' with
the largest prize money
****************************************************/
void largest (Award A[], int n)
{
clrscr();
int max=A[0].prize;
int index=0;
for (int i=0;i<n;i++)
{
if (max<A[i].prize)
{
max = A[i].prize;
index = i;
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-232-320.jpg)
![218
218
cout<<" Name:"<< A[index].Name;
cout<<endl<<" Type: "<<A[index].type;
cout<<endl<<" Prize: "<<A[index].prize;
getch();
}
Example 6.8.3
Using the following structure definition for nested structure ‘Ticket’ given below, Write a
program to enter the details of 5 tickets and calculate the ticket fares for each of the
following depending on distance traveled ( Table Given Below).
Ticket
Members Type
PNR string
DOT date
Passenger person
From string
To string
Traversed int
Solution of 6.8.3
Distance Traveled Fare
< =200 Miles
>200&<300 Miles
>=300 Miles
$5/Mile
$4/Mile
$3/Mile
// Header Files
# include<iostream.h>
# include<conio.h>
// Comments: Use Header File <stdio.h> and use functions gets(), puts() to
// manipulate with strings for better results
struct date
{
int day;
int month;
int year ;
};
struct person
{
char Name[15];
int age;
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-233-320.jpg)
![219
Part I: Chapter 6
Structures
// Structure Definition for structure 'person'
struct person
{
char Name[15];
int age;
};
// Structure Definition for structure 'date'
struct date
{
int day;
int month;
int year;
};
// Structure Definition for structure 'ticket'
struct ticket
{
char PNR[15];
date DOT;
person Passenger;
char From[15];
char To[15];
int Traversed;
};
// Function Headers
void accept( ticket T[], int n);
void fare (ticket T[], int n);
void display(ticket T[], int n);
// Main Function
void main()
{
clrscr();
int choice;
ticket T[50];
int index;
do](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-234-320.jpg)
![221
Part I: Chapter 6
Structures
cout<<"Enter the following details:"<<endl;
for (int i=0; i<n;i++)
{
cout<<"Enter the PNR Number";
cin>>T[i].PNR;
cout<<endl<<"Enter Date:";
cout<<endl<<" Day:";
cin>>T[i].DOT.day;
cout<<endl<<" Month:";
{
cout<<"n PNR:"<<T[i].PNR;
cout<<"n Fare:"<<4*T[i].Traversed;
}
if (T[i].Traversed>=300)
{
cin>>T[i].DOT.month;
cout<<endl<<" Year:";
cin>>T[i].DOT.year;
cout<<endl<<"Enter Passenger Details:";
cout<<endl<<" Name :";
cin>>T[i].Passenger.Name;
cout<<endl<<" Age :";
cin>>T[i].Passenger.age;
cout<<endl<<"Source Station:";
cin>>T[i].From;
cout<<endl<<"Destination Station:";
cin>>T[i].To;
cout<<endl<<"Distance Traversed:";
cin>>T[i].Traversed;
cout<<endl<<"^^^^^^^^^^^^^^^^^^^^"<<endl;
}
}
/***************************************
Name: Fare
Parameters: T[], n
Purpose: To calculate Ticket Fare
**************************************/
void fare (ticket T[], int n)
{
clrscr();
for (int i=0;i<n;i++)](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-236-320.jpg)
![222
222
{
if (T[i].Traversed <=200)
{
cout<<"n PNR:"<<T[i].PNR;
cout<<"n Fare:"<<5*T[i].Traversed;
}
if (T[i].Traversed >200 && T[i].Traversed<300)
cout<<"n Fare:"<<3*T[i].Traversed;
cout<<"n PNR:"<<T[i].PNR;
}
cout<<"--------------";
}
getch();
}
/***************************************
Name: display
Parameters: T[], n
Purpose: To display ticket details
**************************************/
void display(ticket T[],int n)
{
int i;
for ( i=0; i<n;i++)
{
// Entering Ticket Details
cout<<endl<<" Entry :"<<i+1;
cout<<endl<<" PNR Number: "<<T[i].PNR;
cout<<endl<<" Date:";
cout<<endl<<" Day:"<<T[i].DOT.day;
cout<<endl<<" Month:"<<T[i].DOT.month;
cout<<endl<<" Year:"<<T[i].DOT.year;
cout<<endl<<" Passenger Details:";
cout<<endl<<" Name :"<<T[i].Passenger.Name;
cout<<endl<<" Age :"<<T[i].Passenger.age;
cout<<endl<<"Source Station:"<<T[i].From;
cout<<endl<<"Destination Station:"<<T[i].To;
cout<<endl<<"Distance Traversed:"<<T[i].Traversed;
cout<<endl<<"^^^^^^^^^^^^^^^^^^^^"<<endl;
}
getch();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-237-320.jpg)
![224
224
g1.z++;
return g1;
}
void main()
{
clrscr();
graph1 g1;
graph1 g2;
g1.x=1;
g2.x=2;
g1.z=2;
g1.y=0;
g2=increment(g1);
cout<<g2.x<<g2.y<<g2.z;
getch();
}
Q6) Find the error(s) in the following piece of code:
# include<'iostream.h'>
# include<'conio.h'>
structure car
{
char car_name[15];
int car_number
char car_type;
}; diesel, petrol;
main()
{
Program to enter details for structure variables
cout<<" Enter Car Name";
gets(diesel.car_name);
cout<<" Enter Car type";
gets(car_type);
cout<<" Car number";
cin>>petrol.car_type;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-239-320.jpg)
![226
226
struct team
{
char name[20];
int num;
int age;
float avg;
char category;
};
Q12) Write a program to create an array of given structure ‘Train’:
struct Train
{
int Train_No;
char Train_Name[10];
int Train_fare;
};
Write user defined functions to:
a) Sort the array in ascending order of Train_fare, Train_No
b) Print the sorted array in tabular form
Q13) Write a program to define a structure called ‘Scholarship’ with the following data
members.
i) Name of the scholarship: string
ii) Type of scholarship : char type
iii) Number of scholarships : short Int type
Declare an array of the structure with ‘N’ elements write user defined functions
for the following:
a) Input: To read in structure members from the user
b) Display: To display the contents in the array in a tabular form
c) Display_Selective: To display the details of all the Scholarships whose
number is greater than 10.
d) Display: To display the name of the scholarship having the highest
occurrences.
Q14) Declare a structure Air_Canada having the following members:
Flight_Name : A character array
Flight_Fare : A long int
Flight_No: ‘int’ Type:](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-241-320.jpg)
![239Part I: Chapter 7
POINTERS
7.2.2 The ‘size of’ operator
The ‘sizeof’ operator is used to calculate the size of an object/variable belonging to any data
type, structure or class. The ‘sizeof’ operator yields the size of its operand with respect to the
size of type char. It has the following syntax:
Syntax: sizeof ( object/variable_name );
The result of the sizeof operator is of type size_t, an integral type defined in the include file
<STDDEF.H>.
For example,
…
int var;
cout<<sizeof(var);
When these statements are executed the program will display ‘4’ on the screen. The sizeof
operator can also be applied to arrays. In such a case the size of operator will return the size
of the entire memory block occupied by the array passed to it as a parameter.
For example,
…
float var[10];
cout<<sizeof(var);
As expected when the following statements are executed the integer ‘40’ is displayed on the
screen (4 * ‘10’ bytes).
7.2.2.1 Importance of the ‘sizeof’ operator
To understand the importance of the ‘sizeof’ operator look at the following program:
// Header Files
# include <iostream.h>
# include<conio.h>
/
// Main Function
void main()
{
Test test1;
cout<<sizeof(test1);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-254-320.jpg)
![240
}
/ Structure Definition
struct Test{
int var1;
char var2;
};
Now when this program is executed one should expect ‘5’ to be displayed on the screen
as the structure definition for structure ‘Test’ has an integer variable (‘4’ bytes) and a
character variable (‘1’ byte) but to our surprise we notice that a number ‘8’ is displayed
on the screen. Now this is not the compilers fault, but this is because of a concept known
as ‘data wrapping’. A concept introduced briefly in the previous chapter.
What happens is that the compiler always tries to align all the variables to the word
boundary ‘4 or multiples 4 bytes’ for better performance of your programs. So if the
amount of space occupied by a variable is say ‘5’ the compiler then aligns it to the
nearest multiple of ‘4’ which happens to be ‘8’ in this case. The overall understanding of
this concept is however out of the scope of the book but with this example, it should be
clear now as why we need the ‘sizeof’ operator.
7.3 Pointer to an Array
A pointer can also be used to point to an array. A pointer, which stores the address of an
array, is called as a pointer to an array. In fact, there is a close relationship between
pointers and arrays.
An array name can be thought of as a constant pointer. Actually, the identifier of an
array is equivalent to the address of its first element, like a pointer is equivalent to the
address of the first element that it points to. Look at the following piece of code:
Now the array name ‘num’ is a constant pointer. As we already know that arrays are self
reference parameter so while using a pointer to point to an array we do not use ‘The
address of (&) operator’ as we have done in the case of variables earlier in this chapter.
The following statement sets the pointer ‘ptr’ to point to the array ‘num’:
This statement is equivalent to taking the address of the first element of the array ‘num’
as follows:
int num[10];
int *ptr;
ptr = num;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-255-320.jpg)
![241Part I: Chapter 7
POINTERS
Let us now review the comparison between arrays and pointers. At this point the pointer
‘ptr’ and the array ‘num’ are exactly equivalent and they have the same properties, the
only difference is that we could assign another value to the pointer ‘ptr’ as it is a pointer
variable whereas ‘num’ will always point to the zeroth
index of the array ‘num’.
COMMON PROGRAMMING ERRORS
• Ignoring the ‘*’ variable while declaring a pointer variable will result in a logical
error.
• Not dereferencing a pointer to obtain a value to which the pointer points to will
result in a syntax error
.
7.3.1 Pointer expression and pointer arithmetic
Pointer arithmetic is quite different than conventional arithmetic. To begin with, only
addition and subtraction operations are allowed to be conducted, the others make no
sense in the world of pointers. But both addition and subtraction have a different behavior
with pointers according to the size of the data type to which they point. The important
ones which are commonly used are the increment (++) and the decrement (--) operators.
Let’s take up an example to understand the use of these operators with pointer variables.
Assume an array say, int x[5] and say that it is stored at a position 100 in the memory.
Now as the data type is ‘int’ so each variable of this type would occupy 2 bytes of
memory (see Figure 7.3). We can now initialize the pointer ‘ptr’ to point to the array ‘x’
with either of the equivalent statements.
int *ptr=num;
or
int *ptr;
ptr=&num[0];
int *ptr=x;
ptr = x;
ptr = &x[0];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-256-320.jpg)
![242
Figure 7.3(a): Pointer to an array
In simple arithmetic 100 + 2 would give 102 but this is not the case when one is
operating using pointer arithmetic. When using pointer variables as operands then in such
a case they are not just incremented or decremented by the value of an integer but size of
object the pointer refers to times the value of the integer. So in the following statement
ptr+=2;
it would produce final address as 104 (100 + 2*2) and not 102 and hence after this
statement the pointer will now point to X[2].
Figure 7.3 (b): Pointer to an array
An array pointer can also be alternatively referenced with the following pointer
expression.
* ( ptr + 2)
Note: One should remember to use parenthesis as the precedence of operators that is * is
of higher precedence than +. Let us better understand pointer arithmetic with the help of
the following example:](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-257-320.jpg)
![243Part I: Chapter 7
POINTERS
Program 7.3
// Program 7.3: Pointer to an array
// Header Files
# include <iostream.h>
# include <conio.h>
// Main Function
void main ()
{
int x[5];
// Pointer Variable ’ptr’
int * ptr;
ptr = x;
*ptr = 10;
ptr++;
*ptr = 20;
ptr = &x[2];
*ptr = 30;
ptr = x + 3;
*ptr = 40;
ptr = x;
*(ptr+4) = 50;
for (int y=0; y<5; y++)
{
cout << x[y] << ", ";
}
}
10, 20, 30, 40, 50,
The working of this program is explained with the help of Figure 7.4.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-258-320.jpg)
![246
7.4.1 Solved Examples
Example 7.4.1.1
Find the output for the following piece of code:
# include <iostream.h>
# include <conio.h>
# include <ctype.h>
# include <string.h>
void newtext (char string [], int &position)
{
char *pointer = string;
int length = strlen (string);
for ( ; position <length -2; position +=2 , pointer ++)
{
*(pointer + position ) = toupper (*(pointer + position));
}
}
void main ()
{
clrscr();
int location = 0;
char message [] = “Dynamic Act”;
newtext (message, location);
cout<<message<<” # “<<location;
}
Solution of 7.4.1.1
OUTPUT SCREEN
DynAmiC ACt# 10](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-261-320.jpg)
![247Part I: Chapter 7
POINTERS
Example 7.4.1.2
Give the output of the following program segment
# include<iostream.h>
# include<conio.h>
# include<ctype.h>
void main()
{
clrscr();
char *s = "MeTropoliTan";
for ( int x =0; s[x] != '0';x++)
{
if (x%2!=0 && islower(s[x]))
s[x] = toupper(s[x]);
else
if (isupper(s[x]))
s[x] = tolower (s[x]);
else
s[x]= s[x]--;
}
cout<<s<<endl;
}
Solution of 7.4.1.2
In this piece of code‘s’ is declared as a pointer variable pointing to the first character of
the string ‘MeTropoliTan’. The program converts some of the characters to upper case
and some to lower case depending upon the condition provided in the if else loop
construct. The working of this program should be easily understood as we already
observed similar problems above.
OUTPUT SCREEN
mEtRoPoLitaN](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-262-320.jpg)
![248
7.5 Pointer with functions
As we have already studied in the earlier chapters that there are two ways to pass
arguments to a function ‘Call by Value’ and ‘Call by reference’. In C++ ‘Call by
reference’ is practiced by using either the address of operator or by the use of the
pointers. In part I chapter 3 if we recall we were using the address of operator (&) when
we were calling functions by reference. What we were actually doing though was passing
the address of the variable to the function and hence reflecting any changes made in the
formal parameters to the actual parameters. Call by reference can also be replicated by
the use of pointers in C/C++.
The programs 7.5 (a), (b) present similar functions being called by reference by using the
address of operator (&) in the 1st
case and using the indirection operator (*) in the other.
The important thing to note is that both these programs are equivalent and give the same
result.
Program 7.5
(a) (b)
Using Function with Pointers Using Functions as Reference
// Header Files
# include<iostream.h>
# include<conio.h>
# include<string.h>
# include <stdio.h>
/***************************
Name: pointer_example
Arguments: char *ptr, int length
***************************/
void pointer_example( char *ptr, int
length);
// Main Function
void main()
{
// Header Files
# include<iostream.h>
# include<conio.h>
# include<string.h>
# include <stdio.h>
/**********************
Name: reference_example
Arguments: char string[]
**********************/
void reference_example( char string[]);
// Main Function
void main()
{
clrscr();
char string_actual[15];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-263-320.jpg)
![249Part I: Chapter 7
POINTERS
clrscr();
char string[15];
int length_string;
cout<<" Enter String (<15
)"<<endl;
gets(string);
length_string=strlen(string);
// Call to the function
pointer_example
pointer_example(string,
length_string);
getch();
}
// Function Definition and Body
void pointer_example( char *ptr, int
length )
{
for (int i=0; i<length; i++)
{
cout<<*ptr;
ptr++;
}
}
cout<<" Please enter the string (<15
characters>)"<<endl;
gets(string_actual);
// Call to the function
reference_example
reference_example(string_actual);
getch();
}
// Function Definition and Body
void reference_example(char
string_formal[])
{
puts(string_formal);
}
When these programs are executed, one would notice that they result in the same outcome. In
fact, both these programs are equivalent piece of code.
Let us try to understand why is it so? In program 7.4 (a) if we notice we have used a
character pointer in the function definition for function ‘pointer_example’. When the
character array ‘string’ is passed to it as a parameter. The pointer points to the 1st
character of
this character array ‘string’. When we enter the function body we fall into a for loop which
executes till the length of the string passed to it and the pointer ‘ptr’ is incremented every
time in each iteration and hence the whole string is printed on the screen as a result of this
loop (Figure III1.2) loop. In program 7.4 (b) the same thing happens the only difference
being that in this case we have passed the character array as a reference parameter using the
address of operator (&).
PROGRAMING TIPS
Always use parenthesis to avoid compilation and logical
errors when using pointer arithmetic.
Pointer arithmetic is different from conventional arithmetic.
In pointer arithmetic only ‘addition’ and ‘subtraction’ using
operators is allowed.
Pointers and arrays are very similar concepts
Using the address of (&) operator is same as using pointers.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-264-320.jpg)
![250
7.6 New and Delete Operators
To understand the use of the new and delete operator let us look at the following array
program 7.6.
Program 7.6
All this program does is input a name from the user and then print that name in the next
line. What is though more important to observe is that when we declare the character
array string with an index value 15. The computer reserves space for 15 characters.
Now when we input the string say ‘Computer’. It has 8 characters and should occupy 8
characters but instead it would too use up all the space assigned for 15 characters as at
time of declaration for this string we declared it for a set of 15 characters. Hence we are
wasting the rest space for 7 characters and hence the computers memory. C++ does not
allow us to specify the array size at the time of program execution. The compiler
requires the array size to be a constant. Hence, while declaring an array you need to know
how big the array is likely to be which may always not be possible as we just saw in the
previous program. C++ overcomes this problem by providing the ‘new’ and the ‘delete’
operators which allow us to use the computer memory efficiently.
7.6.1 The New Operator
‘New operator’ in C++ returns the addresses of a block of unallocated bytes (depending
on data type a pointer is pointing to). In simple words it helps in obtaining blocks of
// Program Prints the Name of a person
// Header Files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Main Function
void main()
{
clrscr();
char string [15];
cout<<" Enter the Name of the person ";
gets (string);
puts(string);
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-265-320.jpg)
![251Part I: Chapter 7
POINTERS
memory returning a non zero pointer pointing to its starting point.The following program
shows how the new operator can be used in C++.
Program 7.7:
// Header Files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
// Main Function
void main()
{
clrscr();// Clear Screen
char *ptr; // Pointer to char
int size;
cout<<" Enter the number of characters in the persons name";
cin>> size;
ptr = new char [size+1]; // Set aside memory: size
cout<<endl<<" Enter the Name of the person ";
cin>>ptr;
cout<<endl<<ptr;
delete [] ptr; // Frees the allocated block of memory
getch();
}
Let us take a look at this expression:
ptr = new char [size+1];
returns a pointer to a section of memory equal to size bytes just large enough to hold the
name of the person to be inputted. By adding +1 to the size we also make sure that it has
enough space to even include the null operator ‘0’ at the end of the string.
7.6.2 The Delete Operator
As we know that the computers memory is limited therefore while working with these
operators one has to be careful of the amount of memory a program occupies. As if a](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-266-320.jpg)
![252
program uses large chunks of memory then all the available memory will be reserved for
the program and the program will eventually crash. So for safe and efficient use of
memory, the ‘new’ operator has a counter part known as the ‘delete’ operator which
returns all the memory blocks allocated to the program using the ‘new’ operator back to
the operating system.
Syntax:
Let us look at following statement in program 7.7:
delete [] ptr;
This statement returns to the operating system whatever memory pointed to by our
pointer ‘ptr’. The square brackets tell the compiler that we are deleting the entire array
block.
7.7 Array of pointers
An array, who’s each element is a pointer type, is known as an array of pointers. Arrays
of pointers are used to create arrays of strings. Each entry in the array is a string and this
pointer points to the 1st
character of a string. Consider the following definition of a
character array ‘flowers’.
char * flowers[3] = { “ ROSE”, “LILY”, “LOTUS” };
The highlighted part (char *) indicates that each element of the array flowers is a pointer
to char. All entries in the array ‘flowers’ store addresses of these strings Each index
entry of this array points to the 1st
character of their corresponding string (see Figure 7.6).
Figure 7.6: Array of pointers
We can now also look at Program 7.8 which also shows the use of array of pointers in
C++.
delete pointer_name;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-267-320.jpg)
![253Part I: Chapter 7
POINTERS
Program 7.8
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr();
int *a[3]; // Declaring an array of pointers
for (int i=0;i<3;i++)
{
a[i] = new int; // Allocating block of memory for the character array
*a[i]= i * 25;
}
for (i=2;i>=0;i--)
cout<<"*A["<<i<<"]="<<*a[i]<<endl;
for (i=0;i<3;i++)
delete a[i]; // De allocating block of memory for the character array
getch();
}
In this program, a[i] is an array of pointers and each element of this array a[0], a[1], a[2]
points to a memory block of type int. Again each element of this just stores the pointers
which point to their respective memory blocks.
‘Array of pointers’ displays the data structuring capabilities of C++. We will further learn
about them in later chapters.
7.8 Pointer to a structure
A pointer which stores the address of struct type data is known as pointer to structure. An
example of a pointer to a structure is shown in the following program 7.9.
Program 7.9
# include <iostream.h>
# include <conio.h>](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-268-320.jpg)
![257Part I: Chapter 7
POINTERS
Solution of 7.9.2
OUTPUT SCREEN
Example 7.9.3
What will the user see when the following piece of code is executed.
(i)
# include <iostream.h>
# include <conio.h>
struct point
{
int x,y;
}polygon[]={{1,2},{1,4},{2,4},{2,2}};
void main()
{
point *a;
a= polygon;
a++;
a->x++;
cout<<polygon[0].x<<polygon[0].y<<endl;
cout<<a->x<<a->y<<endl;
getch();
}
(ii)
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr();
int x=20, y=30, *b=&x;
(*b)+=15;
cout<<x<<*b<<endl;
i) 10909010
ii) 101192192192
iii)11898978](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-272-320.jpg)
![258
x++;
cout<<*b<<x<<endl;
b=&y;
y+=x;
int *c=b;
b=&x;
y+=12;
x*=4;
cout<<*b<<*c;
getch();
}
(iii)
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr();
int a[]={2,4,6,8,10};
int *k,*p1;
p1=a;
k = new int;
(*p1)+=10;
*k=55;
(*p1)++;
*p1=*k;
(*k)*=2;
cout<<*p1<<*k<<endl;
getch();
}
(iv)
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr();
char *p = "Difficult";
char c;
c=++*p++;
cout<<c<<*p<<endl;
getch();
}
(v)
# include <iostream.h>
# include <conio.h>](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-273-320.jpg)
![260
Example 7.9.4
Write a program which reads in a group of 10 numbers and places them in an array of
type int. It then calculates the sum of all the odd numbers in the set of these 10 numbers
and prints the odd numbers and their sum on the screen. Use pointer notation where ever
possible.
Solution of 7.9.4
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr();
int *ptr[10];
int sumodd=0;
int flag=0;
cout<<" Enter the 10 numbers in the array";
for (int i=0; i<10;i++)
{
ptr[i] = new int;
cin>>*ptr[i];
}
for (i=0; i<10;i++)
{
if (*ptr[i]%2!=0)
{
if (flag==0)
cout<<" The odd numbers are";
sumodd = sumodd + *ptr[i];
cout<<*ptr[i]<<endl;
flag++;
}
}
cout<<" The sum of odd numbers in the array is"<<sumodd;
getch();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-275-320.jpg)
![261Part I: Chapter 7
POINTERS
Example 7.9.5
Declare an array of pointers to strings representing the 5 week days. Also, declare a function
to sort these strings in alphabetical order using bubble sort. Sort the pointers to the string as
and not the actual strings?
Solution of 7.9.5
// Header files
# include <iostream.h>
# include <conio.h>
# include <stdio.h>
# include <string.h>
// Function: b_sort
// Purpose: To sort the array passed to it as a paramter
void b_sort(char *str_pointers[])
{
char *temp;
for (int j=0;j<5;j++)
for (int i=0;i<4-j;i++)
{
if (strcmp (str_pointers[i],str_pointers[i+1])>0)
{
temp = str_pointers[i+1];
str_pointers[i+1]=str_pointers[i];
str_pointers[i]=temp;
}
}
}
// Main Function
void main()
{
char *weekdays[5]={"Monday","Tuesday","Wednesday","Thursday","Friday"};
puts("nWeekdays: Before Sorting");
for (int i=0; i<5; i++)
{
puts(weekdays[i]);
}
b_sort(weekdays);
puts("nWeekdays: After Sorting");
for (i=0; i<5; i++)](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-276-320.jpg)
![262
{
puts(weekdays[i]);
}
getch();
}
Example 7.9.6
Find the output for the following piece of code:
# include <iostream.h>
# include <conio.h>
void main()
{
clrscr ();
int *ptr;
int number = 5;
ptr = &number;
cout<<ptr<<endl;
cout<<&number<<endl;
cout<<*&ptr<<endl;
cout<<*(*&ptr)<<endl;
cout<<*&*&*(*&ptr)<<endl;
getch();
}
Solution of 7.9.6
OUTPUT SCREEN
0x1fa70ffa
0x1fa70ffa
0x1fa70ffa
5
5](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-277-320.jpg)
![263Part I: Chapter 7
POINTERS
7.10 Review Exercise
1 Mark/ 2 Mark Questions
Q1) A pointer stores
a. an address of a variable
b. the values of the variable it points to
c. is a set of unallocated bytes of memory
d. a data type of an address variable
Q2) The expression *post;
a. is a pointer to the variable post
b. holds the address of variable post
c. refers to the contents of the post.
d. refers to the contents stored to the variable to which post points to.
Q3) Will the following piece of code compile without errors:
int ptr;
int *ptr1;
int **ptr2
ptr1=&ptr;
ptr2=*ptr1;
ptr = ptr2;
Q4) Are the following c++ statements equivalent:
i) int ptr[2] and *(int ptr +2)
ii) int *ptr; and int *ptr;
ptr = &add; ptr = *& add;
Q5) Is pointer arithmetic the same as conventional arithmetic. If not how is it different
from conventional arithmetic? Give suitable examples?
Q6) Write c++ statements
i. to input a set of two integers say, ‘a’ and ‘b’ from the user and
ii. swap their values using pointers.
Q7) Assuming the code given below, find the error in each of the program segments a),
b) and c).](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-278-320.jpg)
![264
Code:
int *ptr, number;
int num[5] = { 5,10 ,15 , 20, 25, 30};
ptr = num;
Statements:
a) // Use pointer to get the address of first value of array.
number = ptr;
b) // Change the value ’10’ at the index position to 2 of array num to ‘5’
*(ptr+1)=5;
Q8) Find the output for the following piece of code.
# include <iostream.h>
# include <conio.h>
struct coordinate
{
int x
int y;
};
void display ( coordinate n)
{
cout<<" Row ="<<n.x<<"t Column = "<<n.y<<endl;
}
void main()
{
clrscr();
coordinate A = {20,50}, B;
coordinate *c;
c=&b;
c->x += 20;
display (b);
c = &a;
c->y +=20;
display (a);
getch();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-279-320.jpg)
![265Part I: Chapter 7
POINTERS
Q9) Write a program which reads in a group of 10 numbers and places them in an array
of type int. It then calculates the sum of all the even numbers in the set of these 10
numbers and prints the even numbers and their sum on the screen. Use pointer
notation where ever possible.
Q10) Write a program to input an array of strings. The program should also contain a
function to convert all the characters of the string to upper case and also count the
number of lower case characters. Use pointer notation where ever possible.
3 Mark/ 4 Mark Questions
Q1) Find the output for the following piece of code:
int arr[] = {2,3,4,5};
int *arrptr=array;
int val = *arrptr;
cout<< val<<endl;
val = * arrptr++;
cout<<val<<endl;
val = *arrptr;
cout<<val<<endl;
val = *++arrptr;
cout<<val<<endl;
Q2) Find errors in the following piece of code:
/**************************
The program swaps the values of two integers with the help of pointers
**************************/
// Header Files
# include <iostream.h>
# include <conio.h>
// Main Function
void main ()
{
clrscr ();
int *ptr;
int num1;
int num2;
num1=&ptr;
num2= *ptr;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-280-320.jpg)
![266
Q3) Create a database for 20 employees in a company, the database should include employee
name, employee number, employee designation, employee salary, employee department.
Create separate functions to sort the database in ascending order with employee name,
employee id and employee salary. Also create function Emp_search to search any
employee with employee number as the search key. Use structures and pointer notation
where ever possible?
Q4) Write C++ statements for the following:
i) Initialize a pointer to an integer.
ii) Initialize a pointer to a string.
iii) Initialize a pointer to an array of 10 characters.
iv) Initialize a pointer to a pointer to a character, a floating point.
v) Initialize a pointer to the following structure:
struct graph
{
int x;
int y;
};
Also then initialize x = 5, y =10 using the pointer variable.
Q5) Read a sequence of words from the user. Keep reading the sequence till the point
the user enters the word ‘STOP’. Print these words in the order of sequence do not
print any word starting from the character ‘s’. Also write a function Seq_sort which
returns a sorted list of words to be printed on the screen. Use pointer notation where
ever possible.
Q6) State the difference between an array of characters and a string. How is size of an
array of characters different from a string? What is the size of the array ‘str’?
char str[] = “This is Question 6”;
Declare an array of strings in which the strings contain the names of all the days.
Ask the user to input a set of 10 random strings. Print those days in order of
occurrence. Use pointer notation where ever possible.
7.11 Programming Project
A very popular puzzle which most of us have played when we were young is the ‘10-
puzzle’. This puzzle’s frame can hold 10 tiles. It consists of blocks number from digits 1
to 9 and an empty space. The purpose of the game is to slide the blocks. The goal is to
slide the blocks using the empty space block such that each block is arranged in the
ascending order when observed row wise. For example if ‘X’ represents the empty space](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-281-320.jpg)
![275Part II: Chapter 1
Object Oriented Programming
public:
void input();
void output();
};
:
HUMAN Steve, Jordan ;
In the above example Steve and Jordan are object declarations. Members can be kept in
any of the three accessibility modes private, public and protected. Members in the public
visibility mode of the class are accessible to the objects of the class whereas members in
the private accessibility mode are not accessible to the objects of the class. They are only
accessible inside the class to the member functions of the class.
NOTE: Protected members will be explained later in Chapter 2 entitled ‘Inheritance’.
Program 1.1 illustrates the use of classes and objects:
Program 1.1
/***********************************************************************
Program 1(1).cpp
Explaining the concept of classes and objects
***********************************************************************/
// Header files
# include <iostream.h>
# include <stdio.h>
// Declration of class hospital
class hospital
{
private:
int patientno,wardno; //private data members
char patient_name[20];
public:
void register_patient(); //public member function prototype
void showstatus() //public member function declaration
{
cout<<patientno<<":"<<patient_name<<":"<<wardno<<endl;
}
}; //semicolon is a must](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-290-320.jpg)
![284
1.3.6 Solved Examples
Example 1.3.6.1
Define a class in C++ with the following description
Private members
• A data member Flight Number of type integer
• A data member Destination of type string
• A data member Distance of type float
• A data member Fuel of type float
• A member function CALFUEL() to calculate the value of fuel as per the
following criteria
Distance Fuel
<=1000 500
more than 1000 and 1100
<=2000
more than 2000 2200
Public members
A function FEEDINFO() to allow the user to enter values for flight
number, destination, distance & call function CALFUEL() to calculate the
quantity of fuel
A function SHOWINFO() to allow user to view the content of all the data
members
Solution of 1.3.6.1
class FLIGHT
{
int Fno; // Represents Flight Number
char Destination[20]; // Represents Destination
float Distance; // Represents Distance
float Fuel; // Represents Fuel
void CALFUEL();
public:
void FEEDINFO();
void SHOWINFO();
};
void FLIGHT::CALFUEL()
{
if(Distance<=1000 )
Fuel=500;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-299-320.jpg)
![286
Solution of 1.3.6.2
class TEST
{
int TestCode;
char Description[20];
int Nocandidate;
int Centerreqd;
void CALCNTR();
public:
void SCHEDULE();
void DISPTEST();
};
void TEST::CALCNTR()
{
Centerreqd=Nocandidate/100+1;
}
void TEST::SCHEDULE()
{
cout<<" Test code:";
cin>>TestCode;
cout<<" Description:";
gets(Description);
cout<<"Number:";
cin>>Nocandidate;
CALCNTR();
}
Example 1.3.6.3
WAP (Write a program) to create an array of objects of class stud
class stud
{
int rno;
char nam[10];
float avg;
public:
void in();](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-301-320.jpg)
![287Part II: Chapter 1
Object Oriented Programming
void out();
};
Write a user defined function to sort the array of class stud in ascending order of roll no?
Solution of 1.3.6.3
In this class we will have to create a function retroll() which will return the
Rollno (‘rno’) of the particular objecT
class stud
{
int rno;
char nm[10];
float avg;
public:
void in();
void out();
int retroll ()
{
return rno;
}
};
void stud :: in()
{
cout<<” Roll no :”;
cin>>rno;
cout<<” Enter the name”;
gets(nm);
cout<<” Enter the avg”;
cin>>avg;
}
void stud ::out()
{
cout<<” Roll no:”<<rno;
cout<<” Name:”;
puts(nm);
cout<<”Avg:”<<avg;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-302-320.jpg)
![288
void accept ( stud s[],int n);
void display( stud s[],int n);
void sortroll( stud s[],int n);
void main()
{
stud a[50];
int n;
cout<<”enter the no of students”;
cin>>n;
accept(a,n);
cout<<” The contents of the array”;
display(a,n);
sortroll(a,n);
cout<<” The array in sorted form of roll nos is “;
display(a,n);
}
Example 1.3.6.4
Give the output for the following questions
(a) class parts
{
char pnm[20];
float prc;
public:
void assign(char p[],float pr)
{ strcpy(pnm,p);prc=pr;}
void show()
{cout<<pnm<<”@”<<prc<<endl;}
void price( int l=0)
{ prc=(l==0)?prc+90:prc-90;}
};
void main()
{
parts p,q;
p.assign(“bolts”,900);
q.assign(“clamps”,345);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-303-320.jpg)
![309Part II: Chapter 1
Object Oriented Programming
calculate(SC);
cout<<SC.point<<”;”<<SC.bonus<<endl;
calculate(SC,15);
cout<<SC.point<<”;”<<SC.bonus<<endl;
}
c) OUTPUT SCREEN
d)
// Header Files
# include<iostream.h>
# include<conio.h>
# include<string.h>
# include<stdio.h>
// Declaration of class student
class student
{
char name[15];
int rollno;
public:
student()
{
strcpy(name,"Ankit");
rollno=1;
}
student(int a)
{
strcpy(name,"Neha");
rollno=a;
}
void output();
void increase();
};
void student::output()
21;15
22;25
23;40](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-324-320.jpg)
![314
Question 1.5.4
Define a structure? Declare a structure in c++ with name, roll number and age as its
components.
Answer of 1.5.4
A structure is a collection of variable having different data types. The declaration as
struct student
{
char name[15];
int roll;
int marks;
};
Exercise: Now think how classes and structures are different and similar to each other.
Example 1.5.5
Declare a structure client containing the following members:
The character quantity of the client
The integer quantity called clientid
The integer quantity called no_of_items
The long integer quantity called Client_total
Solution of 1.5.5
struct client
{
char name[15];
int clientid;
int no_of_items;
long client_total;
};
Example 1.5.6
Write a program to add and subtract complex numbers using the structure given below:
struct complex
{
int real;
int imag;
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-329-320.jpg)
![323Part II: Chapter 1
Object Oriented Programming
Solution of 1.5.15
class student
{
int admno;
char sname[20];
float eng,math,science;
float total;
void ctotal();
public:
void takedata();
void showdata();
};
void student::takedata()
{
cout<<” n Enter the admission number of student “;
cin>>admno;
cout<<”n Enter the name of the student”;
gets(sname);
cout<<”n Enter the marks in English”;
cin>>eng;
cout<<”n Enter the marks in Maths”;
cin>>math;
cout<<”n Enter the marks in Science”;
cin>>science;
ctotal();
}
void student ::ctotal()
{
total = eng + math + science;
}
void student::showdata()
{
cout<<”n Name ”<<sname;
cout<<”n Admission no”<<admno;
cout<<”n Total marks”<<total;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-338-320.jpg)
![324
Example 1.5.16
Define a class batsman with the following specifications:-
Private members
bcode (4 digit code number)
bname (20 characters)
innings, notout, runs integer type
batavg it is calculated according to the formula:
batavg = runs/(innings – notout)
calcavg() function to compute batavg
public members
readdata() function to accept values for bcode, name, innings, notout and
invoke the function calcavg()
displaydata() function to display the data members on the screen
Solution of 1.5.16
class batsman
{
int bcode;
char bname[25];
int innings, notout, runs;
float batavg;
float calcavg()
{
return (runs/(innings-notout));
}
public:
void readdata();
void displaydata();
};
void batsman::readdata()
{
clrscr();
cout<<”Enter the batsman code”;
cin>>bcode;
cout<<”Enter the batsman name”;
gets(bname);
cout<<”Enter total innings”;
cin>>innings;
cout<<”Enter total notout”;
cin>>notout;
cout<<”Enter total runs”;
cin>>runs;
batavg=calcavg();
}
void batsman :: displaydata()
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-339-320.jpg)
![327Part II: Chapter 1
Object Oriented Programming
4 Mark questions
Q1) Declare a class employee having the following members:
Data Members:
Employee name
Employee address
Employee number
Member Functions
To read data members
To display the data members
Q2) Define a class teacher with the following specification:
private members of class teacher
name 20 characters
subject 10 characters
basic, hra, da float
netpay float // Obtained by the sum of float variables ‘basic’ ‘hra’ and
‘da’
calculate A function computes and returns the net salary (‘netpay’) a
public member function of class teacher
readdata() Function to accept data values and invoke the calculate function
displaydata() Function to prints the teacher information on the screen
Q3) Declare a structure date having members as
day - a string of size[2]
represents day part of a date
month - a string of size [20]
represents month name of a date](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-342-320.jpg)
![328
year - a string of size[4];
represents year part of a date
Declare a class medicine with the following specifications
private members of a class
medicine_name - a string of size[25]
priceperunit - float
mfgdate - of type date
expdate - of type date
public members of the class
medentry() - A function to accept values of the data members from user
ShowWarning() - A function that displays the name of the medicine along with
warning message as “Use within 2 Months”, if the expiry date is 2 months after
manufacturing date
ShowAll() - A function that displays above details of a medicine properly
formatted in a single line
Q4) Define a class to represent a bank account. Include the following members:
Data members:
Name of the depositor
Type of the account
Account number
Balance amount in the account
Member functions:
To deposit an amount
To withdraw an amount.
To display the name, type of account and the balance of the account holder](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-343-320.jpg)
![335Part II: Chapter 2
INHERITANCE
members and member functions of their own as well as they may use all the functionality
of the base class as well.
Figure 2.1: Inheritance Hierarchy
2.3 Inheritance in C++
Let us look at the Program 2.1 to understand how inheritance is used in C++.
Program 2.1:
class cat_family
{
public: // The data and methods can be accessed directly
char category[10];
char legs;
int age;
cat_family();
~cat_family();
void food_habits();
void place_of_stay();
};
class panther : public cat_family
{
public:
void sex_of_specie()
{
if (strcmp(category,”White Panther”)==0)
{
cout<<” White Panther is an endangered specie”;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-350-320.jpg)
![338
Now that we understand the concept of inheritance let us look at the following examples.
2.3.1 Solved Examples
Example 2.3.1.1
Answer the questions (i) to (iv) on the basis of the following code:
(i) Write the names of data members which are accessible from objects belonging
to class ‘branch’.
(ii) Write the names of all the member functions which are accessible from
objects belonging to the class employee.
(iii) Write the names of all the members which are accessible from member
functions of class ‘employee’.
(iv) How many bytes will be required by an object belonging to the class
‘employee’?
class company
{
char website[12];
char name[20];
protected:
int noemployees;
public:
company();
void register();
void status();
};
class branch:public company
{
long nocomputer;
protected:
float expense;
public:
branch();
void enter();
void show();
};
class employee:private branch
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-353-320.jpg)
![339Part II: Chapter 2
INHERITANCE
int eno;
char ename[20];
float salary;
public:
employee();
void joining();
void showdetail();
};
Solution of 2.3.1.1
(i) None of the data members are accessible from objects belonging to class
branch.
(ii) The names of member functions which are accessible from objects belonging to
class ‘employee’ are joining(), showdetail().
(iii) The members accessible from objects belonging to class ‘employee’ are
Data Members: eno, ename, salary, expense, noemployees
Member Functions: register(), status(), enter(), show(), joining(), showdetails().
(iv) The amount of bytes required by an object belonging to class ‘employee’ is
summation of all the bytes required by the objects of all the base classes and the
class ‘employee’ which comes out to be ‘68’ bytes.
Note: ‘register’ also happens to be a keyword in C. Hence some C/C++ compilers
might give an error in executing the program. Here the author assumes that this is
understood and lays stress only on the concept of inheritance.
Example 2.3.1.2
Given the following class definitions answer the questions that follow:
COMMON PROGRAMMING ERRORS
Assuming that private members of a base class can be
inherited directly by the derived classes and then using them
as a part of the derived class is a logical and a compilation
error.
Declaring the constructors and destructors in the private
access type specifier is an compilation error.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-354-320.jpg)
![340
class publisher
{
char pub[12];
double turnover;
protected:
void register();
public:
publisher();
void enter();
void display();
};
class branch: protected publisher
{
char city[20];
protected:
float employee;
public:
branh();
void haveit();
void giveit();
};
class author: private branch
{
int acode;
char aname[20];
float amount;
public:
author();
void start();
void show();
};
(i) Write the names of data members which are accessible from objects belonging
to class author.
(ii) Write the names of all the member functions which are accessible from objects
belonging to the class branch.
(iii) Write the names of all the members which are accessible from member
functions of class author.
(iv) How many bytes will be required by an object belonging to class author?](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-355-320.jpg)
![343Part II: Chapter 2
INHERITANCE
2.4 Types of Inheritance
Inheritance exists in two forms depending upon the number of parent or base classes a
derived class has.
a) Single Inheritance (Figure 2.2):
When a class inherits only from a single parent or in other words each derived
class has a single parent. A good example of single inheritance is the tree
structure [see Figure II (2.2(b))].
(a) (b)
Figure 2.2: Single Inheritance (SI)
b) Multiple Inheritance
Multiple Inheritances are observed when the derived class inherit properties from
more than one parent or base class. This is a very common practice used in object
oriented programming as it allows each class to inherit and combine all useful
features from a range of classes. Figure 2.3 shows few examples where multiple
inheritances are observed.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-358-320.jpg)
![346
(ii) An object belonging to class ‘yourdata’ will occupy 12 bytes.
(iii) The function definition for ‘catch()’ which displays all the data members
accessible to it is as follows.
void yourdata::catch()
{
cout<<value<<data2<<val<<endl;
}
(iv) The member functions accessible class ourdata are catch(), miss() of ‘yourdata’
and have(), give() of class ‘ourdata’.
Example 2.5.2
Given the following class definitions answer the questions that follow:
class myclass
{
char data1[20];
int data2[5];
public:
void read();
void show();
};
class myfirstclass : private myclass
{
int deriveddata1;
float deriveddata2;
protected:
int standard;
public:
void read_firstderived();
void show_firstderived();
};
class mysecondclass : public myfirstclass
{
float deriveddata3;
public:
void read_secondderived();
void show_secondderived();
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-361-320.jpg)
![349Part II: Chapter 2
INHERITANCE
public:
char equipment_name[15];
int equipment_price;
electronics()
{
cout<<endl<<" An object of type Electronics has been declared";
}
~electronics()
{
cout<<endl<<" The scope of the created object is now over";
}
void enter();
void display();
};
class computer: public electronics
{
public:
computer()
{
cout<<endl<<" Computer is a derived class of electronics";
}
char motherboard;
int modelno;
};
void electronics::enter()
{
cout<<" Enter Equipment Name";
gets(equipment_name);
cout<<endl<<" Enter Equipment Price";
cin>>equipment_price;
}
void electronics:: display()
{
cout<<endl<<" Equipment Name ->"<<equipment_name;
cout<<endl<<" Equipment Price ->"<<equipment_price;
}
void main()
{
clrscr();
computer c1;
electronics e1;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-364-320.jpg)
![351Part II: Chapter 2
INHERITANCE
Q6) Find the error(s) for the constructors in the following code fragments:
i) class sample1 ii) class sample2
{ {
float a,b,c; float a,b,c;
public: sample2();
float sample1(); };
};
Q7) Is it possible for a derived class to access the private members of the base class
indirectly. If yes then create a base class and a derived class then use these classes to
illustrate your point?
4/5 Mark Questions
Q8) Given the following class definitions:
class world
{
char continents[10];
protected:
char oceans[10];
public:
void input (char[]);
void output();
};
class country : private world
{
char cname[20];
protected:
char capital[20];
public:
void indata();
void outdata();
};
class state: public country
{
char sname[20];](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-366-320.jpg)
![359
Part III: Chapter 1
Data Structures
Static Data structure
The data structure in which the numbers of elements are fixed is known
as static data structure. Example: Arrays
Dynamic data structures
The data structures in which the numbers of elements are not fixed are
known as dynamic data structure. Example: Linked List, Dynamic stack
and Dynamic Queue.
1.3 Arrays
Arrays are examples of static homogeneous data structures as the number of
elements in array remain fixed. They are also commonly known as subscripted
variables. We have already studied in previous chapters about how arrays are
stored in consecutive memory locations. In this section, we will learn how we
can determine the address of the memory locations occupied by an array. Let us
begin with One Dimensional arrays.
1.3.1 One Dimensional Array
Suppose we declare a one dimensional array ‘int A[i],’ where ‘i’ can take up
any integer value. Now the number of elements in this array can also be found
by subtracting the last index of the array also known as the ‘Upper Bound’ with
the first index of the array also known as the ‘Lower Bound’. In this example,
we have ‘i’ as the upper bound and ‘0’ as the lower bound. We can use the
following general formula to determine the memory location of any index of a
array.
Memory location of A[i]:
Location (A[i]) = Base_address (A) + W*(i-LB)
Where,
Number of elements = Upper Bound – Lower Bound
= i - 0
= i
Where ‘W’ is the size of the object created.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-374-320.jpg)
![360
1.3.2 Two Dimensional Arrays
As discussed in Part I that the computer memory only supports a linear placement
of data items akin to a one-dimensional array as a result of which the memory
simulates the two/multi dimensional array as a one dimensional array only. Two
dimensional arrays are also stored at contagious memory locations similar to that
of the one dimensional array. The only difference between them being that two
dimensional arrays can be stored in the following two ways:
Row major
Column Major
Let us consider a two dimensional array A[i][j].
Number of elements = i * j = (UB(i) – LB(i) + 1) * (UB(j) – LB(j) +1)
Where UB is the upper bound and LB is the lower bound.
o Row Major
Let us take a two dimensional array A[i][j] of size M*N that is
having M rows and N columns (Figure 1.1).
Figure 1.1: An array having M rows and N columns
The computer always keeps track of the base address only. So,
by knowing the base address (Base (A)) and size of each element
(W) one can calculate the address of any [I,J] element. The
algorithm for a row major two dimensional array is given below:](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-375-320.jpg)
![361
Part III: Chapter 1
Data Structures
Algorithm: For Row Major
For Two dimensional array A[i][j] of size M rows and N
columns.
Base Address : Base (A), Address of the 1st
element of the array
No. of Column s: N = UB (j) – LB(j)
UB(j) is the upper bound of column
LB(j) is the lower bound of the column
No. of Rows: M = UB(i) – LB(j)
UB(i) is the upper bound of the row
LB(j) is the lower bound of the column
Size of object: W is the size of the object
Memory Location: Loc(A[I][J]) = Base(A) + W*(M*I+J)
Column Major
‘Column Major’ is also similar to the ‘Row Major’ order the
only difference being that in the column major the elements are
stored along the column. That is first the elements of the first
column are filled and then the other ones are filled sequentially.
To explain the functionality of column major order let us
suppose an array A[i][j] with M rows and N columns.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-376-320.jpg)
![362
The algorithm given below explains the functionality of column
major order.
Algorithm: For Row Major
For Two dimensional array A[i][j] of size M rows and N columns.
Base Address : Base (A), Address of the 1st
element of the array
No. of Columns: N = UB (j) – LB(j) + 1
UB(j) is the upper bound of column
LB(j) is the lower bound of the column
No. of Rows: M = UB(i) – LB(j) + 1
UB(i) is the upper bound of the row
LB(j) is the lower bound of the column
Size of object: W is the size of the object
Memory Location:
Loc(A[I][J]) = Base(A) + W*(N*(J-LBJ) + (I-LBI))
If we assume LB(i), LB(j) as ‘0’ then we obtain a
much
simplified formula:
Loc (A[I][J]) = Base(A) + W*(N*J+I)
1.3.3 Solved Examples
Example 1.3.3.1
A one-dimensional array p[100] is stored in memory with a base address as 5000. Find
out addresses of p[15] and p[40], if each element of the array requires 4 bytes.
Solution of 1.3.3.1
Given
Base(p) = 5000
W = 4
Loc (p[I]) = Base (p) + W* I
Loc (p[15]) = 5000 + 4*15](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-377-320.jpg)
![363
Part III: Chapter 1
Data Structures
= 5000 + 60
= 5060
Loc (p[40]) = 5000 + 4*40
= 5000 + 160
= 5160
Solved Example 1.3.3.2
A one dimensional array A[-5,…,25] is stored in the memory with each element
requiring 2 bytes. If the base address is 8000, find out the following:
a) Address of A[5] and A[-3]
b) Total no. of elements present in the array
Solution of 1.3.3.2
Given,
Base(A) = 8000
W = 2
LB = -5
Loc(A[I]) = Base(A) + W*(I-LB)
Loc( A[5]) = 8000 + 2*(5-(-5))
= 8000 + 20
= 8020
Loc( A[-3]) = 8000 + 2*(-3-(-5))
= 8000 + 4
= 8004
Total
No of elements = UB – LB + 1 = 25 – (-5) +1 = 31
Solved Example 1.3.3.3
A two dimensional array Q[5][15] is stored in memory along the row with each element
requiring 2 bytes. If the base address is 6500, find out the following:
a) Addresses of q[5][10] and q[3][5]
b) Total no. of elements present in the array
Solution of 1.3.3.3
Given,
Base (q) = 6500
W = 2
M = 5](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-378-320.jpg)
![364
Row major, Loc ( Q [I][J] ) = Base (Q) + W*(M*I+J)
Loc ( Q[5][10]) = 6500 + 2 * (15 *5 +10)
= 6500 +170
= 6670
Loc (Q[3][5]) = 6500 + 2 * (15*3 +5)
= 6500 +100
= 6600
Total number of elements = N * M = 5 * 15 = 75
Solved Example 1.3.3.4
R[-4, …, 4, 7, …,17] is a two dimensional array, stored in the memory along the
column with each element requiring 4 bytes. If the base address is 5000, find out the
following
a) Addresses of R[2][10] and R[3][15]
b) Total no of elements in the array
Solution of 1.3.3.4
Given,
Base( R ) = 5000
W = 4
N = UBI – LBI + 1 = 4 – (-4) +1 = 9
M = UBJ – LBJ + 1 = 17 – 7 + 1 = 11
Column Major,
Loc ( R[I][J]) = Base ( R ) + W * ( N * ( J – LBJ ) + ( I – LBI ))
Loc ( R[2][10]) = 5000 + 4*( 9 * (10 -7 ) + ( 2 – (-4))
= 5000 + 132
= 5132
Loc ( R [3] [15] ) = 5000 + 4*(9*(15-7) + ( 3 – (-4))
= 5316
Total
No of elements = N * M = 9 *11
1.4 Operations on static data structures
1.4.1 Insertion
The insertion operation is performed by inserting an element at a desired location
in an array. However, to add an element in between the array, all elements from](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-379-320.jpg)
![365
Part III: Chapter 1
Data Structures
the location of insertion need to be moved one location forward to accommodate
the new element. This operation is shown in figure 1.2.
Figure 1.2: Insertion of new element '19' in array INSERTION
The program 1.1 gives the function to insert an element at a desired location into an array.
Program 1.1: Function to insert Data at ‘Pos’ in the array ‘arr’.
void insert (int arr[], int &l, int data)
{
if (l<max)
{
for (int c=l;c>pos;c--)
arr[c]=arr[c-1];
arr[pos]=data;
l++ ;// After insertion of an element increases L
}
else
cout<<” Array is Full”<<endl;
}
Note: ‘Pos’ is presumably defined as global variable and ‘pos’ is the position from which
insertion will take place.
1.4.2 Deletion
As the name suggests the ‘deletion’ operation is used to delete an element from a
specified position (‘pos’ in figure 1.3) in the static data structure. When the
selected element is deleted from the data structure, then the elements stored at
subsequent positions in the structure shift either one step forward or shift one step
Location of Element
for insertion
New element
inserted](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-380-320.jpg)
![366
backward towards the position from where the element was deleted as also shown
in the figure 1.3.
Figure 1.3 Element 19 deleted from array INSERTION
The program 1.2 given below enlists the piece of code used to delete an element from an array.
Program 1.2: Function to delete data from POS in the array ‘arr’
void delete (int arr[], int &l, int pos)
{
for (int c=pos;c<l-1;c++)
arr[c]=arr[c+1];
l--;
}
1.5 Solved Examples
Example 1.5.1
The base address of a two dimensional array A with 5 rows and 4 columns is 2000. If each
element requires 4 bytes for its storage, find the address of element a[3][3] if arrangement
is row major?
Solution of 1.5.1
Now we have been given the Base address (A) = 2000
POS
Array INSERTION before
and after DELETION](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-381-320.jpg)
![367
Part III: Chapter 1
Data Structures
Size of each element (w) = 4
Number of columns (N)= 4
Now, the elements have been stored as row major
Address of element A[i][j]= A + w(N*I + j)
= 2000 + 4 * (4*3+3)
= 2060
So, the required address is 2060.
Example 1.5.2
For an array of real numbers Realarr[20][20] find the address of Realarr[10][12] if
Realarr[1][1] is stored in location 1000. Assume each real number requires 4 bytes. Show
the various steps of calculation?
Solution of 1.5.2
We use following formula to compute the address:
A[i][j] = B + w(n(I-LI) + (J-L2))
Now, Base address (B) = 1000
Size of each element = 4
L1=L2=1
A[10][12]= 1000 + 4(20(20-1) + (20-1))
= 1000 + 4(20(19) + 19)
= 1000 + 4(380 +19)
= 1000 + 4(399)
= 1000 + 1596
= 2596
Example 1.5.3
Write an algorithm which finds the location and values of largest and second largest
elements in two dimensional array DATA with N rows and M columns.
Solution of 1.5.3
Let us assume an array A[][] having M rows and N columns.
Step : 1. Max = A[0][0]
2. Min = A[0][0]](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-382-320.jpg)
![368
3. i=0
4. while (i<m) repeat 5 to 10.
5. j=0
6. while (j<N) repeat step 7 to 9
7. if (Max<A[i][j])
Max= A[i][j]
8. if (Max>A[i][j])
Max=A[i][j]
9. j= j+1
10. i= i+1
11. cout<<” Maximum value is”<<Max
12. cout<<” Maximum value is”<<Min
13. End.
Example 1.5.4
Each element in an array DATA[20][50] requires 4 bytes of storage. Base address of
data is 2000, determine the location of DATA [10][10] when the array is stored as
(a) Row major
(b) Column major
Solution of 1.5.4
Base Address = 2000
Size of the element = 4
(i) Row major Order
N=50
DATA[10][10]= 2000 + 4(50(10-1) +(10-1))
= 2000 + 4(50 (9) + 9)
= 2000 + 4(454)
= 2000 + 1816
= 3816
(ii) Column major Order
M = 20
DATA[10][10]= 2000 + 4(20(10-1) + (10-1))
= 2000 + 4(20(9) + 9)
= 2000 + 4(180 +9)
= 2000 + 4(189)
= 2000 + 756
= 2756](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-383-320.jpg)
![369
Part III: Chapter 1
Data Structures
Example 1.5.5
Given an array A[6][16] whose base address is 100. Calculate the location A[2][5] if each
element occupies 4 bytes and array is stored row wise?
Solution of 1.5.5
Base address = 100
W=4
L1 = 1 L2 = 1
N = 16
A[2][5] = B + W(n(I- L1) + (J – L2))
= 100 + 4(16(2-1) + 5-1)
= 100 + 4(16 + 4)
= 100 + 80
= 180
Example 1.5.6
Explain how bubble sort will sort the following array:
11 6 26 9 8 24 18
Solution of 1.5.6
Bubble sort compares adjacent elements and swaps them if necessary and then compares
the next pair of adjacent elements.
The method when applied to the given array gives the following result:-
Pass 1: Swap numbers 11 and 6 to give 6 11 26 9 8 24 18
Swap numbers 26 and 9 to give 6 11 9 26 8 24 18
Swap numbers 26 and 8 to give 6 11 9 8 26 24 18
Swap numbers 26 and 24 to give 6 11 9 8 24 26 18
Swap numbers 26 and 18 to give 6 11 9 8 24 18 26
Pass 2: Swap numbers 9 and 11 to give 6 9 11 8 24 18 26
Swap numbers 11 and 8 to give 6 9 8 11 24 18 26
Swap numbers 24 and 18 to give 6 9 8 11 18 24 26
Pass 3: Swap numbers 9 and 8 to give 6 8 9 11 18 24 26](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-384-320.jpg)
![371
Part III: Chapter 1
Data Structures
Traversal
Linear search
Bubble sort
Merging
Q3) Which of the two (‘bubble’ and ‘selection’) sort is the most efficient and why?
Q4) The base address of a two dimensional array X with 15 rows and 4 columns is 2000.
If each element requires 4 bytes for its storage, find the address of element X[10][4]
if arrangement is row major?
Q5) The following numbers are required to be sorted using bubble and selection sort
showing how the list would appear at the end of each pass.
a) 20 40 30 10 50
b) 10 2 3 4 -2 1 5
Q6) If 2 dimensional array C[5…10,-5…9] is stored using column major representation,
then calculate the address if C[8,-2], if the base address is 10 and each array element
requires 2 bytes of memory.
2/3 Mark Questions
Q7) Each element of an array DATA[20][50] requires 2 bytes of storage. Base address of
DATA is 2000. Determine the location of DATA[10][22] when the array is stored as
(a) Row Major
(b) Column Major
Q8) Given two dimensional array A[10][20], base address of A is 100 and width of each
element is 4 bytes. Find the location of A[12]
Q9) If two dimensional array C[5, …, 9,-4,…, 10] is stored using column major
representation, then calculate the address if C[8,-2], if the base address is 10 and each
array element requires 2 bytes of memory.
4/5 Mark questions
Q10) Write a menu driven program in C++ to sort an array using selection and bubble
sort.
Q11) Write a menu driven program to search the array using linear and binary search?
Q12) Explain the following terms primitive and non primitive data structures and
differentiate between them giving suitable examples.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-386-320.jpg)
![373
Part III: Chapter 1
Data Structures
1.8 Let us revise!
The data structure, which is available in the compiler, is known as a
primitive data structure.
The data structure, which is not available in the compiler, is known as
non primitive data structure.
The data structure in which each element has access to maximum of
one predecessor element and maximum of one successor element is
known as linear data structure. Example: stack, queue, etc.
The data structure in which each element can access any number of
predecessor elements and any number of successor elements known as
non linear data structure. Example: Tree, Graphs, etc.
The data structure in which the numbers of elements are fixed is known
as static data structure. Example: Arrays
The data structures in which the numbers of elements are not fixed are
known as dynamic data structure. Example: Linked List
One dimensional array
Number of elements N= UB – LB + 1
Where LB- lower bound, UB – upper bound
Memory location of A[i]: Loc (A[i]) = Base (A) + W*(i-LB)
Loc (A[i])=Base(A) + W*I
Where N is given as in c++, LB is assumed as 0
Two dimensional array
A two dimensional arrays can be stored in the following two ways.
Row major
Column Major
Number of elements = N * M = (UBI – LBI + 1) * (UBJ – LBJ +1)](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-388-320.jpg)
![384
strcut Employee
{
char Name[15];
int Age;
Employee *ptr;
};
Solution of 2.2.2.1.1
/************
Solved Example: 2.2.2.1.1
*****************************************/
// Header Files
# include<iostream.h>
# include<conio.h>
# include<stdio.h>
/****************
Self Referencial structure
Name: Employee
Data Members: character array Name[]
integer variable 'Age'
pointer variable 'ptr'
******************************************************/
struct Employee
{
char Name[15];
int Age;
Employee *ptr;
};
// Main Function
void main()
{
clrscr();
int i=1;
// Pointer variable pointing to memory block of type Employee
Employee *E1 = new Employee ;
// Pointer variable l2 of type Employee](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-399-320.jpg)
![392
Figure 2.19: Deleting a data value from a stack
We can also develop a C++ code to carry this procedure. The program 2.3 given below
represents stack as an array.
Program 2.3
// Header Files
# include<iostream.h>
# include<conio.h>
// Maximum No. of data entries allowed in the stack
const int max=8
/*******************************************
Function Name: Push
Purpose: To insert a data entry in the array
*******************************************/
void Push(int s[],int &t)
{
if (t<max)
{
cout<<" Data to be entered";
cin>>s[++t];
cout<<endl<<"Data has been entered";
cout<<endl<<"_____________________";
}
else
cout<<"n The stack is full"<<endl;
}
/*******************************************
Function Name: Pop
Purpose: To delete a data entry from the array](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-407-320.jpg)
![393
Part III: Chapter 2
Linked Lists
*******************************************/
void Pop(int s[],int &t)
{ if (t!=-1)
{
cout<<endl<<" The data entry "<<s[t]<<" has been
deleted";
t--;
}
}
/*******************************************
Function Name: display
Purpose: To display data entries in the stack
*******************************************/
void display(int s[],int &t)
{
cout<<endl<<"Displaying data entries in the stack";
for (int i=t;i>=0;i--)
cout<<endl<<s[i];
}
//Main Function
void main()
{
clrscr();
int static_stack[max], top=-1;
int ch;
do
{ cout<<endl<<"n 1) Push";
cout<<"n 2) Pop";
cout<<"n 3) Display";
cout<<"n 4) Quit";
cin>>ch;
switch(ch)
{
case 1:Push(static_stack,top);
break;
case 2:Pop(static_stack,top);
break;
case 3:display(static_stack,top);
break;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-408-320.jpg)
![394
} while (ch!=4);
getch();
}
Let us closely examine this program. In this program, we have declared a one
dimensional array ‘static_stack’ which can hold 8 integer data values represented by the
constant variable ‘max’. The variable ‘Top’ is initialized to value ‘-1’ and is used to keep
track of the position from where data values are inserted and deleted. The function
‘Push’ is given below:
Function ‘Push’
void Push(int s[],int &t)
{
if (t<max)
{
cout<<" Data to be entered";
cin>>s[++t];
cout<<endl<<"Data has been entered";
cout<<endl<<"_____________________";
}
else
cout<<"n The stack is full"<<endl;
}
This function is used to insert data entries into the stack. The linear array and the
variable ‘Top’ are passed to it as reference parameters. Which means that any changes
made to the formal parameters ‘s’ and ‘t’ are also reflected back to the actual
parameters. The function inserts the data entry at the index ‘t+1’ of the linear array.
This is also the reason why the variable ‘top’ is initialized to -1 when it is being
declared.
This process carries on as long as the number of data entries inserted is less than the
maximum number of data entries allowed in the stack.
Function ‘Pop’
void Pop(int s[],int &t)
{
if (t!=-1)
{
cout<<endl<<" The data entry "<<s[t]<<" has been deleted";
t--;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-409-320.jpg)
![395
Part III: Chapter 2
Linked Lists
}
}
This function is used to delete or pop data entries from the static stack. It also
receives the static array and the variable ‘top’ as reference parameters which are
used to pop a data entry from the stack. This process can carry on as long as the
stack has data entries in it or in other words as long as the value of ‘t’ is not -1.
The important thing to note in this function deletes only one data entry at a time
and it always deletes the last data entry.
Function ‘Display’
void display(int s[],int &t)
{
cout<<endl<<"Displaying data entries in the stack";
for (int i=t;i>=0;i--)
cout<<endl<<s[i];
}
All this function does is that it prints all the data entries present in the stack on
the screen.
2.3.2 Dynamic Stacks
Stacks which are implemented as arrays are very useful if we know the number of data
entries from before, in other words they are useful when a fixed amount of data is to be
accessed. To get an idea of this imagine a recursive loop. A menu driven stack program
might be a good example. Now in this loop, we do not know how many times this loop
might recur or how many data entries the user is willing to enter. The decision entirely
rests on the user and might vary every time the program is executed. Using a stack
represented as an array might be a poor solution as one will have to declare a very large
block of memory so that there is no danger of the program running out of memory. This
can waste a significant amount of computer’s memory if the loop only repeats itself a few
times during the execution of the program. How do we solve this problem?
In such a case, we have to use the concept of dynamic memory allocation. As stacks are
represented as arrays, they can also be represented as pointers by creating a linked list
with the use of which one can create ‘Dynamic stacks’, whose size can increase or
decrease during the execution of the program.
Program 2.4 demonstrates a ‘Dynamic Stack’ with its various operations.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-410-320.jpg)
![411
Part III: Chapter 2
Linked Lists
the same index of the array. Let us now observe, what happens if we insert a data entry
say ‘5’ into the queue?
As we have learnt before that in a queue all the data insertions takes place from the
‘Rear’ end so the data entry ‘5’ is assigned to the 1st
index of the array ‘static_queue[]’
(see Figure 2.33), which is the index the ‘Rear’ end points to. At this point, one might
wonder, why is the data entry added to the 1st
index of the array and not to the zeroth
index of the array? This question will be answered later in this section. For now just try to
understand how stacks work.
Figure 2.33: Insertion of a data element in a queue
We can further use this operation to enter more data entries like ‘10’ and ‘15’ in the
queue. Figure 2.34 shows the effect of this operation on the queue.
(a)](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-426-320.jpg)
![413
Part III: Chapter 2
Linked Lists
Program 2.5: Static circular queue
/***********************
Program 2.5
Static Circular Queue
***********************
/* Header Files */
# include <iostream.h>
# include <conio.h>
const int max=10;
/*******************
Name: qinsert
Return Type: void
Purpose: To enter data into the queue
**************************************/
void qinsert (int q[], int &r, int f)
{
if ((r+1)%max != f);
{
r = (r+1)%max;
cout<<"n Data";
cin>>q[r];
cout<<"n Data has been entered";
}
}
/*******************
Name: qinsert
Return Type: void
Purpose: To delete data from the queue
**************************************/
void qdelete( int q[], int r, int &f)
{
if (r!=f)
{
f=(f+1)%max;
cout<<endl<<q[f]<<"has been Deletedn";
}
else
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-428-320.jpg)
![414
cout<<" Queue is empty n"<<endl;
}
}
/*******************
Name: qdisplay
Return Type: void
Purpose: To display the contents of the queue
**********************************************/
void qdisplay(int q[], int r, int f)
{
int cn=f;
cout<<"n Displaying the contents of the queue n";
while (cn!=r)
{
cn = (cn+1)%max;
cout<< q[cn]<<endl;
}
}
// Main Function
void main()
{
clrscr();
// Initialization steps
int queue[max], rear=0, front=0;
int ch;
do
{
cout<<"n === MENU ===";
cout<<"n 1: Insertn";
cout<<" 2: Deleten";
cout<<" 3: Shown";
cout<<" 4: Quitn";
cout<<" !: Enter choice";
cin>>ch;
switch(ch)
{
case 1: qinsert(queue,rear,front);
break;
case 2: qdelete(queue,rear,front);
break;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-429-320.jpg)
![415
Part III: Chapter 2
Linked Lists
case 3: qdisplay(queue,rear,front);
break;
}
}
while (ch != 4);
getch();
}
Let us closely observe this program. In this program, we have declared a one dimensional
array ‘queue’ which can hold 10 integer data values represented by the constant variable
‘max’ in the program. We have also declared two integer variables ‘rear’ and ‘front’
which are position variables. The variable ‘rear’ is the position variable which directs to
the index of the array from where all the data entries are inserted and the other position
variable ‘front’ directs to the index of the array from where all the dequeue operation is
executed. This program also consists of the following functions to insert, delete and
display the contents of the queue:
Function ‘qinsert’
void qinsert (int q[], int &r, int f)
{
if ((r+1)%max != f);
{
r = (r+1)%max;
cout<<"n Data";
cin>>q[r];
cout<<"n Data has been entered";
}
}
This function is used to insert data entries into the queue. The function receives the linear
array ‘queue’, position variable ‘front’ as variable parameters and the position variable
‘rear’ as a reference parameter which are represented by ‘q’, ‘f’ and ‘r’ in the function.
The function inserts the data entry at the index ‘r’ of the linear array provided the
position variable ‘r’ and ‘f’ do not direct to the same index of the array or in other words
it enters data in the queue until the queue is full. Let us take an example, say this function
is called for the first time by the program to enter the first data entry then the expression
‘(r+1)%max’ would yield the value ‘1’ which is not equal to ‘f’ as ‘f’ was initialized to
‘0’ and hence the program would enter the ‘if’ loop. In the ‘if’ loop the function would
insert the data entered by the user at ‘(r+1)%max’ index of the array, which happens to be
the 1st
index of the array.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-430-320.jpg)
![416
Function ‘qdelete’
void qdelete( int q[], int r, int &f)
{
if (r!=f)
{
f=(f+1)%max;
cout<<endl<<q[f]<<"has been Deletedn";
}
else
{
cout<<" Queue is empty n"<<endl;
}
}
This function is used to delete or pop data entries from the static queue. It receives the
static array ‘queue’, the position variable ‘rear’ as variable parameters and the other
position variable ‘front’ as a reference parameter. The entries are deleted or dequeued
until the queue is not empty which is performed by the if condition ‘r!=f’. To delete data
entries the position variable ‘front’ represented by the formal parameter ‘f’ in the
function is incremented by unity every time the program enters the if loop deleting the
entry at this index of the array. Let us analyze figure 2.36.
Figure 2.36: static array queue[]
Assume this is the state of the static array ‘queue’ when this function is called for the
first time by the program. First, before entering the ‘if’ loop it will check for equality
between position variables ‘f’ and ‘r’. As ‘f=0’ is not equal to ‘r=8’ it will enter the ‘if’
conditional loop. After it enters the ‘if’ conditional loop, the position variable ‘f’ is
assigned the value of ((f+1)%max, (0+1)%10=1) deleting the data entry ‘10’ at this
index of the static array.
Function ‘qdisplay’
void qdisplay(int q[], int r, int f)
{
int cn=f;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-431-320.jpg)
![417
Part III: Chapter 2
Linked Lists
cout<<"n Displaying the contents of the queue n";
while (cn!=r)
{
cn = (cn+1)%max;
cout<< q[cn]<<endl;
}
}
All this function does is that it prints all the data entries present in the queue on the
screen.
2.4.2 Dynamic Queue
The concept of dynamic queues is similar to that of dynamic stacks. As queues can be
represented as arrays they can also be represented with the use of pointers by creating a
link list. Dynamic queues overcome the limitations present with static circular queues as
they can grow and shrink during program execution, thus, save memory. Program 2.6
implements a ‘Dynamic Queue’.
Program 2.6: Dynamic Queue
/**************************
Program
Dynamic Queue
**************************/
/* Header Files */
# include <iostream.h>
# include <conio.h>
/**************************
Self referencial structure
Name: node
Members: 'data' of type integer
: 'next' of type node
********************************/
struct node
{
int data;
node *next;
};
/******************************
Name: queue](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-432-320.jpg)
![430
else
{
rear->next=temp;
rear=temp;
}
}
Example 2.4.2.2.2
Write a function in C++ to perform insert operation on dynamically allocated queue
containing names of students.
struct node
{
char name[20];
node *link;
};
class queue
{
node *rear,*front;
public:
queue();
void insert();
void delete();
};
void queue::insert()
{
node *temp;
temp = new node;
gets(temp->name);
temp->link=NULL;
if (front==NULL)
{
rear = temp;
front =temp;
}
else
{
rear->link=temp;
rear=temp;
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-445-320.jpg)
![433
Part III: Chapter 2
Linked Lists
struct node
{
char Emp_name[20];
int Emp_code;
int Emp_basic;
node *link;
};
Solution of 2.5.4
# include<iostream.h>
# include<conio.h>
# include<stdio.h>
struct node
{
char Emp_name[20];
int Emp_code;
int Emp_basic;
node *link;
};
class queue
{
node *rear,*front;
public:
queue();
void insert();
void delete();
};
void queue::insert()
{
node *temp;
temp = new node;
cout<<" Enter Employee Name";
gets(temp->Emp_name);
cout<<endl<<" Enter Employee Code";
cin>>temp->Emp_code;
cout<<endl<<" Enter Basic Salary";
cin>>temp->Emp_basic;
temp->link=NULL;
if (front==NULL)](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-448-320.jpg)
![436
Q14) Create a menu driven program to ‘create’ a link list using the structure given below.
The program should also ‘display’ and ‘sort’ the link list in ascending order of the
structure member ‘Emp_Basic’.
struct node
{
char Emp_Name[20];
int Emp_No;
int Emp_Basic;
node *ptr;
};
Q15) Create two link lists storing integers say ‘A’ and B’, once this is done write a
function merge to which these link lists ‘A’ and ‘B’ are passed to it as parameters.
This function should merge these two link lists ‘A’ and ‘B’ and produce a third
linked ‘C’ sorted in ascending order of integers.
Q16) Write a function in C++ to perform insert and delete operation on a dynamically
allocated queue using the following structure.
struct node
{
char Book_Name[20];
char Book_Author[20];
int Book_Price;
node Book_ptr;
};
Q17) Given the following structure of a node that represents a clinic in a dynamic stack.
struct clinic
{
char bednumber[10];
char patientname[20];
char disease[30];
char doc_attending[20];
int patientno;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-451-320.jpg)
![444
created/opened using the following statement. What these statements do is create objects
of the ‘ostream’ or the ‘fstream’ class depending upon which set of statements are used
and then the object is used to create or open an existing text file. For example,
ofstream fil(“ABC.txt”);
ofstream filout (“Notes.txt”);
These statements create two objects of class ofstream ‘fil’ and ‘filout’. These objects are
then used to create the ‘ABC.txt’ and ‘Notes.txt’ text files. We can reach the same outcome
using the ‘fstream’ class as shown below. Where the statement
‘fil.open(“ABC.txt”,ios::out);’ signifies that we are creating/opening a text file ‘ABC.txt’
in the ‘out’ mode which allows us to write on the file.
fstream fil;
fil.open(“ABC.txt”,ios::out);
Now that we have learnt how to create and open existing text files let us now learn how to
write on a file. Program 3.1 given below creates a text file “ABC.txt” and allows the user to
write or insert data into this file.
Program 3.1: Writing on a text file
/* Header Files */
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
void main()
{
clrscr();
ofstream fil;
fil.open("ABC.txt");
char lin[80];
char ch;
do
{
cout<<"n Enter lines of text in the text file";
gets(lin);
fil<<lin<<endl;
cout<<" Do you want to enter more lines of text (y/n)";](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-459-320.jpg)
![446
When these statements are executed the file ‘ABC.txt’ is opened for reading purposes.
Now that we know how to open text files for reading purposes let us learn how to display
the contents stored in a text file. Program 3.2 given below opens the text file ‘ABC.txt’ for
reading purposes and displays its contents on the screen.
Program 3.2: Reading a text file
/* Header Files */
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
// Main Function
void main()
{
clrscr();
fstream fil; // Creates an object of class fstream
fil.open("ABC.txt",ios::in); // Opens the file "ABC.txt" in Read Mode
char lin[80]; // String buffer
while (fil.getline(lin,80))
{
puts(lin);
}
getch();
}
In this program we have used the ‘fstream’ class and read the file from it one line at a time
using the getline() function which is a member function of ‘fstream’ or the ‘ifstream’ class.
This function reads characters, including blank spaces until it encounters the ‘n’ operator
and places the resulting string in the buffer supplied to it as an argument. The maximum
size of the buffer is passed as the second argument. This process is repeated until we reach
the end of the file displaying the contents stored in the buffer each time the condition is
checked for. In our program this process is carried out by the following statements.
while (fil.getline(lin,80))
{
puts(lin);
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-461-320.jpg)
![447Part III: Chapter 3
Streams & Files
Where the character array ‘lin’ acts as the buffer string having a maximum size of 80
characters. This is only one of the ways in which the contents of a file can be read and
displayed program 3.4 illustrates another method of performing this function.
Program 3.3: The EOF function
/* Header Files */
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
// Main Function
void main()
{
clrscr();
fstream fil; // Creates an object of class fstream
fil.open("ABC.txt",ios::in); // Opens the file "ABC.txt" in Read Mode
char lin[80]; // String buffer
while (!fil.eof())
{
fil.getline(lin,80);
puts(lin);
}
getch();
}
Always try to use the ‘fstream’ class to declare stream objects
as one can then use the same class for opening the file in
input/output modes and the syntax happens to be very similar
to what we will use for binary files as well
Whenever a file is opened for writing command using the
‘ostream’ class or the ‘fstream’ class using the ‘out’ specifier
all the previous content of the file gets deleted. In order to
keep our previous data intact we should use the ‘app’
specifier.
Programming Tips](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-462-320.jpg)
![449Part III: Chapter 3
Streams & Files
3.3.4 Solved Examples
Example 3.3.4.1
Write a function to transfer the content of a text file to another text file
Solution of 3.3.4.1
void transfer()
{
char fn1[20],fn2[20];
fstream fil1,fil2;
fil1.open(fn1,ios::in);
fil2.open(fn2,ios::out);
char lin[80];
while(fil1.getline(lin,80))
{
fil2<<lin<<endl;
}
fil1.close();
fil2.close();
}
Example 3.3.4.2
Write a function to transfer the content of a text file to another text file after changing the
cases to uppercase with each line reversed.
Solution of 3.3.4.2
void newchange()
{
char fn1[20],fn2[20];
fstream fil1,fil2;
fil1.open(fn1,ios::in);
fil2.open(fn2,ios::out);
char lin[80];
while (fil1.getline(lin,80))
{
for (int i=strlen(lin)-1;i>=0;i--)
{
lin[i]=toupper(lin[i])
}
fil2<<lin<<endl;
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-464-320.jpg)
![450
fil1.close();
fil2.close();
}
Example 3.3.4.3
Write a menu driven program to create a text file ‘Experiment.txt’. The program should
also have an option to display the contents of the text file and also count the number of
words starting from the letter‘s’.
Solution of 3.3.4.3
/* Header Files */
# include<fstream.h>
# include<conio.h>
# include<stdio.h>
# include<string.h>
int counter=0;
/*************
Function Name: create
Purpose: creates a text file 'Experiment.txt'
**********************************************/
void create()
{
fstream fil;
fil.open("Experiment.txt",ios::out);
char choice;
char lin[80];
do
{
cout<<"nn Enter the lines of text";
gets(lin);
fil<<lin<<endl;
cout<<"n Do you want to enter more lines of text (y/n)";
cin>>choice;
}
while (choice!='n');
fil.close();
}
/*************
Function Name: display
Purpose: displays the text file 'Experiment.txt'](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-465-320.jpg)
![451Part III: Chapter 3
Streams & Files
**********************************************/
void display()
{
fstream fil;
fil.open("Experiment.txt",ios::in);
char lin[80];
while (fil.getline(lin,80))
{
puts(lin);
}
fil.close();
}
/*************
Function Name: count
Purpose: counts the number of words beginning with the letters 's'
****************************************************************
***/
void count()
{
counter=0;
fstream fil;
fil.open("Experiment.txt",ios::in);
char lin[80];
int i=0;
while (fil.getline(lin,80))
{
if (lin[0]=='s')
{
counter++;
}
for (i=0;i<strlen(lin);i++)
{
if (lin[i]==' ' & lin [i+1] =='s')
{
counter++;
}
}
}
cout<<"nn The number of words starting from letter 's' are "<<counter;
fil.close();
}
// Main function](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-466-320.jpg)
![453Part III: Chapter 3
Streams & Files
int counter=0;
/*************
Function Name: create
Purpose: creates a text file with a user defined name
********************************************************/
void create(char filename[])
{
fstream fil;
fil.open(filename,ios::out);
char choice;
char lin[80];
do
{
cout<<"nn Enter the lines of text";
gets(lin);
fil<<lin<<endl;
cout<<"n Do you want to enter more lines of text (y/n)";
cin>>choice;
}
while (choice!='n');
fil.close();
}
/*************
Function Name: display
Purpose: displays the text file created
**********************************************/
void display(char filename[])
{
fstream fil;
fil.open(filename,ios::in);
char lin[80];
while (fil.getline(lin,80))
{
puts(lin);
}
fil.close();
}
/*************
Function Name: count_vowels
Purpose: counts the number of vowels in the text file
********************************************************/](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-468-320.jpg)
![454
void count_vowels(char filename[])
{
int vowels=0;
fstream fil;
fil.open(filename,ios::in);
char lin[80];
int i=0;
while (fil.getline(lin,80))
{
for (i=0;i<strlen(lin);i++)
{
if (lin[i]=='a' || lin [i] =='e'|| lin [i] =='i'|| lin [i] =='o'|| lin [i] =='u')
{
vowels++;
}
}
}
cout<<"nn The number of vowels are "<<vowels;
fil.close();
}
// Main function
void main()
{
clrscr();
int choice;
char filename[15];
cout<<"Enter the file name";
gets(filename);
do
{
clrscr();
cout<<"nn------------- MENU -----------n";
cout<<"1. Create the text filen";
cout<<"2. Count the number of vowels n";
cout<<"3. Display the text filen";
cout<<"4. Quitn";
cout<<"! Enter Choicen";
cin>>choice;
switch(choice)
{
case 1:
create(filename);
break;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-469-320.jpg)
![455Part III: Chapter 3
Streams & Files
case 2:
count_vowels(filename);
getch();
break;
case 3:
display(filename);
getch();
break;
}
}
while (choice !=4);
getch();
}
Example 3.3.4.5
Write a menu driven program using separate function to perform the following operations:
Create a text file with a user defined name.
Copy the contents of this text file into another text file ‘Copy.txt’ with all the blank
spaces in the text file converted to ‘$’.
Display both the text files.
Solution of 3.3.4.5
/***************
Header Files
***************/
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
# include <string.h>
/*****************
Function Definitions
********************/
/*****************
Function: create
Parameters: filename[]
Return Type: void
Purpose: Creates a text file with a user defined name
*******************************************************/
void create(char filename[])
{
fstream fil;
fil.open(filename,ios::out);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-470-320.jpg)
![456
char lin[80];
char ch;
do
{
cout<<"nEnter lines of text";
gets(lin);
fil<<lin<<endl;
cout<<"nDo you want to enter more lines of text (y/n)n";
cin>>ch;
}
while (ch!='n');
fil.close();
getch();
}
/*****************
Function: display
Parameters: filename[]
Return Type: void
Purpose: Displays the text file entered by the user
*******************************************************/
void display(char filename[])
{
fstream fil;
fil.open(filename, ios::in);
char lin[80];
while (fil.getline(lin,80))
{
puts(lin);
}
fil.close();
getch();
}
/*****************
Function: copy
Parameters: filename[]
Return Type: void
Purpose: Creates the text file 'Copy.txt' with all the
blank spaces converted to '$' sign.
******************************************/
void copy(char filename[])
{
fstream fil1,fil2;
fil1.open(filename,ios::in);
fil2.open("Copy.txt",ios::out);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-471-320.jpg)
![457Part III: Chapter 3
Streams & Files
char lin[80];
while(fil1.getline(lin,80))
{
for(int i=0;i<strlen(lin);i++)
{
if (lin[i]==' ')
{
lin[i]='$';
}
}
fil2<<lin<<endl;
}
cout<<"n The text file 'Copy.txt' has been created";
fil1.close();
fil2.close();
}
// Main Function
void main()
{
clrscr();
char filename[15];
char file[15];
int choice;
do
{
cout<<"nn ----------- MENU ----------- n";
cout<<" 1. Creating the text file n";
cout<<" 2. Displaying the text filen";
cout<<" 3. Create the text file 'Copy.txt' n";
cout<<" 4. Exitn";
cin>>choice;
switch(choice)
{
case 1:
cout<<"nn Enter Filename:";
gets(filename);
create(filename);
break;
case 2:
cout<<"Enter the name of the file to be displayednn";
gets(file);
display(file);
break;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-472-320.jpg)
![459Part III: Chapter 3
Streams & Files
For example,
Example:
fstream fil;
fil.open(“Binary_File.DAT”,ios::binary|ios::out);
OR
char fname[20];
cout<<” Enter Filename:”;
gets(fname);
fstream fil;
fil.open(fname,ios::binary|ios::out);
The set of statements given above would create a binary file ‘Binary_File’. The
‘ios::binary’ statement in the piece of code given above signifies that we are opening a
binary file where as the proclamation ‘ios::out’ tells the compiler that we are opening the
binary file for writing purposes. This syntax should seem similar to the one we learnt for
opening text files as binary files are nothing but a more general form of text files.
3.4.2 Opening binary files in input mode (For reading from a file)
In the previous section we learnt about how to open binary files for writing purposes. Let
us now learn how to open binary files for reading purposes. For opening binary files in
input mode we again use the ‘fstream’class using the following syntax.
Syntax:
fstream<FileObject>;
<FileObject>.open (<Filename>,ios::binary |ios::in);
The example given below opens the binary file ‘Binary_File.DAT’ in input mode.
Example:
fstream fil;
fil.open(“Binary_File.DAT”,ios::binary|ios::in);
OR
char fname[20];
cout<<” Enter Filename:”;
gets(fname);
fstream fil;
fil.open(fname,ios::binary|ios::in);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-474-320.jpg)
![460
The only change we observe from the previous section is in the statement
‘fil.open(“Binary_File.DAT”,ios::binary|ios::in);’. Where we have changed the
proclamation from ‘ios::out’ to ‘ios::in’ which signifies we are opening the binary file for
reading purposes.
3.4.3 Opening binary files in APPEND mode (For writing on a file)
When we are opening the binary file in output mode what was happening was that all the
previous data stored in the file was being deleted. In simple terms we were left with a black
file. The advantage of opening a file in append mode is that the earlier content if existing,
is not over writen and the file write pointer moves to the end of the file for adding new
lines at the bottom of the file. This can be done using the following syntax.
Syntax:
fstream<Fileobject>;
<Fileobject>.open(<Filename>,ios::binary|ios::app);
For example for opening the text file ‘ABC.txt’ in append mode we would use the
following statements.
Example:
fstream fil;
fil.open(“Binary_File.DAT”,ios::binary|ios::app);
OR
char fname[20];
cout<<” Enter Filename:”;
gets(fname);
fstream fil;
fil.open(fname,ios::binary|ios::in);
3.4.4 Creation of binary files
When creating binary files inputting and outputting data using operators like << and >>
and functions like ‘getline’ does not make too much sense although it is possible to use
them. The file streams include two member functions which are specially designed for
input and output of data sequentially. They are the ‘write’ and the ‘read’ functions. The
‘write()’
function is a member function of the ‘ostream’ class and is inherited by ‘ofstream’ where
as the function ‘read()’ is a member function of the ‘istream’ class and is inherited by
ifstream. Both these functions can be accessed by creating objects of ‘fstream’ class. These
functions have the following prototypes.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-475-320.jpg)
![461Part III: Chapter 3
Streams & Files
write ( char * buffer, streamsize size );
read ( char * buffer, streamsize size );
The function prototype takes two arguments. The first one being ‘buffer’ which is the
address of the memory block where the data is inserted by the user or the data to be written
on the binary file is stored temporarily where as the second argument defines the maximum
allowed size that can be fed into the buffer. Binary files can be created in C++ with the
help of structure variables or objects of a class. Program 3.4 given below declares a
function which creates a user defined binary file with a single record.
Program 3.4: Insertion of a record in a binary file using structures and classes
Using Structures Using Classes
struct stock
{
int Ino;
char Item[20];
float qty;
};
void create(char fname[])
{
fstream fil;
stock s;
fil.open(fname,ios::binary|ios::out);
cout<<”Item No. ”;
cin>>s.Ino;
cout<<”Item Name”;
gets(s.Item);
cout<<”Quantity”;
cin>>s.qty;
fil.write((char *)&s,sizeof(s));
fil.close();
}
class stock
{
int Ino;
char Item[20];
float qty;
public:
void input()
{
cout<<”Item No.”;
cin>>Ino;
cout<<”Item Name”;
gets(Item);
cout<<”Quantity”;
cin>>qty;
}
void output()
{
cout<<Ino<<”:”<<Item<<”:”<<qty;
}
int Rino()
{
return Ino;
}
};
/***********
Function: create
Parameters: fname](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-476-320.jpg)
![462
****************/
void create( char fname[])
{
fstream fil;
stock s;
fil.open(fname,ios::binary|ios::out);
s.input();
fil.write((char*)&s,sizeof(s));
fil.close();
}
The entire structure of this program might seem very seimilar to the one we have already
dealt in text files. The only difference one will observe is in the writing and the reading of
binary files in this program we have used the following piece of code to write on the binary
file.
Let us see how this statement works. For this purpose let us look at it its general form.
stream_object.write((char*)&(structure_variable/class_object),sizeof(structure
_variable/class_object));
The first part of the statement ‘(char*)&(structure_variable/class_object)’ typcasts the
structure or the class to character type. Where as the second part of the statement
‘sizeof(structure_variable/class_object)’ allocates the number of bytes to be fed into the
binary file. For example, in our statement given above. The first part of the statement
‘(char*)&s’ typcasts the class ‘s’ so that data from it can be inserted into the binary file and
the second part of the statement ‘sizeof(s)’ allocates the amount of data interms of bytes to
be fed into the file at one time. This is the procedure one has to follow in order to insert a
single record into a binary file but what happens if we want to insert multiple records and
not just a single record ?
Program 3.5 declares a function ‘create’ which proposes a solution to the given problem.
Program 3.5: Creation of a binary file with multiple records
struct stock
{
int Ino;
char Item[20];
float qty;
};
class stock
{
int Ino;
char Item[20];
float qty;
public:
fil.write((char*)&sizeof(s));
STREAMS&FILES](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-477-320.jpg)
![463Part III: Chapter 3
Streams & Files
3.4.5 Reading a binary file
In this section of the chapter we will learn how to dislay already created binary files. We
have already learnt in the previous section that binary files can be displayed using the
function ‘read()’ which is a member function of the ‘istream’ class’ inherited by ifstream
and the fstream classes. The process of reading data from binary files is very similar to the
process of writing on them. Program 3.6 given below displays the piece of code required to
display a single record.
void create (char Fname[])
{
fstream fil;
stock S;
fil.open
(Fname,ios::binary|ios::out);
char choice;
do
{
cout<<” Item No?”;
cin>>S.Ino;
cout<<”Item Name”;
gets(S.Item);
cout<<”Quantity”;
cin>>S.qty;
fil.write((char
*)&S,sizeof(S));
cout<<”More(Y/N) ?”;
cin>>choice;
}
while (choice==’y’);
fil.close();
}
void input()
{
cout<<”Item No.”;
cin>>Ino;
cout<<”Item Name”;
gets(Item);
cout<<”Quantity”;
cin>>qty;
}
void output()
{
cout<<Ino<<”:”<<Item<<”:”<<qty;
}
int Rino()
{
return Ino;
}
};
void create(char Fname[])
{
fstream fil;
stock s;
fil.open(Fname,ios::binary|ios::out);
char choice;
do
{
s.input();
fil.write((char*)&s,sizeof(s));
cout<<” More(Y/N) ?”;
cin>>choice;
}
while (choice == ‘y’);
fil.close();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-478-320.jpg)
![464
Program 3.6:
Using Structures Using Classes
struct stock
{
int Ino;
char Item[20];
float qty;
};
void display (char Fname[])
{
fstream fil;
stock S;
fil.open(Fname,ios::binary|ios::in);
fil.read((char*)&S,sizeof(S));
cout<<”ItemNo:”<<S.Ino;
cout<<”Item_Name:”;
gets(S.Item);
cout<<”Item quantity”;
cin>>S.qty;
fil.close();
}
class stock
{
int Ino;
char Item[20];
float qty;
public:
void input()
{
cout<<”Item No.”;
cin>>Ino;
cout<<”Item Name”;
gets(Item);
cout<<”Quantity”;
cin>>qty;
}
void output()
{
cout<<Ino<<”:”<<Item<<”:”<<qty;
}
int Rino()
{
return Ino;
}
};
void display(char Fname[])
{
fstream fil;
stock S;
fil.open(Fname,ios::binary|ios::in);
fil.read((char*)&S,sizeof(S));
S.output;
fil.close();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-479-320.jpg)
![465Part III: Chapter 3
Streams & Files
As it is clear we have used the 'read’ function to display records from the file. The function
‘read’ works in the same way as the function ‘write()’ we learnt in the last section the only
difference being that this time data is being fed out of the file. Rest of the program should
be easily understood as it follows the same logic as present in program 3.5. It is also
possible to read various records from a binary file. Program 3.8 given below shows how
this is done.
Program 3.7: Reading Records from a binary File
struct stock
{
int Ino;
char Item[20];
float qty;
};
void display(char Fname[])
{
fstream fil;
stock S;
fil.open(Fname,ios::binary|ios::in);
while(fil.read((char*)&S,sizeof(S)))
{
cout<<”Item No:”<<S.Ino<<endl;
cout<<”ItemName:”<<S.Item<<”Qty:”<<S.qty<<en
dl;
}
fil.close();
}
class stock
{
int Ino;
char Item[20];
float qty;
public:
void input()
{
cout<<”Item No.”;
cin>>Ino;
cout<<”Item Name”;
gets(Item);
cout<<”Quantity”;
cin>>qty;
}
void output()
{
cout<<Ino<<”:”<<Item<<”:”<<qty;
}
int Rino()
{
return Ino;
}
};
void display(char Fname[])](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-480-320.jpg)
![467Part III: Chapter 3
Streams & Files
(iii) Accept a disease name from the user and display the details of all such
patients who are suffering of that particular disease
// class definition of class 'clinic'
class clinic
{
char bednumber[10];
char patientname[20];
char disease[30];
char doc_attending[20];
public:
void getdetails()
{
gets(bednumber);
gets(patientname);
gets(disease);
gets(doc_attending);
}
void showdetails()
{
cout<<"Bed Number"<<bednumber;
cout<<"Patient Name"<<patientname;
cout<<"Disease"<<disease;
cout<<"Doctor Attending"<<doc_attending;
}
char*retdisease()
{
return disease;
}
};
Solution of 3.4.6.1
(i)
void store()
{
fstream fout;
fout.open("clinic.dat",ios::out|ios::binary);
clinic ob;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-482-320.jpg)
![468
char ch;
do
{
ob.getdetails();
fout.write((char*)&ob,sizeof(ob));
cout<<"More Records...(y/n)";
cin>>ch;
}
while (toupper(ch)=='y');
fout.close();
}
(ii)
void readrecords()
{
fstream fin;
fin.open("clinic.dat",ios::in|ios::binary);
clinic ob;
while(fin.read((char*)&ob,sizeof(ob)))
{
ob.showdetails();
}
fin.close();
}
(iii)
void diseasedetails()
{
char dis[30];
cout<<"Enter Disease name";
gets(dis);
fstream fin;
fin.open("clinic.dat",ios::in|ios::binary);
clinic ob;
while (fin.read((char*)&ob,sizeof(ob)))
{
if (!strcmp(ob.retdisease(),dis))
{
ob.showdetails();
}
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-483-320.jpg)
![469Part III: Chapter 3
Streams & Files
fin.close();
}
Example 3.4.6.2
Write a function in C++ to add new objects at the bottom of a binary file "student.dat"
assuming the binary file is containing the objects of the following class:
class stud
{
int rno;
char name[20];
public:
void enter()
{
cin>>rno;
gets(name);
}
void display()
{
cout<<rno<<name<<endl;
}
};
Solution of 3.4.6.2
void addnew()
{
fstream fil;
fil.open("student.dat",ios::binary|ios:
:app);
stud s;
char ch;
do](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-484-320.jpg)
![470
{
s.enter();
fil.write((char*)&s,sizeof(s));
cout<<"More(Y/N)?";
cin>>ch;
}
while(ch!='y');
fil.close();
}
3.5 Solved Examples
Example 3.5.1
Write a program in C++ to create a text file with a user defined name. The program should
also display the created text file and count the number of times the letters ‘‘g’‘h’‘o’‘s’‘t’’
appear together ?
Solution of 3.5.1
/**********
Header Files
**********/
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
# include <string.h>
/********
Function Definitions
***********/
/*****************
Function: create
Parameters: filename[]
Return Type: void
fstream fil;
fil.open(filename,ios::out);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-485-320.jpg)
![471Part III: Chapter 3
Streams & Files
char lin[80];
char ch;
do
{
cout<<"nEnter lines of text";
gets(lin);
fil<<lin<<endl;
cout<<"nDo you want to enter more lines of text (y/n)n";
cin>>ch;
}
while (ch!='n');
fil.close();
getch();
}
/*****************
Function: display
Parameters: filename[]
Return Type: void
Purpose: Displays the text file
******************************/
void display(char filename[])
{
fstream fil;
fil.open(filename, ios::in);
char lin[80];
while (fil.getline(lin,80))
{
puts(lin);
}
fil.close();
getch();
}
/*****************
Function: count
Parameters: filename[]
Return Type: void
Purpose: counts the number of times the letters 'g''h''o''s''t' encountered](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-486-320.jpg)
![472
void count(char filename[])
{
fstream fil;
fil.open(filename, ios::in);
char lin[80];
int count=0;
while (fil.getline(lin,80))
{
for (int i=0;i<strlen(lin);i++)
{
if (lin[i]=='g'&lin[i+1]=='h'&lin[i+2]=='o'&lin[i+3]=='s'&lin[i+4]=='t')
count++;
}
}
cout<<" No. of times "<<count;
fil.close();
getch();
}
// Main Function
void main()
{
clrscr();
char filename[15];
int choice;
do
{
cout<<"nn ---------------------- MENU ----------------------- n";
cout<<" 1. Creating the text file n";
cout<<" 2. Displaying the text filen";
cout<<" 3. Number of times the letters 'g''h''o''s''t' are encountered together";
cout<<"n 4. Exitn";
cin>>choice;
switch(choice)
{
case 1:
cout<<"nn Enter Filename:";
gets(filename);
create(filename);
break;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-487-320.jpg)
![473Part III: Chapter 3
Streams & Files
Example 3.5.2
Write a program to create a user defined text file and display alternate lines of the file?
Solution of 3.5.2
case 2:
display(filename);
break;
case 3:
count(filename);
break;
}
}
while (choice!=4);
getch();
}
/**********
Header Files
**********/
# include <fstream.h>
# include <conio.h>
# include <stdio.h>
# include <string.h>
/********
Function Definitions
***********/
/*****************
Function: create
Parameters: filename[]
Return Type: void
Purpose: Creates a text file with a user defined name
*******************************************************/
void create(char filename[])
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-488-320.jpg)
![474
fstream fil;
fil.open(filename,ios::out);
char lin[80];
char ch;
do
{
cout<<"nEnter lines of text";
gets(lin);
fil<<lin<<endl;
cout<<"nDo you want to enter more lines of text (y/n)n";
cin>>ch;
}
while (ch!='n');
fil.close();
getch();
}
/*****************
Function: display
Parameters: filename[]
Return Type: void
Purpose: Displays alternate lines of the text file entered by the user
**********************************************************************/
void display(char filename[])
{
fstream fil;
fil.open(filename, ios::in);
char lin[80];
int even=0;
while (fil.getline(lin,80))
{
if (even%2==0)
{
puts(lin);
}
even++;
}
fil.close();
getch();
}
// Main Function
void main()](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-489-320.jpg)
![520
2.7 Let us revise!
Here are the general forms of the statements discussed in this chapter, plus some extra
important ones (explanations given). REMEMBER that all of these statements may or
may not be available on your system, so check documentation regarding availability:
ALTER TABLE <TABLE NAME> ADD|DROP|MODIFY (COLUMN
SPECIFICATION[S]...see Create Table); --allows you to add or delete a column or
columns from a table, or change the specification (data type, etc.) on an existing column;
this statement is also used to change the physical specifications of a table (how a table is
stored, etc.), but these definitions are DBMS-specific, so read the documentation. Also,
these physical specifications are used with the Create Table statement, when a table is
first created. In addition, only one option can be performed per Alter Table statement--
either add, drop, OR modify in a single statement.
CREATE [UNIQUE] INDEX <INDEX NAME>
ON <TABLE NAME> (<COLUMN LIST>); --UNIQUE is optional; within brackets.
CREATE TABLE <TABLE NAME>
(<COLUMN NAME> <DATA TYPE> [(<SIZE>)] <COLUMN CONSTRAINT>,
...other columns); (also valid with ALTER TABLE)
--where SIZE is only used on certain data types (see above), and constraints include the
following possibilities (automatically enforced by the DBMS; failure causes an error to be
generated):
1. NULL or NOT NULL (see above)
2. UNIQUE enforces that no two rows will have the same value for this column
3. PRIMARY KEY tells the database that this column is the primary key column
(only used if the key is a one column key, otherwise a PRIMARY KEY (column,
column, ...) statement appears after the last column definition.
4. CHECK allows a condition to be checked for when data in that column is
updated or inserted; for example, CHECK (PRICE > 0) causes the system to
check that the Price column is greater than zero before accepting the
value...sometimes implemented as the CONSTRAINT statement.
5. DEFAULT inserts the default value into the database if a row is inserted without
that column's data being inserted; for example, BENEFITS INTEGER
DEFAULT = 10000
6. FOREIGN KEY works the same as Primary Key, but is followed by:
REFERENCES <TABLE NAME> (<COLUMN NAME>), which refers to the
referential primary key.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-535-320.jpg)
![521Part IV: Chapter 2
Structured Query Language
CREATE VIEW <TABLE NAME> AS <QUERY>;
DELETE FROM <TABLE NAME> WHERE <CONDITION>;
INSERT INTO <TABLE NAME> [(<COLUMN LIST>)]
VALUES (<VALUE LIST>);
SELECT [DISTINCT|ALL] <LIST OF COLUMNS, FUNCTIONS, CONSTANTS,
ETC.>
FROM <LIST OF TABLES OR VIEWS>
[WHERE <CONDITION(S)>]
[GROUP BY <GROUPING COLUMN(S)>]
[HAVING <CONDITION>]
[ORDER BY <ORDERING COLUMN(S)> [ASC|DESC]]; --where ASC|DESC
allows the ordering to be done in ASCending or DESCending order
UPDATE <TABLE NAME>
SET <COLUMN NAME> = <VALUE>
[WHERE <CONDITION>]; --if the Where clause is left out, all rows will be updated
according to the Set statement](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-536-320.jpg)
![527Part IV: Chapter 3
Connecting Databases with C++
#include <windows.h>
#include <sqlext.h>
#include <stdio.h>
#include <iostream.h>
void main()
{
HENV hEnv = NULL; // Handle from SQLAllocEnv()
HDBC hDBC = NULL; // Handle for Connection
HSTMT hStmt = NULL; // Statement handle
UCHAR szDSN[SQL_MAX_DSN_LENGTH] = "ABC"; // Data Source Name
UCHAR* szUID = NULL; // User ID buffer
UCHAR* szPasswd = NULL; // Password buffer
UCHAR szModel[128]; // Model buffer
SDWORD cbModel; // Model buffer bytes received
UCHAR szSqlStr[] = "Select Model From Makes Where Make=''Suzuki";
// SQL string
RETCODE retcode; // Return code
// allocate memory for ODBC Environment handle
SQLAllocEnv (&hEnv); // Allocate memory for the connection handle
SQLAllocConnect (hEnv, &hDBC);
// connect to the data source "ABC" using userid and password.
retcode = SQLConnect (hDBC, szDSN, SQL_NTS, szUID, SQL_NTS, szPasswd,
SQL_NTS);
if (retcode == SQL_SUCCESS || retcode == SQL_SUCCESS_WITH_INFO)
{
// allocate memory for the statement handle
retcode = SQLAllocStmt (hDBC, &hStmt);
// prepare the SQL statement by assigning it to the statement handle
retcode = SQLPrepare (hStmt, szSqlStr, sizeof (szSqlStr));
// execute SQL statement handle
retcode = SQLExecute (hStmt);
// Project only column 1 which is the models
SQLBindCol (hStmt, 1, SQL_C_CHAR, szModel, sizeof(szModel), &cbModel);
// Get row of data from the result set defined above in the statement
retcode = SQLFetch (hStmt);
while (retcode == SQL_SUCCESS || retcode == SQL_SUCCESS_WITH_INFO)
{
cout ("t%sn", szModel); // Print row (model)
retcode = SQLFetch (hStmt); // Fetch next row from result set
}
// Free the allocated statement handle
SQLFreeStmt (hStmt, SQL_DROP);
// Disconnect from datasource
SQLDisconnect (hDBC);](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-542-320.jpg)
![532
shift in ctrl-O 15 0F SI
data line escape ctrl-P 16 10 DLE
device control 1 ctrl-Q 17 11 DC1
device control 2 ctrl-R 18 12 DC2
device control 3 ctrl-S 19 13 DC3
device control 4 ctrl-T 20 14 DC4
Neg acknowledge ctrl-U 21 15 NAK
synchronous idel ctrl-V 22 16 SYN
End of xmit block ctrl-W 23 17 ETB
Cancel ctrl-X 24 18 CAN
end of medium ctrl-Y 25 19 EM
Substitute ctrl-Z 26 1A SUB
Escape ctrl-[ 27 1B ESC
file separator ctrl- 28 1C FS
group separator ctrl-] 29 1D GS
record separator ctrl-^ 30 1E RS
unit separator ctrl-_ 31 1F US
A.2 PRINTABLE CHARACTERS
The ASCII table also includes of 96 printable characters which are given in table A.2. For
example, the capital letter ‘A’ is always represented by the order number ‘65’,
Table A.2: ASCII TABLE
Dec Hex Char Dec Hex Char Dec Hex Char
32 20 Space 64 40 @ 96 60 `
33 21 ! 65 41 A 97 61 a
34 22 " 66 42 B 98 62 b
35 23 # 67 43 C 99 63 c
36 24 $ 68 44 D 100 64 d
37 25 % 69 45 E 101 65 e
38 26 & 70 46 F 102 66 f
39 27 ' 71 47 G 103 67 g
40 28 ( 72 48 H 104 68 h
APPENDIXA](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-547-320.jpg)
![533Appendix A
APPENDIXA
41 29 ) 73 49 I 105 69 i
42 2A * 74 4A J 106 6A j
43 2B + 75 4B K 107 6B k
44 2C , 76 4C L 108 6C l
45 2D - 77 4D M 109 6D m
46 2E . 78 4E N 110 6E n
47 2F / 79 4F O 111 6F o
48 30 0 80 50 P 112 70 p
49 31 1 81 51 Q 113 71 q
50 32 2 82 52 R 114 72 r
51 33 3 83 53 S 115 73 s
52 34 4 84 54 T 116 74 t
53 35 5 85 55 U 117 75 u
54 36 6 86 56 V 118 76 v
55 37 7 87 57 W 119 77 w
56 38 8 88 58 X 120 78 x
57 39 9 89 59 Y 121 79 y
58 3A : 90 5A Z 122 7A z
59 3B ; 91 5B [ 123 7B {
60 3C < 92 5C 124 7C |
61 3D = 93 5D ] 125 7D }
62 3E > 94 5E ^ 126 7E ~
63 3F ? 95 5F _ 127 7F DEL
A.3 EXTENDED CHARACTER SET
The Extended ASCII Character Set (table A.3) also consists of 128 decimal numbers and
ranges from 128 through 255 (using the full 8-bits of the byte) representing additional
special, mathematical, graphic, and foreign characters.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-548-320.jpg)
![536
536
hash_multiset, hash map, and hash_multimap. Each class in STL is a template, and can
be instantiated to contain any type of object. For example, you can use a vector<int> in
the same way as you would use an ordinary C array, except that if you use vector you
don’t need to manage dynamic memory allocation by hand.
vector <int> v(3); // Declare a vector of 3 elements.
v[0] = 1;
v[1] = v[0] + 2;
v[2] = v[0] + v[1]; // v[0] == 1, v[1] == 3, v[2] == 4
The data stored in containers is manipulated through a large collection of algorithms
included in STL, e.g. to reverse the order of elements in a vector, reverse algorithm
described below can be used:
reverse(v.begin(), v.end()); //result will now be that v[0] == 4, v[1] == 3, v[2] == 1
You need to observe two important points in reverse algorithm: 1) it is a global function,
not a member function and 2) it takes two arguments which define a range of elements.
Range can be a few selected elements or the entire container v.
STL algorithms are decoupled from the STL container classes. This means that reverse
can be used not only to reverse elements in vectors, but also to reverse elements in lists,
and even elements in arrays defined in the C++. The following program shows how
reverse algorithm works equally well on a array:
double A[4] = { 1.0, 2.4, 3.6, 1.5};
reverse(A, A + 6); //range is denoted [A, A + 6);
for (int i = 0; i < 4; ++i)
cout << "A[" << i << "] = " << A[i];
B.3 Iterators
Iterators are the mechanism that makes it possible to decouple algorithms from containers.
Algorithms are templates, and are parameterized by the type of iterator, so they are not
restricted to a single type of container. Let us revisit the reverse algorithm in section D
1.2. You may note that arguments to reverse are iterators, which are a generalization of
pointers. Pointers themselves are iterators, which is why it is possible to reverse the
elements of an array using the same reverse algorithm. In the example above, the type
returned by v.begin() and v.end() is vector<int>::iterator. There are also some iterators,
such as istream_iterator and ostream_iterator, that aren't associated with containers.
Consider, for example, how to write an algorithm that performs linear search through a
range (see Table B.1).](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-551-320.jpg)
![553Appendix D
Program D.1
// p3_1.cpp - An application creating a main window with the caption “trial”. It can
//resize, move, minmize, maximize, and close the generated window.
#include <owlapplicat.h>
// This header file has definition of Tapplication which is //derived from TModule
and includes owlmodule.h.
int OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TApplication("trial").Run();
}
It is necessary to derive a new class from TApplication to introduce more functionality in the
application. Initializing an ObjectWindows application requires the following steps:
i. Constructing the application object
Member functions are InitApplication, InitInstance, and InitMainWindow. One
can override any of these members to customize initialization of one's application.
The constructor for the TrialApp class derived from TApplication shown in the
following example takes application name as only argument. The default value is
zero indicating no name. The application name is referenced by a char far*
member of the TModule base class called Name.
Program D.2
// p3_2.cpp - Creates and shows a window entitled, “ Trial”.
#include <owlapplicat.h>
class TrialApp: public Tapplication
{
public:
TrialApp(const char far* name = 0): TApplication(name) {}
};
int
OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TrialApp("trial").Run();
}](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-568-320.jpg)
![555Appendix D
int OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TrialApp("trial").Run();
/* The application name given in the TApplication constructor is used only if you do not
provide a main window. As this program has used SetMainWindow function -- the
caption bar is titled "First Main Window" and not "trial". */
}
Doc/View Classes
In Doc/View model data is contained in and accessed through a document object and
displayed and manipulated through a view object. It comprises of TDocManager,
TDocument and TView classes.
Miscellaneous Classes
ObjectWindows miscellaneous classes include menus and clipboard classes.
Once an application is initialized, the application object’s MessageLoop starts running. All
incoming messages from Windows are processed by the MessageLoop. One can override the
default message handling functions and introduce additional message processing capability in the
application. TAppliaction includes functions, such as MessageLoop, IdleAction, PreProcessMenu,
and ProcessAppMsg which provide message handling functionality for any ObjectWindows
application. TAppliaction also includes functions that enable one to load the Borland Custom
Control Library (BWCC.DLL for 16- and 32- bit applications) and Microsoft 3-D Control Library
(CTL3DV2.DLL and CTL3D32.DLL for16- and 32- bit applications, respectively. Function
TApplication::EnableBWCC is used to open and close the Borland Custom Control Library.
When one passes True to EnableBWCC, the function loads the DLL if it is not already loaded.
Similarly, if one passes False to EnableBWCC it unloads the DLL. EnableCTL3D function works
in the same way as EnableBWCC for loading and unloading Microsoft 3-D Control Library.
Response tables are used to handle all events in an ObjectWindows application. These tables are
used in the following way:
Define the response table
ObjectWindows provides DEFINE_RESPONSE_TABLEX macro to help a user to define
response tables.
DEFINE_RESPONSE_TABLE1( TrialFrame, TframeWindow)
EV_WM_LBUTTONDOWN,
EV_WM_LBUTTONUP,
EV_WM_MOUSEMOVE,
EV_WM_RBUTTONDOWN,
END_RESPONSE_TABLE;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-570-320.jpg)
![558
and message "Left mouse button pressed" appears. Attribute MB_OK will put an "OK" button.*/
MessageBox(" Left mouse button pressed", "Message Sent", MB_OK);
}
void TrialWindow::EvRButtonDown(uint, TPoint&)
{
MessageBox(" Right mouse button pressed", "Message Dispatched", MB_OK);
}
class TrialApp : public TApplication
{
public:
TrialApp() : TApplication() {}
void InitMainWindow()
{
SetMainWindow(new TFrameWindow(0, " Trial program", new TrialWindow));
}
};
int OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TrialApp().Run();
}
D.2.1 Graphics Device Interface
It refers to the graphics output function of Windows and isolates a Windows program from the
physical output device such as display or printer. In addition to producing output on the physical
devices it also supports output on two pseudo devices bitmaps and metafiles. Bitmaps refer to a
rectangular array of pixels and metafiles. Most of the PC output devices like video display
adapters, dot-matrix printers and laser printers are raster devices and represent images as a pattern
of dots, i.e., bitmaps. The important GDI functions can be grouped in following categories:
Vector drawing functions to draw graphical objects such as lines, circles and rectangles;
bitmap manipulation functions.
Text output functions
Palette management functions to display colors supported by a display adapter. Palette
management in systems with super VGA & other enhanced graphics adapters, supporting
more than 16 colors, is useful.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-573-320.jpg)
![564
int OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TDrawApp().Run();
}
D.2.6 GDI Coordinate System
The GDI supports two coordinate systems referred to as physical(or device) and logicalcoordinates.
The physical coordinate system is fixed for a device, e.g., for a window on the screen the origin
is at the upper left corner of a window's client area, the positive x-axis extends to the right, and the
positve y-axis extends downwards. In all device coordinate systems, units are in erms of pixels.
Windows supports several logical coordinates systems and maps each one of them to the physical
systems before displaying any graphics output.
Figure D.2: Mapping of logical coordinates to physical coordinate system
Windows supports eight mapping mode identifiers specifying logical units as 0.001 inch, 0.01 inch,
0.01 millimeter, or 0.1 millimeter and the positive y-axis extending downward or upwards. One](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-579-320.jpg)
![576
Pen = new TPen(Color, PenSize); //TColor::Black defined in owlcolor.h
}
void TrialWindow::SetPen(TColor& newColor, int penSize)
{
// If penSize isn't the default (0), set PenSize to the new size.
if (penSize)
PenSize = penSize;
Color = newColor;
delete Pen;
Pen = new TPen( Color, PenSize); // pickup a pen of new color
}
bool TrialWindow::CanClose() // file save feature not implemented
{
return MessageBox("Do you want to save?", "Not implemented",
MB_YESNO | MB_ICONQUESTION) == IDNO;
// insert logic for saving the file
}
void TrialWindow::EvLButtonDown(uint, TPoint& point)
{
if (!CommonDC)
{
SetCapture();
CommonDC = new TClientDC(*this); //'this' is a local variable avialable
// in the body of a nonstatic function
CommonDC->SelectObject(*Pen); //select pen in device context
CommonDC->MoveTo(point);
}
}
/* Let us learn to output text on the window. Output mouse position when right button of mouse
is pressed. */
void TrialWindow::EvRButtonDown(uint, TPoint& point)
{
char str[40];
short posx = point.x;
short posy = point.y;
TClientDC CommonDC(*this);
wsprintf(str, "Mouse Position (%d,%d) ", posx, posy);
CommonDC.SetTextColor(RGB(255,0,0)); // set text color to red
CommonDC.SetTextAlign(TA_BOTTOM | TA_LEFT); //Text Alignment](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-591-320.jpg)
![577Appendix D
CommonDC.TextOut(20, 20, str, strlen(str));
}
void TrialWindow::EvMouseMove(uint, TPoint& point)
{
if (CommonDC)
CommonDC->LineTo(point);
}
void TrialWindow::EvLButtonUp(uint, TPoint&)
{
if ( CommonDC)
{
ReleaseCapture();
delete CommonDC;
CommonDC = 0;
}
}
void TrialWindow::CmPenSize() //Processes menu command Pen Size
{
GetPenSize();
}
void TrialWindow::CmPenColor()
{
TChooseColorDialog::TData colors; //Processes menu command Pen Color
static TColor custColors[16] =
{
0x010101L, 0x101010L, 0x202020L, 0x303030L,
0x404040L, 0x505050L, 0x606060L, 0x707070L,
0x808080L, 0x909090L, 0xA0A0A0L, 0xB0B0B0L,
0xC0C0C0L, 0xD0D0D0L, 0xE0E0E0L, 0xF0F0F0L
};
colors.Flags = CC_RGBINIT;
colors.Color = TColor::Black;
colors.CustColors = custColors;
if (TChooseColorDialog(this, colors).Execute() == IDOK)
SetPen(colors.Color);
}
void TrialWindow::GetPenSize()](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-592-320.jpg)
![578
{
char inputText[6];
wsprintf(inputText, "%d", PenSize);
if ((TInputDialog(this, "Line Thickness","Input a new thickness:",
inputText,sizeof(inputText))).Execute() == IDOK)
{
int newPenSize = atoi(inputText);
if (newPenSize < 0)
newPenSize=1;
SetPenSize(newPenSize);
}
}
void TrialWindow::SetPenSize(int newSize)
{
delete Pen;
PenSize = newSize;
Pen = new TPen(Color, PenSize);
}
void TrialWindow::CmClearWindow()
{
Invalidate(); // clears the entire client area of the window
}
void TrialWindow::CmAbout()
{
TDialog(this, IDD_ABOUT).Execute();
}
void TrialWindow::CmClose() // terminate the program
{
EvClose(); // shuts down the main window and ends the application
}
class TDrawApp : public TApplication
{
public:
TDrawApp() : TApplication() {}
void InitMainWindow();
};](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-593-320.jpg)
![579Appendix D
void TDrawApp::InitMainWindow()
{
SetMainWindow(new TFrameWindow(0, "Freehand Drawing", new TrialWindow));
GetMainWindow()->AssignMenu("COMMANDS");
}
int OwlMain(int /*argc*/, char* /*argv*/ [])
{
return TDrawApp().Run();
}
Resource file for Program D.6
//----------------------------------------------------------------------------
// Resource file for program D.6
//----------------------------------------------------------------------------
#include <owl/inputdia.rc>
#include <owl/except.rc>
#define CM_ABOUT 205
#define CM_PENSIZE 206
#define CM_PENCOLOR 207
#define CM_CLEARWINDOW 208
#define IDD_ABOUT 211
#define CM_CLOSE 212
#ifdef RC_INVOKED
#ifndef WORKSHOP_INVOKED
# include <windows.h>
#endif
COMMANDS MENU
{
POPUP "&Setup"
{
MENUITEM "Pen &Size", CM_PENSIZE
MENUITEM "Pen &Color", CM_PENCOLOR
MENUITEM "&CleanUp", CM_CLEARWINDOW
}
POPUP "&Help"
{
MENUITEM "&About", CM_ABOUT
}
POPUP "&Exit"
{](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-594-320.jpg)
![594
Program F.1: Tower Of Hanoi
int towers (int n, int start, int goal, int temp)
{
if ( n==0)
return 0 ; // Base Case
towers ( n-1, start, temp, goal );
cout << "Move from "<<start << " to " << goal ;
towers ( n-1,temp , goal , start) ;
}
F.3.2 Binary Search
Recall the binary search algorithm covered in ‘Part I Chapter 5, page 155’. We
implemented the binary search algorithm using an iterative procedure. The same
algorithm can also be implemented using an recursive algorithm as shown below.
Algorithm:
o The element to be searched for is taken from the user say the ‘Key’.
o The given data array is taken. Any arbitrary lower limit or the lower bound
for the data values is taken. Similarly one defines the upper limit or the upper
bound. Then we obtain the middle value using.
Middle = (lower bound + upper bound)/2;
o The search ‘key’ is compared to the Middle value if ‘Key=Middle’ then
the search is complete. If Key<Middle, the program continues and the search
is confined to the first half only. On the other hand if Key>Middle then also
the program continues and the search is confined to the latter half. This
process is continued till the time ‘Key = Middle’.
Program F.2 implements the algorithm given above using an recursive function.
Program F.2: Binary Search
int binsrch(list x[],int srch ,int left ,int right)
{
int mid = (left + right)/2 ;](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-609-320.jpg)
![595Appendix F
if (x [ mid]== srch)
return (mid) ;
else if(x(mid)>srch && mid>left)
return binsrch (x , srch , left ,mid-1);
else if (x(mid)<srch && right>mid)
return binsrch (x , srch , mid+1 , r i g h t )
else
return UNSUCCESSFUL;
}
F.4 Recursion Vs Iteration
It is a popular computer science belief that any algorithm that can be coded with
recursion can also be coded with an iterative control structures. Both approaches achieve
the same goal, but the question arises which one is the best to use among them?
Although a recursive algorithm might be easier to understand but there are several
reasons why use of recursion in curbed. Recursive algorithms are certainly less efficient
than corresponding iterative algorithms. Each time a function is called, the system incurs
an overhead that is not necessary with a loop. Also, in some cases it is possible the
algorithm design for an iterative procedure might be simpler.
The speed and amount of memory available to modern computer today however
decreases the impact of recursion so much that its inefficiency is no longer a strong
argument against it. Today the choice for a recursive or an iterative algorithm is just a
matter of design.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-610-320.jpg)
![APPENDIX G
XML & C++
G.1 Overview of XML
Extended Markup Language (XML) is a subset of the Standard Generalized Markup Language
(SGML), a complex standard for describing document structure and content. XML is a language
for organizing - not merely presenting – data. XML is intended to be a metalanguage, i.e., a
language in which you can describe another language. XML, therefore, empowers you to design
your own markup.
Hyper Text Markup Language (HTML) most popular in making web pages has following
limitations:
• It’s only a presentation format and does not describe contents.
• HTML search engines can carry out key word searches and yield high number of
irrelevant results as they do not have capacity to search based on context.
• Links are hard coded into a document when it is created.
• HTML does not permit link to an element or to multiple locations; XML does.
XML defines structure of data in an open and self describing manner facilitating transfer over a
network and consistence processing by the receiver.
G.1.1 Data, Information and Knowledge
XML has potential to tag data so that it can be interchanged among applications, such as
business-to-business e-commerce, directory services, financial transactions and wireless devices.
The tags are context related and instantly turn data into information. Here context refers to the
identity of, and relationships between a data set’s entities. A set of numbers [B 75 120 R] alone
may not make any sense but if we associate it with say, type, passenger capacity, maximum
permissible speed and color respectively of a railway coach – it becomes information.
Information, in fact is union of data and context whereas knowledge is union of information
with specific issues. XML has potential to play a key role in knowledge management as it is
designed to do just this, i.e., to make information self-describing.
G.1.2 XML – Syntax and Organization
XML defines both document syntax and organization. This aspect is exploited to create data
markup tags, schemas (Collection of tags and data structuring rules), and document type
definitions (DTDs).](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-612-320.jpg)
 const
{
if (index < 0 || index >= numElements)
throw "Out of Bounds";
return this->elements[index];
}
Further you can also create a special ‘IndexOutOfBounds’ class and throw a temporary
instance of it. The advantage is that you have a type specific to the nature of the
exception in this case.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-619-320.jpg)
 const
{
if (index < 0 || index >= numElements) {
throw IndexOutOfBounds(index, numElements);
}
return this->elements[index];
}
H.6 Catching Exceptions
Catching exceptions is the C++ lingo for handling exceptions. If you don’t catch an
exception that has been thrown, your program will terminate. The basic control structure
for exception handling in C++ is the try/catch block. A ‘try/catch’ block that handles
‘IndexOutOfBounds’ exceptions might look like this:
try {
// code that might throw an IndexOutOfBounds exception
} catch (IndexOutOfBounds iobe) {
// code designed to handle any exception of type IndexOutOfBounds
}
Any exceptions thrown within the ‘try’ block will be handled by a matching exception
handler in the ‘catch’ block. You can catch different types of exceptions within one
‘try/catch’ block, and you can catch exceptions of any type by using an ellipsis (...) as the
exception type. This is also called a catch-all handler, and is demonstrated below:
try {
// Call some code here...
} catch (char* str) {
// Handle character string exceptions here...
} catch (IndexOutOfBounds& boundsError) {
// Handle IndexOutOfBounds exceptions here...
} catch (...) {
// Handle all other exceptions here...
}
The order in which catch blocks are placed is important. The first matching Exception
handler will be invoked, so if you placed the catch-all exception handler first, no other
exception handler would be called.
One should never throw and catch an actual object— always use a reference, a pointer, or
a built-in as the exception type. You can choose to filter out some exceptions by handling
them in a particular ‘try/catch’ block and allowing an outer ‘try/catch’ block to handle the
ones you don’t want to handle. This can be done by re-‘throw’ing the exceptions you
don’t want to handle from the ‘catch’ block.](https://image.slidesharecdn.com/tailieu-vncty-com-cforbeginners-masters2007-130613105157-phpapp02/85/Tailieu-vncty-com-c-for-beginners-masters-2007-620-320.jpg)
Tailieu.vncty.com c++ for beginners......masters 2007
- 5. Copyright © 2007, NewAge International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to rights@newagepublishers.com PUBLISHING FOR ONE WORLD NEWAGE INTERNATIONAL(P) LIMITED, PUBLISHERS 4835/24,Ansari Road, Daryaganj, New Delhi - 110002 Visitusatwww.newagepublishers.com ISBN (13) : 978-81-224-2628-1
- 6. PPRREEFFAACCEE C++ is a general-purpose object oriented programming language with features, such as, economy of expression, modern flow control and data structures, and a rich set of operators. C++ is not tied to any particular hardware or system and it is easy to write programs that will run without change on any machine that supports C++. The purpose of this book is to provide an introductory text for understanding the C++ language and to empower the reader to write C++ programs. The book also introduces reader to the paradigm of object oriented programming. The main strength and USP of this book is that it is written by a student for students but will be equally useful for intermediate level programmers, and software development professional. The author is in the best position to identify and address issues and areas where a student needs maximum help in understanding the C++ concepts and developing programming skills. The book is written in an easy to comprehend format and includes inline programs, illustrations, number of well chosen completely solved and tested examples to explain difficult concepts associated with OOP and unsolved exercises to ensure that reader understands both C++ language and underneath algorithms. The book also covers database concepts including SQL, data structures and explains how to interface C++ programs with any database using Open database connectivity (ODBC); basic windows graphics programming and a write up on Standard Template Library (STL) which is now part of standard C++. One salient feature of this book is not to fathom the depths of topics such as the Databases, SQL, ODBC, Windows API, the Windows Graphics Device Interface GDI, STL, or any other hardcore technical syntax. Instead, we endeavor to show you how to speedily build real-world applications with only basic knowledge of the Windows architecture. We do know that these are sticky subjects that take many pages to cover appropriately. However, an interested reader can always refer to many other volumes dedicated to such topics. Author
- 8. AACCKKNNOOWWLLEEDDGGEEMMEENNTTSS I would like to deeply thank the various people who, during the several months in which this endeavor lasted, provided me with useful and helpful assistance. Without their care and consideration, this book would not be possible. First, I would like to thank my teachers and professors ‘Dr T.E. Doyle’ ‘Lecturer for department of Electrical and Computer Engineering, Mc Master University’, ‘Mrs. Romi Sharma’, Delhi Public School, New Delhi and ‘Dr. Peter Smith’ ‘Associate Dean of Engineering’, McMaster University for introducing me to the world of computers and should I say a world of ‘C++’. Second, I would like to give a very special thanks to Susan Visser (IBM DB2 Publishing Manager, Toronto, On, Canada) for the help, encouragement and guidance in completion of the book. Third, on a personal level I would like to commend the interest, encouragement provided by my mother (‘Dr. Geeta Asthana’) and sister (‘Deepti Asthana’) whose support, patience and understanding made this possible. Forth, I would really like to thank all my friends Chris, Avi, Fahad, Rohan, Jaimy, Helen, Neha and Samina. Most important, to my dad ‘Dr R.G.S. Asthana’ who put up his weekends, working hours and his patience throughout the editing and cross-reviewing process which constitutes a rather difficult balancing act. Author
- 10. CCOONNTTEENNTTSS Preface…….…………………………………………………………………………….V Acknowledgements…………………………………………………………………..VII Part I: INTRODUCTION TO C++………….…………………………... 1 Chapter 1: INTRODUCTION………..……………………………………………….. 3 1.1 Evolution of Computers and Programming Languages 4 1.2 Chronology of Development of Programming Languages 5 1.3 Object - Oriented Programming 11 1.4 C++ Programming Languages 12 1.5 Structure of the book 13 Chapter 2: FUNDAMENTALS OF C++………………………..…………………..….15 2.1 Introduction 16 2.2 Our first C++ program 16 2.3 Comments 17 2.4 Tokens in C++ 18 2.5 Escape Sequences 26 2.6 The cin statement 28 2.7 Mathematical expressions in C++ 29 2.8 Relational Operators 32 2.9 Logical Operators 34 2.10 Types of operator 35 2.11 Output Manipulators 36 2.12 The ‘conio.h’ header files 38 2.13 Errors 38 2.14 Type Conversion & Type Casting 38 2.15 Unsigned Data types 42 2.16 Postfix and Prefix Operators 43 2.17 What is an algorithm? 46 2.18 Flow-charts 47 2.17 Building Your Own Programs 52 2.18 Review Exercise 61
- 11. X 2.19 Programming Project 65 2.20 Let us revise! 66 Chapter 3: CONTROL STRUCTURES…………….………………………………...... 69 3.1 Introduction 70 3.2 Conditional structure: if then else 70 3.3 Repetitive structures or loops 73 3.4 Bifurcation of control loops 81 3.5 The Selective Structure: switch 83 3.6 Review Examples 87 3.7 Review Exercise 104 3.8 Let us revise! 108 Chapter 4: FUNCTIONS………………………………..…………………………........111 4.1 Introduction 112 4.2 Function Definition 115 4.3 Accessing a function 118 4.4 Default Arguments 120 4.5 Constant Arguments 121 4.6 Arguments as Reference Parameters 122 4.7 Inline functions 126 4.8 Scope of variables 127 4.9 Character functions in C++ 131 4.10 Mathematical functions in C++ 132 4.11 Recursive functions 133 4.12 Review Examples 133 4.13 Review Exercise 138 4.14 Programming Project 141 4.15 Let us revise! 142 Chapter 5: ARRAYS…………………………….………………………………………. 143 5.1 Introduction 144 5.2 Declaring Arrays 144 5.3 Passing arrays as parameters to functions 149 5.4 Traversing of Arrays 151 5.5 Searching element in an array 152 5.6 Merging of arrays 157 5.7 Sorting with arrays 159 5.8 Arrays as strings 161 5.9 Two-dimensional arrays 170 5.10 Solved Examples 175 5.11 Review Exercise 183 5.12 Programming Project 185 5.13 Let us revise! 186
- 12. XI Chapter 6: STRUCTURES…………….…...………………………………………… 187 6.1 Introduction 188 6.2 Structure Definitions 188 6.3 Structure Variables 189 6.4 Accessing Members of a Structure 192 6.5 Structure Arrays 198 6.6 Structures with Functions 203 6.7 Nested Structures 208 6.8 Solved Examples 210 6.9 Review Exercise 223 6.10 Programming Project 227 6.11 Let us revise 228 Chapter 7: POINTERS……….……………………………………………………..…..229 7.1 Introduction 230 7.2 Addresses and Pointers 230 7.3 Pointer to an Array 240 7.4 Pointer to Pointers 245 7.5 Pointers with Functions 248 7.6 The New and Delete operators 250 7.7 Array of Pointers 252 7.8 Pointer to a Structure 253 7.9 Solved Examples 255 7.10 Review Exercise 263 7.12 Programming Project 266 7.13 Let us revise 268 Part II: OBJECT ORIENTED PROGRAMMING (OOP)…………..….…………..269 Chapter 1: OBJECTS ORIENTED PROGRAMMING…….…….………………..….271 1.1 Introduction 272 1.2 Basic Features of OOP 273 1.3 Classes and Objects 273 1.4 Constructors and Destructors 290 1.5 Review Examples 303 1.6 Review Exercise 325 1.7 Programming Project 329 1.8 Let us Revise 331 Chapter 2: INHERITANCE………………………………………………………..…...333 2.1 Introduction 334 2.2 Base and Derived Classes 334 2.3 Inheritance in C++ 335 2.4 Types of Inheritance 343 2.5 Solved Examples 345 2.6 Review Exercise 350
- 13. XII 2.7 Programming Project 352 2.8 Let us revise 353 Part III: DATA STRUCTURES & FILES……………...…………………………….355 Chapter 1: DATA STRUCTURES…………………………….…………………...……357 1.1 Introduction 358 1.2 Data Structures 358 1.3 Arrays 359 1.4 Operations on static data structures 364 1.5 Solved Examples 366 1.6 Review Exercise 370 1.7 Programming Project 372 1.8 Let us revise! 373 Chapter 2: LINKED LISTS…..…………….……………………………………………375 2.1 Introduction 376 2.2 Linked lists 376 2.3 Linked lists as stacks 390 2.4 Linked lists as queues 409 2.5 Solved examples 431 2.6 Review exercise 434 2.7 Programming project 438 2.8 Let us revise! 439 Chapter 3: Streams and Files ..…….……………………...……………………………441 3.1 Introduction 442 3.2 Streams 442 3.3 Text files 443 3.4 Binary files 458 3.5 Solved examples 470 3.6 Review exercise 480 3.7 Programming project 483 3.8 Let us revise 485 Part IV: DATABASES & C++………………………………………………….….…..487 Chapter 1: DATABASE CONCEPTS……………….….………………………………489 1.1 Introduction 490 1.2 Databases 490 1.3 Database Management System 490 1.4 Purpose of Databases 491 1.5 DBMS Models 492 1.6 Relational Database Terminology 493 1.7 Relational Algebra 494
- 14. XIII 1.8 Normalization 496 1.9 Review Examples 499 1.10 Let us revise 501 Chapter 2: STRUCTURED QUERY LANGUAGE………………...….………………503 2.1 Introduction 504 2.2 Capabilities of SQL 504 2.3 SQL Data types 504 2.4 SQL Commands and Functions 505 2.5 Review Examples 514 2.6 Review Exercise 518 2.7 Let us revise 520 Chapter 3: OPEN DATABASE CONNECTIVITY .…………….………….…..….... 523 3.1 Open Database Connectivity (ODBC) 524 3.2 Setting up ODBC data source 524 3.3 Connecting from Delphi/C++ Builder using the ODBC data source 525 3.4 Connecting MS Visual C++ 5.0 to Databases 526 3.5 Code to Connect C/C++ to the ODBC data source 526 Appendices……………………….……………………………………………….........529 A: ASCII Table 531 A.1 Control characters 531 A.2 Printable characters 532 A.3 Extended character set 533 B: Standard Template Library (STL) 535 B.1 Introduction 535 B.2 Containers and Algorithm 535 B.3 Iterators 536 B.4 Concepts and Modeling 538 B.5 Refinements 539 B.6 Other parts of the STL 539 B.7 Links to other resources 544 C: Compilers…………………………………………………………………………545 C.1 Introduction 545 C.2 Borland C++ Builder 545 C.3 Microsoft Visual C++ 546 D: Graphics Programming in C++ ……………………………………………...... 549 D.1 Introduction 549 D.2 Windows Graphics Programming 551 E: Number System ………………………………………………………………....583 F: Recursion ……………………………………………………………………......589 F.1 Introduction 589 F.2 Recursive Functions 589
- 15. XIV F.3 Practical Applications of recursion 592 F.4 Recursion versus Iteration 595 G: XML and C++ ………………………………………………………………. 597 G.1 Overview of XML 597 G.2 C++ and XML 599 H: Exceptions and Exception Handling ……………………………………….. 603 H.1 Exceptions 603 H.2 Conventional error handling methods 603 H.3 Exceptions in C++ 604 H.4 Types of exceptions 604 H.5 Throwing of exceptions 604 H.6 Catching Exceptions 605 H.7 Resource Management 606 H.8 Exceptions Specifications 606 H.9 Efficiency Concerns 607 H.10 Exception in constructors 607 H.11 Throwing exceptions in destructors 607 I: Namespaces ……………………………………………………………………609 I,1 Introduction 609 I.2 Using the names 610 I.3 Using Directive 611 I.4 Example of name clashes 614 I.5 Friend and Extern declarations 615 I.6 Unnamed Namespaces 615 Index ……..……………………..………………………………………………… 617
- 16. 1 PPAARRTT II IINNTTRROODDUUCCTTIIOONN TTOO CC++++
- 18. Part I CHAPTER 1 INTRODUCTION AIM OUTLINE • To enlighten the origin of computers and programming languages • To elucidate the development of programming languages • To discuss Object-oriented Languages. • To explain purpose and plan of the book 1.1 Evolution of Computers and Programming Languages 1.2 Chronology of Development of Programming Languages 1.3 Object - Oriented Programming 1.4 C++ Programming Language 1.5 Structure of the book
- 19. 4 1.1Evolution Of Computers and Programming Languages We've come a long way since the invention of Charles Babbage's ‘difference engine’ in 1822 and the period when computers used wires and punch cards for programming purposes. The evolution in the history of computers though began in the early 50’s of the nineteenth century. Since then both computers and programming languages written to work on them have expanded manifolds both due to development of new languages and also due to the slow death of older, outdated ones. Computer languages were first composed only of a series of steps to wire a particular program. These resulted into a series of steps keyed into the computer which were then executed one by one. With time, these languages attained many additional features and capabilities, such as, logical branching and objects orientation and this is how we know these languages today. In the beginning, Charles Babbage's difference engine was made to execute tasks by changing the gears which executed the calculations. Thus, we can say that the earliest form of a computer language was described by physical motion. The next major development occurred when the US Government built the ‘ENIAC’ in 1942 which used electrical signals instead of physical motion. In 1945, John Von Neumann developed two important concepts: 1. The first was known as the "shared-program technique" stating that the actual computer hardware should be simple and need not be hand-wired for each program, instead, complex instructions should be used to control the simple hardware, allowing it to be reprogrammed much faster. 2. The second concept was referred to as "Conditional control transfer" which defined the concept of ‘subroutines’. ‘Subroutines’ are small blocks of code that can be jumped to in any order, instead of a single set of instructions for the computer to take. Subroutines were used to branch from one instruction to another in a program based on logical statements such as the ‘IF (expression) THEN’, and FOR looping construct. This was also one of the major milestone paving the way for the development of ‘libraries’ of functions commonly used in variety of programming applications. Libraries are used in almost every programming language today. The first computing language though was developed in 1949 and was called ‘Short Code’. It was used for electronic devices and it required the programmer to change its statements into 0's and 1's by hand. This concept gave birth to machine language. Machine language, in fact, is a system of codes directly understandable by a computer's CPU. Machine code is composed only of the two binary digits 0 and 1. Every CPU has its own machine language and programming in it was the only way to work on these machines. It was time taking and cumbersome experience to program in machine language, therefore, development efforts were put to find ways to enable working on these computers without knowing the machine language. This showed the
- 20. 5Part I: Chapter 1 INTRODUCTION way for development of ‘assembly languages’. Assembly languages are nothing but a human-readable notation for the machine language that specific computer architecture uses. Machine language, a pattern of bits encoding machine operations, is made readable by replacing the raw values with symbols called mnemonics. This process of conversion of machine to assembly language is made possible with the help of an ‘assembler’. The next milestone was achieved in programming history with the development of ‘compilers’ in 1951. A compiler is a program that converts another program from some source language (or a programming language) to machine language (object code). The compiler allowed automatic conversion to machine language, i.e., 0’s and 1’s so that statements could be interpreted and executed by the computer. This resulted in faster programming as the programmer no longer had to work by hand for converting statements into 0’s and 1’s. A compiler is distinguished from an assembler by the fact that each input statement does not, in general, correspond to a single machine instruction. A compiler may support such features as automatic allocation of variables, arbitrary arithmetic expressions, control structures such as FOR and WHILE loops, variable scope, input/ouput operations, higher-order functions and portability of source code. With the advances in technology it became viable to introduce computers in business and scientific application, thereby, enhancing importance and power of both computers and programming languages. The following section outlines the salient events in history in chronological order which lead to the development of various programming languages. 1.2 Chronology of Development of Programming Languages This section outlines the salient events in history in chronological order which lead to the development of various programming languages. 1.2.1 FORTRAN Fortran also known ‘FORTRAN’ is one of the oldest programming language. The language was originally developed in the 1950s and is still seriously used, particularly in parallel programming applications. The language derives its name from the words ‘Formula Translator/Translation’. Today, this language is considered outdated as it only included the ‘IF’, ‘DO’, and ‘GOTO’ statements, but at the time, these commands were a huge step forward. The basic types of data types used today were derived from FORTRAN, which included logical variables, integer, real, and double-precision numbers. 1.2.2 COBOL Though FORTRAN was good at manipulating numbers, it was not so good at handling input and output requirements, an aspect important to business computing. This lead to the development of ‘COBOL’ in 1959. ‘COBOL’s name
- 21. 6 is an acronym, for ‘Common Business Oriented Language’, defining its primary domain in business, finance, and administrative systems for companies and governments. COBOL statements had very English-like grammar making it quite easy to learn. These features made it easier to program the business applications and COBOL became the most popular language for the businessmen. 1.2.3 LISP In 1958, John McCarthy of Massachusetts Institute of Technology (MIT) created the List Processing (or LISP) language. It was primarily designed for Artificial Intelligence (AI) research. The main difference between LISP and other languages was that the basic and only type of data supported was the list, denoted by a sequence of items enclosed in parentheses. LISP is in use even today in certain specialized applications due to its highly specialized and abstract nature. 1.2.4 Algol Algol was created for scientific use in 1958. In fact, Algol is the root of the tree that not only guided but resulted in the development of languages such as Pascal & C/C++. It was also the first language with a formal grammar, known as ‘Backus-Naar Form’ or BNF. Algol was first language to introduce recursive calling of functions but its next version called ‘Algol 68’ was packed with too many features and thus became difficult to use. These limitations of ‘Algol’ lead to the development of simple and compact languages like Pascal. 1.2.5 APL APL is acronym for “A Programming Language” or “Array Processing Language”. It was invented in 1962 by Kenneth E. Iverson. Iverson received the Turing Award in 1979 for his work. APL has evolved over time and has changed significantly from the original language described by Iverson in his book. 1.2.6 Simula In 1962, Simula introduced the object-oriented programming paradigm and thus can be considered the first and true object-oriented programming language and a predecessor to Smalltalk, C++, Java, and all modern class-based object-oriented languages. As its name implies Simula was, primarily, designed for facilitating development of application program for carrying out simulations, and had the framework for many of the features of object oriented languages present today. 1.2.7 Basic Kurtz and John Kemeny co-developed BASIC in 1964. BASIC belongs to a family of high-level programming languages. Originally devised as an easy-to-use
- 22. 7Part I: Chapter 1 INTRODUCTION tool, it became very popular on home microcomputers in the 1980s, and remains popular to this day. BASIC’s acronym stands for Beginner's All-purpose Symbolic Instruction. 1.2.8 PL/I In 1968, PL/I ("Programming Language One", pronounced "pee el one") emerged in the market. It was designed for scientific, engineering, and business applications. The language syntax was English-like and hence suited for describing complex data formats, with a wide set of functions available to verify and manipulate them. The principal domain of PL/1 was data processing. PL/I supported both recursion and structured programming. 1.2.9 Pascal Development of Pascal began in 1968 by Niklaus Wirth. Pascal derived best features of the languages in use at the time, viz., COBOL, Fortran, and Algol. Pascal also improved the "pointer" data type, a very powerful feature of any language that implements it. It also added a ‘Case’ statement that allowed instructions to branch like a tree in such a manner: Pascal also supported dynamic variables, which could be created while a program was being run through the NEW and DISPOSE commands. The combination of features, input/output and solid mathematical features made Pascal a highly successful language. Pascal did not implement dynamic arrays, or groups of variables, which proved to be needed and led to its downfall. Niklaus Wirth tried to bridge the gap and created Modula-2 - a successor to Pascal, but by the time it appeared, C had already gaining popularity and had become a established language. 1.2.10 Ada ‘Ada’ is a structured and statically typed programming language. It was designed by Jean Ichbiah of ‘Cii Honeywell Bull’ in the 1970s. It is similar to C or C++. ‘Ada’ was named after ‘Ada, Lady Lovelace’, often thought to be the first computer programmer. ‘Ada’ targeted at embedded and real-time systems. In Case expression of expression-value-1: statements to execute... expression-value-2: statements to execute... End
- 23. 8 1983, the Department of Defense, U.S.A. directs that all new "mission-critical" applications be written in Ada. The Ada 95 revision included better support for systems, numerical, and financial programming. 1.2.11 SQL Structured Query Language (SQL) is the most popular computer language used to create, modify and query databases. Technically, SQL is a declarative computer language for use with "quasi-relational databases". 1.2.12 C The C programming language was developed in the early 1970s by Ken Thompson and Dennis Ritchie. C was originally developed for and implemented on the UNIX operating system, on a PDP-11 computer. Ritchie restructured the language and rewrote the compiler and gave his new language the name "C” in 1972. In fact, 90% of UNIX was then written in C. C uses pointers extensively and was built to be fast and powerful but in the process it became cryptic and hard to read. C code is portable, i.e., it is not tied to any particular hardware/computer or operating system. C combines the elements of high-level languages with the functionality of assembly language. C makes it easy to adapt software for one type of computer to another. C was a direct descendant of the language B. The language B was developed by Ken Thompson in 1970 for the new UNIX OS. By the mid 90's, the Internet had become part of our society. With the start of the new millennium, the Internet is as common as the television, and with it comes new languages designed specifically for its use. 1.2.13 Java In the early 1990's, interactive TV was believed to be the technology of the future. Sun Microsystems decided to develop a new language for the interactive TV which should be portable (i.e., it can run on many types of machines). This language led to the development of Java. In 1995, Netscape licensed Java for use in their internet browser. This development became a milestone in Java attaining the status of the language of the future. Java was easy-to-learn and use, therefore, it easily became popular among faculty members, students, and everyone else. 1.2.14 Visual Basic Even with all the new languages introduced, use of BASIC continues to grow. Visual Basic is often taught as a first programming language today as it is based on the BASIC language developed in 1964. Microsoft has extended BASIC in its Visual Basic (VB) product. The heart of VB is the form, or blank window on which you drag and drop components such as menus, pictures, and slider bars.
- 24. 9Part I: Chapter 1 INTRODUCTION These items are known as "widgets." Widgets have properties (such as its color) and events (such as clicks and double-clicks) and are central to building any user interface today in any language. 1.2.15 Perl Perl has often been described as the "duct tape of the Internet," because it is most often used as the engine for a web interface or in scripts that modify configuration files. It has very strong text matching functions which make it ideal for these tasks. Perl was developed by Larry Wall in 1987. 1.2.16 C# An object-oriented and type-safe programming language supported by Microsoft for use with the .NET Framework. C# (pronounced "see-sharp") was created specifically for building enterprise-scale applications using the .NET Framework. It is similar in syntax to both C++ and Java and is considered by Microsoft as the natural evolution of the C and C++ languages. Programming languages have been under development for years and this process will continue for many years to come. They got their start with a list of steps to wire a computer to perform a task. These steps eventually found their way into software and began to acquire newer and better features. The first major languages were intended for special purposes while the languages of today are differentiated by the way they are programmed in, as they can be used for almost any purpose and perhaps the languages of tomorrow will be more natural with the invention of quantum and biological computers. The entire evolution of programming languages is also shown in table 1.1 in chronological order. Table 1.1: Evolution of Programming Languages Year Name of the Language Brief Description 1957 FORTRAN Mathematical FORmula TRANslating system developed by John Backus and his team. 1958 FORTRAN II Upgraded to handle subroutines and links to assembly language. 1959 LISP 1.5 COBOL Created by the Conference on Data Systems and Languages (CODASYL). 1960 ALGOL 60 It was First block-structured language. 1960 - 1962 APL Kenneth Iverson begins work on APL (A Programming Language). It uses a specialized character set that, for proper use, requires APL- compatible I/O devices. 1964 APL360 Implemented BASIC BASIC was invented by Prof. John G. Kemeny and Thomas E. Kurtz at Dartmouth College in 1964 to
- 25. 10 give students a simple programming language that was easy-to-learn. 1966 FORTRAN 66 Comes in Market LOGO LOGO is best known for its "turtle graphics." 1968 ALGOL 68 ALGOL 68 proves difficult to implement. 1970 SMALLTALK Work on Prolog and SMALLTALK begins. 1972 C Dennis Ritchie produces C. 1975 Tiny BASIC Bob Albrecht and Dennis Allison (implementation by Dick Whipple and John Arnold) runs on a microcomputer in 2 KB of RAM. Bill Gates and Paul Allen write a version of BASIC that they license to Micro Instrumentation and Telemetry Systems on a per-copy royalty basis. RATFOR-- RATional FORTRAN It is a preprocessor that allows C-like control structures in FORTRAN. RATFOR is used in Kernighan and Plauger's "Software Tools," which appears in 1976. 1978 AWK A text-processing language named after the designers, Aho, Weinberger, and Kernighan. UCSD Pascal Kenneth Bowles makes Pascal available on PDP-11 and Z80-based computers. FORTRAN 77 The ANSI standard appears. 1980 Smalltalk- and Modula-2 Emerge C++ Bjarne Stroustrup develops a set of languages -- collectively referred to as "C With Classes" -- that serve as the breeding ground for C++. 1982 ISO Pascal & PostScript Emerge 1983 Smalltalk-80 Goldberg et al publish its implementation. Ada Its name comes from Lady Augusta Ada Byron, the first computer Programmer because of her work on Charles Babbage's analytical engine. 1983-1984 C Microsoft and Digital Research release the first C compilers for microcomputers. C++ The name C++ is coined by Rick Mascitti and its implementation emerges. Borland's Turbo Pascal Comes in the market and becomes extremely popular. 1986 Turbo Prolog Borland releases Turbo Prolog. 1986 Smalltalk/V The first widely available version of Smalltalk for microcomputers Object Pascal Apple releases it for the Mac. 1987 Perl Perl was developed by Larry Wall and has very
- 26. 11Part I: Chapter 1 INTRODUCTION strong text matching functions. 1989 ANSI C The specification is published. C++ 2.0 The 2.0 version adds features such as multiple inheritance and pointers to members. 1990 C++ 2.1 An Annotated C++ Reference Manual by B. Stroustrup et al, is published. This adds templates and exception-handling features. 1991 Java Java is an object-oriented programming language developed primarily by James Gosling and colleagues at Sun Microsystems. The language, initially called Oak (named after the oak trees outside Gosling's office), was intended to replace C++, although the feature set better resembles that of Objective C 1993 Object-oriented COBOL ANSI releases the first-draft proposal. 1998 C++ The C++ programming language standard was ratified in 1998 as ISO/IEC 14882:1998, the current version of which is the 2003 version, ISO/IEC 14882:2003. 2000 C# C# (pronounced “See-Sharp” was developed by Microsoft fo ruse with its .net platform. 1.3 Object-Oriented Programming In the late 1970's, a new programming paradigm known as ‘Object Oriented Programming’, or OOP was developed. Objects are pieces of data that can be packaged and manipulated by the programmer. Bjarne Stroustroup liked this method and developed extensions to C known as "C with Classes." This set of extensions developed into the full-featured language C++, which was released in 1983. OOP languages are the natural choice for implementation of an Object-Oriented Design because they directly support the object notions of classes, inheritance, information hiding, and dynamic binding. An object-oriented system programmed with an OOP language results in least complexity in the system design and implementation, which can lead to an increase in maintainability. Object-oriented applications can be written in conventional languages but they are much easier to write in languages especially designed for OO programming. OOP language can be divided into two categories: hybrid languages and pure OO languages. Hybrid languages are based on some non-OO model that has been enhanced with OO concepts. C++ (a superset of C), Ada 95, and CLOS (an object-enhanced version of LISP) are sometimes referred to as hybrid languages. Pure OO languages are based entirely on OO
- 27. 12 principles; Smalltalk, Eiffel, Java, and Simula are pure OO languages. The emergence of high level languages and their relationship with the hardware is shown in figure 1.1. Figure 1.1: Emergence of high level languages 1.4 C++ Programming Language C++ (pronounced "see plus plus") is a general-purpose computer programming language. It supports procedural programming, data abstraction, object-oriented and generic programming. Prior to 1983, Bjarne Stroustrup added features to C and formed what he called "C with Classes". He combined the Simula's use of classes and object-oriented features with the power and efficiency of C. Further, enhancements started with the addition of classes, followed by, among many features, virtual functions, operator overloading, multiple inheritance, information hiding, polymorphism, dynamic binding, templates and exception handling. During the 1990s, C++ became one of the most popular commercial programming languages. In particular, "ARM C++" added exceptions and templates, and ISO C++ added RTTI, namespaces, and a standard library. C++ was originally designed for the UNIX system environment. With C++, programmers could improve the quality of code they produced and reusable code was easier to write. The C++ programming language standard was ratified in 1998 as ISO/IEC 14882:1998, the current version of which is the 2003 version, ISO/IEC 14882:2003. C++ was designed to organize the raw power of C using OOP, but maintain the speed of C and be able to run on many different types of computers. In terms of number of applications, the most popular OO language in use is C++. One advantage of C++ for commercial use is its syntactical familiarity to C, which many programmers already know and use; this lowers training costs. One disadvantage of C++ is that it lacks the level of polymorphism and dynamics most OO programmers expect.
- 28. 13Part I: Chapter 1 INTRODUCTION 1.5 Structure of the book The book has been divided into four parts. Part I includes seven chapters which introduce the basics of procedural programming in C++. Part II of the book comprises of two chapters which introduce the concept of object oriented programming and its applications. Part III covers ‘Data structures’ and ‘Files’ in C++ and Part IV of the book comprises of basic concepts of databases; SQL also known as ‘Structured Query Language’ - A database sublanguage used in querying, updating, and managing relational databases; SQL also known as ‘Structured Query Language’. The book includes appendices, viz., ASCII table which is a standard code used for information interchange and communication between data processing systems; the Standard Template Library (STL) - a general-purpose C++ library of algorithms and data structures, container classes, and iterators; the list of recommended compilers to be used when dealing with C++, graphics programming in C++ and number systems. To enable beginners to grasp and master complex concepts in topics, such as, Pointers and data structure, we have provided step by step pictorial representation of the logic. The unsolved exercises at the end of each chapter have been created from a set of very common and unique questions to test how much you have learned from the chapter. These are followed by the ‘Programming Projects’ and the ‘Let us revise’ sections to enhance the books usefulness in the class room environment. The book is also accompanied with electronic C.D. which consists of source code of all solved examples and an animated tutorial to facilitate reader to understand data structures better.
- 29. 14 Notes
- 30. Part I CHAPTER 2 FUNDAMENTALS OF C++ AIM OUTLINE 2.1 Introduction • Introduce the fundamentals of C++. • Learn about types of expressions and operations used in programming languages. • Learn the concept of program creation and optimization. 2.1 Introduction 2.2 Our first C++ program 2.3 Comments 2.4 Tokens in C++ 2.5 Escape Sequences 2.6 The cin statement 2.7 Mathematical expressions in C++ 2.8 Relational Operators 2.9 Logical Operators 2.10 Types of operators 2.11 Output Manipulators 2.12 The ‘conio.h’ header file 2.13 Errors 2.14 Type Conversion & Type Casting 2.15 Unsigned Datatypes 2.16 Postfix and Prefix Operators 2.17 What is an algorithm? 2.18 Flow-charts 2.19 Building your own programs. 2.20 Review Exercise 2.21 Programming Project 2.22 Let us revise!
- 31. 16 2.1 Introduction As we already know, C++ was developed by Bjarne Stroustrup of AT&T Bell Laboratories in the early 1980's, and was based on the C language. The name is a pun - "++" is a syntactic construct used in C (to increment a variable). The language ‘C’ is a subset of C++ written before C++ came into being. In this chapter, we will introduce the fundamental concepts of C++ and then slowly move towards creating our own C++ programs. 2.2 Our first C++ program Look at the program 2.1. This program prints the message ‘My first C++ program’ on the screen. Program 2.1 # include<iostream.h> void main() { cout<<" My first C++ Program"; } Let us observe this piece of code carefully. This program consists of a function called ‘main’. Functions are one of the fundamental building blocks of C++. We will learn more about them in later chapters but for now all we need to know is that every C++ program has a ‘main’ function. It is from this function where the execution of the program starts. The round brackets ‘()’ after the word ‘main’ are known as ‘parenthesis’ and they tell the compiler that this is a function and not a variable. They are followed by ‘{}’ curly braces. The first curly bracket ‘{‘begins the function and its counterpart ‘}’ is used to delimit the function. They can be compared to the BEGIN and the END keywords in some other programming languages. Functions have the following general syntax. Syntax: return_type main() { …………. body of the function ...………. }
- 32. 17 Part I: Chapter 2 Fundamentals of C++ As of now, we will always be using the key word ‘void’ as the return type when referring to function ‘main’. Now let us come to the piece of code inside the function or to the function body. In the body of the function, we have used the program statement ‘cout<<" My first C++ Program”;’. Program statements instruct the computer to execute a desired operation. Both these statements tell the compiler to display the quoted text on the screen. So when this program is executed we will observe the following footage on the screen: OUTPUT SCREEN The semicolon after every program statement marks the end of the program statement. Forgetting the semi colon will result in a syntax error. So we should always remember to use them. Now let us look at the first line of the program ‘# include<iostream.h>’. This statement is a preprocessor directive. A preprocessor directive is an instruction to the compiler. A part of the compiler known as the ‘preprocessor’ deals with these directives before the actual compilation process of the program begins. The preprocessor directive ‘#’ include tells the compiler to insert another file into the source file. As a result of this the compiler replaces the ‘# include’ directive with the contents of the file indicated. There exist many such preprocessor directives; all preprocessor directives begin with the '#’ sign. The type of file included by ‘#include’ is defined as a header file, ‘iostream’ is an example of a header file and deals with the basic input/output operations and contains information about ‘cout’ that is necessary for our program. Header files usually have the ‘.h’ extension. They are predefined in the compiler but we can also create our own header files. 2.3 Comments Comments make programming simple. They not only help us to understand a piece of code written by some one else better but also help in writing ones own program specially in large programs where one needs to keep track of what’s going on. Comments help in increasing the program readability and increase program’s maintainability. In C++ comments can be given in two ways: a) Single line comments: They start with // (double slash) symbol and terminate at the end of the line. My first C++ Program FUNDAMENTALSOFC++
- 33. 18 E.g.: int a ; // declares the variable ‘a’ of integer type b) Multi line comments: Start with a /* symbol and terminate with a */ symbol. For example, /* This is a multi line comment */ 2.4 Tokens in C++ A Token is a group of characters that logically belong together. It is the smallest individual unit in a program. For example, look at the following statement: The statement declares ‘y’ to be an integer variable and assigns a value ‘10’ to it. Here ‘int’, ‘y’, ‘=’, ‘10’ and semicolon (;) are all referred to as tokens. C++ supports the following types of tokens. 2.4.1 Identifiers It is the symbolic name given by a programmer for any data item or function. The identifier is a sequence of characters taken from the C++ character set. The rules for the formation of an identifier are: Can consist of alphabets, digits and only one special character i.e. underscore “_”. Can start with an alphabet or underscore, but not with a digit. C++ is case sensitive language Variables are also a kind of identifiers. They are the most fundamental part of any language. A variable is a location in the computer memory which can store data and is given a symbolic name for easy reference. Its value can change during the program execution. The syntax for declaring a variable is given below: < data type > < variable name >; In the syntax given above, the ‘data type’ parameter has to be any one of the data types present in C++. The basic data types available in C++ are given in Table 2.1. int y=10;
- 34. 19 Part I: Chapter 2 Fundamentals of C++ Table 2.1: Basic Data Types in C++ Data Type Range Bytes required Bits (8 bits = 1 byte) Required Usage int -32768 to 32767 2 16 For storing numbers without decimal long -2,147,483,648 to 2,147,4836,47 4 32 For storing integers. Has higher range than ‘int’. char 0 to 255 1 8 For storing characters float -3.4 * 1038 to 3.4 * 1038 4 32 For storing floating point numbers. Its has seven digits of precision double 1.7e + / - 308 (15 digits) 8 64 It is used to store double precision floating point numbers. The size acquired by a variable in the computer memory depends upon whether we are storing it as an integer, a character or a decimal. In other words, a variables size depends upon its data type. The computer memory is organized in the form of ‘bytes’. Each ‘byte’ is further a collection of even smaller units called ‘bits’. We will now discuss the fundamental data types available in C++. 2.4.1.1 Integer variables An integer variable exists in variable sizes. In Microsoft DOS also known as MSDOS it is 16 bits or 2 bytes long whereas in operating systems like Windows 2000/NT it is found to be 32 bits (4 bytes) long. Integer variables represent integers like 1, 2 and -100. Integer variables as the name suggests do not represent decimals or have a fractional part. They were built to express the idea of ‘1’ using integers and not ‘1.5’. Program 2.2 given below declares two integer variables and prints them on the screen. FUNDAMENTALSOFC++
- 35. 20 Program 2.2: To declare two integer variables # include <iostream.h> void main() { int var1=10; int var2=20; cout<<" Var 1:"<<var1; cout<<" Var 2:"<<var2; } This is the output one will observe on the screen when this program is executed. All C++ programs run in such MSDOS windows. Output Screen Let us now try to understand how this program actually works. We again begin with the header file ‘iostream.h’ as it contains the necessary definition for using the basic I/O operations in the language. Then we enter the ‘main’ function from where the actual execution of the program begins. If we notice carefully we have declared two integer variables ‘var1=10’ and ‘var2=20’. As variables are also identifiers, we have to keep in mind to use the guidelines given earlier when declaring variables. To declare variables one uses the basic syntax given below: < data type > < variable name >; int var1=10;
- 36. 21 Part I: Chapter 2 Fundamentals of C++ Then we have used ‘cout’ to display these variables on the screen. The important thing to note is that ‘cout’ knows how to handle integers and characters. So if an integer is fed into it than it prints the integer as a number and if an alphabet is fed into it, it prints it like a character. For example, cout << “The total of 7 and 10 is ” << (7 + 10); This statement sends the ‘string’ and ‘value’ to cout String: “The total of 7 and 10 is” Value: value of the expression 7 + 10 Display produced: The total of 7 and 10 is 17 2.4.1.2 Character Variables Character variables are used to store integers from the range of (-128 to 127) they are also used to store ASCII characters. ASCII which stands for ‘American Standard for Information Interchange’ is the numerical representation of a character such as 'a' or '@' or an action of some sort. These numbers range from 0 to 127. ASCII character set includes every key present on a keyboard of a computer in numerical form so that it can be read by the computer. For example, the alphabet ‘a’ is represented by the number ‘97’ in ASCII format. When the C++ compiler encounters a character it translates to the corresponding ASCII code. The entire ASCII table is given in the appendix A of the book. Character variables are stored as character constants and it occupies a byte of the computers memory. Which means a character variable storing the character ‘1’ and an PROGRAMMING TIPS Using commas, decimal points, and special signs with integer entries is not allowed and will result in syntax errors. For example, o Valid: 15 -110 +235 243 -643 +16 o Invalid: $255.62 23,521 6,282 14.89 The ‘cout’ program statement is defined under the ‘iostream.h’ header file and it is used to display data fed into it using the standard output device FUNDAMENTALSOFC++
- 37. 22 integer variable ‘1’ are not equivalent. Figure 2.1 given below shows how character variables are stored in the computers memory. Figure 2.1: Character 'a' stored in the computers memory Character variables are also declared using the syntax used for declaring integer variables the only difference being that this time the ‘data type’ changes to ‘char’. Program 2.3 given below declares two character variables and prints them on the screen. Program 2.3: To declare two character variables // Header Files # include <iostream.h> // Function ‘main’ void main() { // Declaring variables char var1='10'; char var2='20'; cout<<" Var 1:"<<var1; cout<<" Var 2:"<<var2; } In program 2.3, we have declared two character variables ‘var1’ and ‘var2’ and assigned them the characters ‘10’ and ‘20’ using the assignment statement. If we observe carefully we have used single codes (‘ ‘) when assigning characters to ‘var1’ and ‘var2’, this is done so that the compiler can differentiate between character constants and integer values.
- 38. 23 Part I: Chapter 2 Fundamentals of C++ 2.4.1.3 Floating Point Variables We have already learnt about how the compiler stores integers and characters using the ‘int’ and the ‘char’ data types. In this section, we will learn about floating point variables and understand how they are stored in the computers memory. Floating points represent numbers with decimals or they represent those numbers which have a fractional part attached to them for example, 10.2, -10.2 or 2.10. Floating point variables occupy 4 bytes of the computers memory and have a range of about 3.4 * 1038 to 3.4 * 1038 . Figure 2.2 given below shows how floating variables are stored in the computers memory. Figure 2.2: Storing floating variables Program 2.4 given below declares two floating variables ‘var1’ and ‘var2’ and displays them on the screen. Program 2.4: To declare two floating variables // Header File # include <iostream.h> // Main Function void main() { // Floating variables float var1=-10.2; float var2=10.2; // Printing Variables cout<<" Var 1:"<<var1; cout<<" Var 2:"<<var2; } FUNDAMENTALSOFC++
- 39. 24 2.4.2 Keywords in C++ A keyword is a reserved word that has a predefined meaning and purpose in the language. It cannot be used as an identifier by the user in his program e.g. float, int and so on. Every language has its own set of key words. The list of the keywords present in C++ is given in Table 2.2. Table 2.2: Keywords in C++ asm float register auto for return break friend short case goto signed catch huge sizeof cdecl If static char inline struct COMMON PROGRAMMING ERRORS Not declaring all variables used in a program will lead to a syntax error. Attempting to store one data type in a variable declared for a different type might result in a syntax error.
- 40. 25 Part I: Chapter 2 Fundamentals of C++ class int switch const interrupt template continue long this default near typedef do new union double operator unsigned else pascal virtual enum private void extern protected volatile far public while 2.4.3 Constants in C++ A constant is an expression whose value doesn’t change during program execution. A constant can be declared as an integer, floating point or character. Constants can be declared in C++ using any of the following procedures. One can define a constant identifier by using the ‘ #define’ preprocessor directive. It has the following syntax. Syntax: #define identifier value For example, #define PI 3.14159265 #define length 100 FUNDAMENTALSOFC++
- 41. 26 Once these constants are declared they can be used in the rest of the program as shown in program 2.5. Program 2.5: Calculating circumference of a circle // Header File # include <iostream.h> // Declaring constant PI # define PI 3.14 // Main Function void main() { // Declaring Variables float circle_circumference; float radius=3.0; circle_circumference=(2*PI*radius); // Calculating circumference cout<<" The circumference of the circle is "<<circle_circumference; } There is yet another method of declaring constants using the ‘access modifier’. It has the following syntax: Syntax: const <data type of the constant > < name of the constant identifier >; For example, const float pi = 3.14; const int length=100; Once a constant is declared any attempt to alter the value of the constant variable will elicit an error message from the compiler. We can use either of the two methods describe above to declare and use constant variables.
- 42. 27 Part I: Chapter 2 Fundamentals of C++ 2.5 Escape Sequences Escape sequences are a combination of characters preceded by a code-extension character (also called as escape character). The code-extension character indicates that the succeeding characters are interpreted differently. Escape sequences are normally used to control printed or displayed output. Some of them will be used very regularly when we start writing programs. The list of various common escape sequences is given in table 2.3. Table 2.3: Escape sequences in C++ Escape Sequence Meaning a Alarm b Back space f Form feed (for the printer) n New line r Carriage return t Tab v Vertical tab Backslash ? Question mark The most commonly used escape sequences are the ‘n’ and the ‘t’ sequences. The ‘n’ escape sequence is used to shift the cursor to the next line where as the ‘t’ sequence is used to shift the cursor by eight spaces when printing lines of text. It acts like the ‘tab’ key on our keyboard. PROGRAMMING TIPS Variables with the same data type can be grouped together and declared in one statement Format: dataType variableList; For example, int A1, A2, A3; A good programming practice is to declare each initialized variable on a line by itself. C++ also allows us to create same named variables, functions belonging to different data types. We will learn how to do that in later chapters. FUNDAMENTALSOFC++
- 43. 28 2.6 The cin Statement Just as we have the ‘cout’ statement to display data on the screen, the ‘cin’ statement is used to extract data from the user or insert data into a variable. The statement is used with the insertion operators (<<). It is also defined under the <iostream.h> header file and has the following syntax: Syntax: cin>>variable1>>variable2…………….; You may observe that we have used the insertion operator repeatedly ‘>>’ in the syntax. This is perfectly valid and is known as cascading of operators. For example, cin>>variable1>>variable2; Cascading the insertion operators allow us to input multiple variables using a single ‘cin’ statement. This is the same as using the ‘cin’ statement twice. cin>>variable1; cin>>variable2; Cascading can also be done with the extraction operators (<<) used with output statement ‘cout’. To understand how the ‘cin’ statement works let us create a program which enters two integers from the user and displays their sum on the screen. Program 2.6: To find the sum of two integers // Header File # include <iostream.h> // Main Function void main() { int a,b;
- 44. 29 Part I: Chapter 2 Fundamentals of C++ cout<<" Enter the two integers"; // Entering the integers cin>>a>>b; // Displaying the sum of the integers cout<<" The sum of the two integers "<<a<<" and "<<b<<" is "<<a+b; } Note: Both ‘cin’ and ‘cout’ refer to the basic input/output devices usually keyboards and monitors. 2.7 Mathematical expressions in C++ A mathematical expression is any expression containing arithmetic operators. The arithmetic operators available in C++ are given in the table 2.4. Table 2.4: Arithmetic operators in C++ Operator Usage + Used for addition - Used for subtraction * Used for multiplication / Used for division % This operator is called the remainder or the modulus operator. It is used to find the remainder when the integer value/variable is divided by another value/variable. This operator cannot be used with floating type variables. The piece of code given below illustrates use of arithmetic operators: int a = 10, b=3; cout << ” Sum = ” << a + b << endl; cout << ” Difference = “<< a – b << endl; cout << ” Product = “<<a * b <<endl; cout << ”Quotient = “<<a /b << endl; cout << ” Remainder = “<< a %b << endl; Output Screen Sum=13 Difference=7 Product=30 Quotient=3 Remainder=1 FUNDAMENTALSOFC++
- 45. 30 2.7.1 Precedence of operators The multiplication (*), division (/) and the remainder operation (%) are given precedence over addition (+) and subtraction (-) operations. All operations of the same precedence are performed from left to right. Parentheses have the highest precedence among all the operators. Table 2.5 further enlightens the concept of precedence of operators. Table 2.5: Precedence of Mathematical Pperators S.NO OPERATOR PRECEDENCE 1. ( ) Highest precedence given to parentheses. In case of several pairs of parentheses without nesting, they are evaluated from left to right. 2. *, /, % Multiplication, division and remainder operators are evaluated after the parentheses. In case of several such operators, they are evaluated from left to right 3. +, - Plus and minus are evaluated last. If there are several such operators then they are evaluated from left to right COMMON PROGRAMMING ERRORS Not using the multiplication operator (*) when required results in an error. For Example, (x+2)(x) this statement is illegal. Using the ‘%’ operator with float variables will result in a syntax error.
- 46. 31 Part I: Chapter 2 Fundamentals of C++ Let us look at the following examples to understand how we use the precedence of operators. 2.7.1.1 Solved Examples Example 2.7.1.1.1 Assuming the following declaration, evaluate the value stored in the variable ‘result’ using precedence of operators. int a=5,b=20,c=10,d=5,e=7; int result=0; (a) result= a * b + c – d % e; (b) result= (a * b) + c – d % e; (c) result= a * b + (c – d) % e; (d) result= a * ( b + c) – d % e; Solution of 2.7.1.1.1 (a) The value ‘105’ is stored in variable ‘result’ after the first statement is executed. Figure 2.3 illustrates how this expression is evaluated by the compiler. Step1: The first operator we encounter is the multiplication (*) operator with ‘a=5’ and ‘b=20’ as its operands. By looking at table 2.5, we see that the (*) operator has higher precedence than the addition operator (+) which happens to be the next operator in the expression. Hence the expression (a*b) is evaluated to obtain ‘100’ as the result. Step 2: Now if we again look at table 2.5 we can see that addition and subtraction operators have the same precedence but as the addition operator comes first in the expression the expression ‘100+c’ is calculated and replaced in the overall expression. result = a * b + c – d % e; result = 100 + c – d % e; FUNDAMENTALSOFC++
- 47. 32 Step 3: So now we are left with the statement ‘110 - d % e’. As the (%) operator has higher precedence over the subtraction operator (-). Hence the modulus operator with ‘d’ and ‘e’ as operands is executed even though the subtraction operator comes first in the expression. Step 4: As a result of the preceding steps the value of ‘105’ is stored in the variable ‘result’. Figure 2.3: Solution of 2.7.1.1.1(a) Now solve the remaining parts of the example to obtain the following results: (b) result = 105 (c) result = 105 (d) result = 145 2.8 Relational Operators An operator is a symbol or letter used to indicate a specific operation on variables or values in a program. In order to carry out a comparison between two expressions (values/variables) one needs to use the relational operators. ANSI C++ standards specify that the result of a relational expression is binary. That means that the result of a comparison operation is either ‘True’ or ‘False’. Relational Operators are generally used in control structures and loops which we will study in later chapters. A list of relational operators supported in C++ is given in table 2.6. COMMON PROGRAMMING ERRORS Using a variable in an expression before the variable is assigned a value will lead to unexpected result as unintialized variables may have junk value. Dividing a variable with zero will lead to an over flow error. Forgetting to separate individual data streams passed to ‘cout’ with an insertion (“put to”) symbol will result in a syntax error. result = 110 – d % e; result = 105
- 48. 33 Part I: Chapter 2 Fundamentals of C++ Table 2.6: Relational operators in C++ Operator Usage Example Explanation < Less than A<B A is less than B > Greater than A.>B A is greater than B <= Less than or equal to A<=B A is less than or equal to B >= Greater than or equal to A>=B A is greater than or equal to B != Not equal to A != B A is not equal to B == Equality A == B Compares the two variables ‘A’ and ‘B’ and checks for equality between these variables. If the variables seem to have equal values stored in them then the expression returns the integer value ‘1’ otherwise it returns the integer value ‘0’. Note: The values ‘1’ and ‘0’ are not any random values but are ‘Boolean digits’ where ‘1’ corresponds to the expression being true and ‘0’ corresponds to the expression being false. For Example, Relational Expression Result (2 == 4) The expression would check for the equality of the two integers ‘2’ and ‘4’. As they are not equal the expression would return ‘false’ as the result of the statement. FUNDAMENTALSOFC++
- 49. 34 2.9 Logical Operators The logical operators are used to combine multiple conditions formed using relational or equality expressions. Table 2.7 describes the logical operators present in the language. Table 2.7: Logical operators in C++ Operator Usage Example && AND The compound condition evaluates to true, if both the conditions in the compound condition evaluate to true ( (a>b) && (a>c)) || OR The compound statement evaluates to true, if any or both the conditions in the compound condition evaluate to true ( ( a>b) || (a>c)) ! NOT It negates the condition That is: • If the condition evaluates to true, it makes it false • If the condition evaluates to false, it makes to true !(a>b) For Example, (1 > 4) As the integer ‘1’ is not greater than ‘4’ the expression would return ‘false’ as the result of the expression. (3 != 0) As the two operands ‘3’ and ‘0’ are not equal the expression would return ‘true’ as the result of the statement. Logical Expression Result ((2 == 4) && (7>5)) The expression would evaluate to false as we are using the ‘and’ operator. For which both the conditions have to evaluate to true but the relational expression (2==4) is not true.
- 50. 35 Part I: Chapter 2 Fundamentals of C++ 2.10 Types of operators Besides the logical and relational operators, we can also divide operators into different groups depending upon the number of operands applicable with an operator. In C++, there exist mainly three data types when divided using this criteria. 2.10.1 Unary operators They are the operators which operate on a single operand. Examples of the Unary operators are the ‘increment’ and the ‘decrement’ operators or the logical ‘not’ operator as shown below: X++; // Increases the value stored in X by ‘1’ (Increment operator) Y--; // Decreases the value stored in Y by ‘1’ (Decrement operator) 2.10.2 Binary operators They are the operators which operate on two operands. For example, all arithmetic operators and all relational operators AND (&&) and OR (||) are the binary operators as shown below: ((X>Y) && (X>Z)) // Binary ‘and’ operator ((X>Y) || (X>Z)) // Binary ‘or’ operator !(1 > 4) The expression would evaluate to true as ‘1’ is not greater than ‘4’ hence the condition in the parentheses will evaluate to ‘false’ but as we are using the not operator ( ! ) the resultant condition will evaluate to true. (!(False) = true) ((2 > 0) || (5<0)) The result of this statement is true. As in this statement we are using the ‘or’ operator and if any of the two relational conditions evaluate to true so will the resultant expression. FUNDAMENTALSOFC++
- 51. 36 2.10.3 Ternary operators As the name suggests, they are the operators which are applicable on three operands. C and C++ both have only one conditional ternary operator. It is commonly known as short hand if-else statement (?:). It is generally used where conditional looping is required. The syntax for this operator is as follows: Syntax: Expression3 = ((condition)? Expression1: Expression2); If the ‘condition’ is true than ‘Expression3’ is assigned the value stored in ‘Expression1’. On the other hand, if the condition evaluated is false then ‘Expression3’ is assigned the value of ‘Expression2’. Example, int y = 3; int x; x = ((y > 10) ? 1 : 0); // Since y is lesser than 10, x is assigned a value of 0 2.11 Output Manipulators Output manipulators are operators used with the insertion operators (<<) to control or manipulate how data is displayed. Manipulators are the most common way to control output formating. They are used to make the printed data look appealing and make programs user friendly. All the ouput manipulators in C++ are defined under the ‘iomanip.h’ header file. We will discuss some of the most commonly used output manipulators in the follwing sub-sections. 2.11.1 The ‘setw’ manipulator The ‘setw ‘manipulator is used to set the field width for the next insertion operator. It applies only to the next element inserted in the output. We can use either the left or the right direction to justify the data appropriately in the field. Output is right justified by default. The default field width is just wide enough to hold the expression. The ‘setw’ operator overcomes this problem by allowing a fixed field width. Let us look at program 2.7 to understand how ‘setw’ overcomes the problems posed by default field widths. Program 2.7: Default Field width // Header Files # include<iostream.h> # include<iomanip.h> // Main Function FUNDAMENTALSOFC++
- 52. 37 Part I: Chapter 2 Fundamentals of C++ void main() { int Durban=4000, Moscow=5000, Sydney=10, Paris=1; cout<<" City "<<"t No. of Churches"; cout<<"n Durban t"<<Durban; cout<<"n Moscow t"<<Moscow; cout<<"n Sydney t"<<Sydney; cout<<"n Paris t"<<Paris; } The output window of this program is shown along with the program. As one might observe, it is hard to read and compare these numbers. It would be better if they are right justified. Program 2.8 uses the ‘setw’ manipulator to eliminate the problem by specifying field widths to the respective columns. Program 2.8: Setting field widths using the ‘setw’ manipulator // Header Files # include<iostream.h> # include<iomanip.h> // Main Function void main() { int Durban=4000, Moscow=5000, Sydney=10, Paris=1; // Using ‘setw’ manipulator cout<<setw(10)<<" City "<<setw(30)<<" No. of Churches n"; cout<<setw(10)<<" Durban"<<setw(20)<<Durban<<"n"; cout<<setw(10)<<" Moscow"<<setw(20)<<Moscow<<"n"; cout<<setw(10)<<" Sydney"<<setw(20)<<Sydney<<"n"; cout<<setw(10)<<" Paris"<<setw(20)<<Paris<<"n"; } 2.11.2 The ‘endl’ manipulator The ‘endl’ operator works exactly like the “n” escape sequence. The only difference being that we do not use quotes (“ “) while using this operator, Just like ‘setw’, it is also defined under the ‘iomanip.h’ header file. When this operator is used, it shifts the cursor to the next line so that lines of text are displayed from the new line. FUNDAMENTALSOFC++
- 53. 38 2.12 The ‘conio.h’ header file The conio.h header file contains a large set of functions that are used to perform input and output operations. Functions defined under this header file can directly access the standard input or output devices. Table 2.8 describes some of the commonly used functions defined under the ‘conio.h’ header files. Table 2.8: Functions under 'conio.h' 2.13 Errors There are three types of errors encountered in any programming language. a) Syntax Errors The errors which are traced by the compiler during compilation, due to wrong grammar of the language used in the program, are called as syntax errors. For Example, cin<<a; // instead of extraction operator the insertion operator is used b) Run–Time Errors: The errors encountered during the execution of the program due to unexpected calculation or input or output, are called as runtime errors. For Example, a=n/0; // Division by zero c) Logical Errors: These errors are encountered when the program does not give the desired output, due to wrong logic of the program. For Example, a = b + c; // The programmer used addition operator though // intended to use multiplication operator 2.14 Type Conversion & Type Casting In program 2.9, we have calculated the value of the ‘double’ variable ‘c’ using an integer variable ‘a’ and a float variable ‘b’. One might wonder, whether this program compiles successfully or not, as in this program we are using variables of different data types in a single prorgam statement. Functions Purpose clrscr() Clears the screen getch() Extracts a character from the user. Can also be used to pause programs.
- 54. 39 Part I: Chapter 2 Fundamentals of C++ Program 2.9: Automatic Type Conversion # include <iostream.h> # include <conio.h> void main() { clrscr(); int a=10; float b=4.23; double c; c=a*b; cout<<" c: "<<c; } C++ is very flexible when treating mixed datatypes. In C++, it is perfectly normal to perform any arithmetic operations on variables of different data types. This is achieved by a process called ‘Automatic Type Conversion’. Such a conversion takes place when two operands of different data types are encountered in the same expression. In such a case the variable of lower data type is automatically converted to the variable with the higher data type. Table 2.9, gives the order of data types in C++ according to which type conversions take place. Table 2.9: Order of data types in C++ Look at the following program to understand how type casting really works. Data Type Priority long double double float long int char Highest Lowest FUNDAMENTALSOFC++
- 55. 40 Program 2.10: Automatic Type Conversion # include <iostream.h> # include <conio.h> void main() { clrscr(); int A=10; float B=4.2; float C=A/B; cout<<"C:"<<C; } So what happens when this program is executed. As we have already learnt that when the compiler confronts with such mixed expressions in same statement it converts the variable belonging to the lower data type to the variable belonging to the higer data type. That is in our program the variable ‘A’ is converted to a float variable and then divided by the float variable ‘B’ so that the it can be assigned to the float variable ‘C.’ If suppose variable ‘C’ was of type ‘double’ then both these variables ‘A’ and ‘B’ would get converted to type ‘double’. The entire process is also explained in figure 2.4. Figure 2.4: Automatic Type Conversion These conversions are automatically done by the compiler but it is also possible to exercise data conversion manually as opposed to automatic type conversion provided by the compiler. This process of exercising data conversion manually is known as ‘type casting’. We will learn about this process in the next section.
- 56. 41 Part I: Chapter 2 Fundamentals of C++ 2.14.1 Type Casting Type casting is required as while programming occasionly we might want to convert a value from one data type to another. If in such a case the compiler doesnot do it automatically or without warnings it might be handy if we can perfrom the process manually. A cast, is special programming instuction which specifies what data type to treat a variable as in a given expression. Type casting can be done using the any of the following syntax. Syntax: type(expression); Where ‘type’ refers to the data type to which the variable/value needs to be casted To understand how type casting is useful study program 2.11. Program 2.11: Type casting vs. Type conversion Type Casting Automatic Type Conversion // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { clrscr(); // Variable Declarations int A=10; int B=4; float C; // Type Casting C = (float (A))/B; // Printing C on the screen cout<<"C:"<<C; } OUTPUT SCREEN //Header Files # include <iostream.h> # include <conio.h> void main() { clrscr(); // Variable Declarations int A=10; int B=4; float C; C = A/B; cout<<"C:"<<C; } OUTPUT SCREEN FUNDAMENTALSOFC++
- 57. 42 As we can observe both these programs give different outputs. The compiler in this case does not automattically type cast the following expression and one need to type cast it manually. 2.14.2 Solved Examples Example 2.14.2.1 Find the value stored in the variables using the declaration given below: int var1, var2, var3; double var4, var5, var6; var1 = 120; var2 = 14; var3 = 12.9; var4 = 10.1; (a) var1= var2+ var3; // Find the value stored in var1 (b) var3 = var3/var4; // Find the value stored in var3 (c) var2=var1/var3; // Find the value stored in var2 (d) var6=var1/var3; // Find the value stored in var6 (e) var6=var4*(var2/var1); // Find the value stored in var6 (f) var3=(int)((double)var1*(int) var3 + 1.0)/var4; // Find the value stored in var3 Solution of 2.14.2.1 (a) var1 = 26 (b) var3 = 1 (c) var2 = 10 (d) var6 = 10 (e) var6 = 0 (f) var3 = 142 2.15 Unsigned Data types Unsigned data types store non negative (0 and positive) data entries. Unsigned data types provide us with a range of positive values double that of unsigned counterparts. Table 2.9 given below depicts the unsigned versions of the data types as we explained in Table 2.1. C = A/B;
- 58. 43 Part I: Chapter 2 Fundamentals of C++ Table 2.9: Unsigned data types Unsigned data types are used when ever quantities represented by variables are non negative, e.g., when representing day, months or years of a particular date. 2.16 Postfix and Prefix Operators One unique feature of unary operators, such as, increment/decrement operators is that these operators can be used either as ‘prefix’ or as ‘postfix’ operators. In the ‘prefix’ notation as the word ‘pre’ suggests, the operator precedes the variable (++ sample_variable;) where as in ‘postfix’ notation the variable comes after the operator (sample_variable++;). So how does it matter whether we use the prefix operator with a variable or the postfix operator? Program 2.12 explains the difference. Program 2.12: Prefix, Postfix Notations Prefix Notation Postfix Notation int a= 10; int b= 11; int c; c = a + ++b; int a= 10; int b= 11; int c; c = a + b++; What is the value stored in ‘C’ in each case? When using the prefix notation the variable is first incremented and then this incremented value is used in the rest of the expression. Where as when we are using the postfix expression the variable is first used in the expression it is in and then it is incremented. Looking at the program we created, when we are using the ‘prefix notation’ with variable ‘b’, the variable is first incremented to ‘12’ and then this incremented value ‘12’ is used in the expression ‘c = a + ++b’ the result is ‘22’. On the other hand when variable ‘b’ is incremented using the postfix operator, the variable is first used in the program statement it is in and then its value is incremented by 1 as a result of which a different value for variable ‘c’ (21) is obtained. Figure 2.5 also explains how postfix and prefix operators differ from each other. Keyword Range Bytes Occupied Bits Occupied unsigned int 0 to 65,535 2 16 unsigned long 0 to 4,294,967,295 4 32 unsigned char 0 to 255 1 8 FUNDAMENTALSOFC++
- 59. 44 Figure 2.5: The increment operator 2.14.1 Solved Examples Example 2.14.1 .1 Determine the output for the following piece of code: // Header Files # include<iostream.h> # include<conio.h> void main() { // Clears the screen clrscr(); // Variable Declarations int a =10, b=4, c=0; c = a % b; cout<<" a "<<++a; cout<<"n a "<<( ++a + (a++) + b); cout<<"n a "<<( a++ + (a++) + (b++ * c)); } FUNDAMENTALSOFC++
- 60. 45 Part I: Chapter 2 Fundamentals of C++ Solution of 2.14.1.1 OUTPUT SCREEN This is the output we will observe on our screen when this program is executed. Let us see how we arrive at this output. We begin with ‘a=10’, ‘b=4’ and ‘c=0’. We can refer to these variables as rectangular boxes shown below: Then we assign ‘c’ the value obtained by the expression ‘a%b’. In this expression ‘a%b’ one might observe that we are using the ‘remainder of operator’. As ’10/4’ leaves ‘2’ as the remainder, so ‘2’ is the value assigned to variable ‘c’. Next, the compiler reads the statement ‘cout<<" a "<<++a;’. As we are using the prefix operator with variable ‘a’, so ‘a’ will first be incremented to ‘11’ and then displayed on the screen. Similarly, the compiler then reaches the second ‘cout’ statement “(++a + (a++) + b);”. Remember that the compiler reads the program statement from right to left as a result of which the compiler will evaluate this expression to ‘28’. Once this is done, the compiler would then encounter the last ‘cout’ statement ‘cout<<"n a "<< (a++ + (a++) + (b++ * c)); ’. The compiler will first read the expression ‘(b++ *c). FUNDAMENTALSOFC++
- 61. 46 The result of this expression would be the value we get by multiplying b and c, i.e., ‘4 * 2 = 8’ after which variable ‘b’ will be incremented to ‘5’. The remaining expression would then look like ‘a++ + (a++) + 8;. Next in this expression, we are using the postfix increment operator with variable ‘a’ twice as a result of which the expression would then evaluate to ‘35’. 2.17 What is an algorithm? A popular television network broadcasts a cookery show. In one of the episodes the host demonstrated how to bake bread. For doing so, he gave his viewers a list of instructions (called ’recipe’). Algorithm Example: Recipe for baking bread Preparing mixture of ingredients o 1/2 cups sugar o 2 cups of dry yeast mixture o 1 tablespoon salt o 2 packages of dry yeast o 1/4 cup shortening o 3 cups warm water o Knead 5 minutes. Final Preparations o Punch down. o Let rise 30 minutes. o Knead 2 minutes. o Shape into 4 loaves. o Place loaves into greased bread pans. o Let rise until doubled. o Bake at 350 degrees F for 30 minutes.
- 62. 47 Part I: Chapter 2 Fundamentals of C++ In the process, what the host has just demonstrated is an interesting metaphor to understand the concept of algorithm design. An algorithm is basically a set of instructions to solve a given problem or in other words, an algorithm is any well-defined computational procedure that takes some value, or set of values, as input and produces some value, or set of values as output. For example, in our analogy, ‘packets of dry yeast’, ‘sugar’, ‘water’ and ‘flour’ serve as a part of our inputs where as the ‘baked bread’ is the manufactured output when the definite procedure is followed. Let us take a simpler example, say we wish to calculate the area of a circle given the radius of the circle. For doing so, we would follow the following algorithm: 1. Get a value for r, the radius of the circle 2. Set the value of area to (pi * r2 ) 3. Print the value of area So again, in this example, ‘r: radius of the circle’ serves as an input and the ‘area: area of the circle’ turns out to be the output for our algorithm. As it might be evident by now, algorithms act as a road map for accomplishing a given, well-defined tasks. Algorithms are usually written in easy to understand language and are independent of any programming languages. Now, let us learn about another important concept required for quality control before we get started with building our own programs. 2.18 Flow-charts In simple words, flow-charts can be defined as maps or pictorial representations of a well defined process where the various steps in a process are depicted with symbolic shapes and the flow of the process is indicated with arrows connecting these symbolic shapes. Flow-charts are important as they provide a pictorial view of the program flow and hence provide a better understanding of the complete process or algorithm. While implementing a complex algorithm, different people may have different interpretation of how a process/algorithm is implemented. A flow-chart helps in gaining an agreement about the sequence of steps and how they are implemented. Let us take a simple example to understand the concept of flow-charts. Figure 2.6 shows the flow chart for calculating the area of the circle. Each symbol in flow-chart signifies a different operation (see Figure 2.6). The ‘terminal’ symbol ‘1: flow-chart key’ is used to mark the starting or the ending point of a program, symbol ‘4’ is used to represent any input/output operations taking place for example in the flow chart given in Figure 2.6 the first input/output symbol is used to read the radius of the circle ‘r’, where as the latter is used to output the area of the circle. A process in a flow-chart is represented by a rectangle (symbol ‘2’:flow-chart key). This symbol usually indicates a process such as a mathematical computation or a variable assignment, e.g., calculation of the area of the circle in our ecample. FUNDAMENTALSOFC++
- 63. 48 Figure 2.6: Flow-chart: Calculating area of a circle The entire list of the design elements used to generate a flow-chart along with their pictorial representation and description is given in table 2.10.
- 64. 49 Part I: Chapter 2 Fundamentals of C++ Table 2.10 Elements of a Flow-chart FLOW-CHART DESIGN ELEMENTS Symbol Description This symbol is used for the beginning and the end of a program or of a module of a program. The words START (alternatively BEGIN) or STOP (END) would typically be entered in this symbol. This symbol is used for all processes in a program. A process is generally used for arithmetic/computational operations. Used to test for a given condition. Testing whether one quantity is larger than another is one example of a testing/decision operation. Similarly, testing whether a specific quantity is positive or zero is another typical test. 'IF...THEN...ELSE statements would typically characterize this test operation This symbol is used for input or output operations. For example, if we want to input the radius of a circle or output the calculated area of the circle.
- 65. 50 2.18.1 Types of structured flow-charts We described sequence flow-chart in the previous section, in this section we will learn about the other two types of widely used structured flow-charts. Let us start with decision structured flow-charts. The decision structured flow-charts are implemented when the flow of the program depends upon the inputs to the program. Let us take a simple example, say we want to determine whether the integer(X) entered by a user is a positive or a negative integer. The solution to this problem is shown in the flow- chart given in figure 2.7. Figure 2.7: Decision structured flow-chart
- 66. 51 Part I: Chapter 2 Fundamentals of C++ Note the use of the decision structure ‘*’ in figure 2.7. The flow of the program depends upon the value of the integer ‘X’ read into the program. If the value of the integer ‘X’ happens to be greater than zero, then the program follows ‘path 1’ of the flow-chart where as if the value of the integer ‘X’ is negative the program follows ‘path 2’ until it reaches the terminal ‘End’ symbol. The general form of this type of flow-chart is given in figure 2.8. Figure 2.8: Type of structured flow-charts Let us now look at the structured flow-chart, known as the ‘repetitive structured flow- chart’. We take an example involving these flow-chart to understand the concept behind it. Say, we want to display the string, “I am learning about flow-charts” ten times on the screen. For doing so, we would require a repetitive structured flow-chart, the general structure for this type of flow-chart is given in figure 2.8. Customizing this structure we can now form a solution to our problem as illustrated in figure 2.9. FUNDAMENTALSOFC++
- 67. 52 A repetitive structure repeats itself until the test expression evaluates to a true value. Figure 2.9: Decision structured flow-chart As shown in figure 2.9, the program written using this flow-chart will print “We are learning flow-charts” until the counter X is less than ’10’ once this condition evaluates to false the program will terminate. 2.19 Building Your Own Programs Till now we have learnt the basics of the language. In this section, we will learn how to create and write C++ programs. We will take a few programs in this section to teach you to think like computer scientists. Let’s begin with the following example:
- 68. 53 Part I: Chapter 2 Fundamentals of C++ Example 2.19.1 Write a program to enter marks of 5 subjects of a student and print the ‘%’ marks scored by the student on the screen. Solution of 2.19.1 Before writing a program it is very helpful if we can make an outline of how the program would look like in our head. Once we have the outline of the program ready we can use it to create our program. Let us begin our job in a step wise manner. Step 1: We need to enter the marks of a student in 5 subjects. Things to do Declare 5 integer variables say Mark1, Mark2. . . Mark5. Use the ‘cin’ function defined under ‘iostream.h’ header file to enter these 5 variables Step 2: Add all the five variables to find the total marks secured by the student. Things to do Declare a variable ‘Total’ of type int. Initialize it to ‘zero’ to get rid of junk values. Assign the value of (Mark1+Mark2…+Mark5) to ‘Total’. Step 3: Find the percentage of marks scored by the student and display it on the screen. Things to do Declare a variable percentage of type float. Assign the value obtained by (Total/5). Display this using the ‘cout’ function defined under the header file ‘iostream.h’. These are the steps which we need to perform in order to create our program. By noting down these steps (things to do not included) what we have done is that we have developed a ‘pseudo code’ for our program. A ‘pseudo code’ is an outline of a program, written in a form that can easily be converted into real programming statements. Pseudo code can neither be compiled nor executed, and there is no real formatting or syntax rules. The benefit of pseudo code is that it enables the programmer to concentrate on the algorithms without worrying about all the syntactic details of a particular programming language. In fact, you can write pseudo code without even knowing what programming language you will use for the final implementation. Getting back to our example, let us now create our program using the pseudo code we developed. FUNDAMENTALSOFC++
- 69. 54 // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { /* Function clrscr(), to clear the screen */ clrscr(); // To clear the screen /* Variable Declrations */ int Mark1, Mark2, Mark3, Mark4, Mark5; int Total=0; float Percentage=0; cout<<" Enter the marks scored in five subjects"<<endl; cin>>Mark1>>Mark2>>Mark3>>Mark4>>Mark5; /* To calculate the percentage*/ Total = Mark1 + Mark2 + Mark3 + Mark4 + Mark5; Percentage = (Total/5); cout<<" The percentage scored by the student is "<<Percentage<<"%"; } Example 2.19.2 Write a program which asks the user for the radius of a circle and then uses it to find the area of the circle? Solution of 2.19.2 Let us begin the program by writing the pseudo code. Pseudo code Step 1: Define a constant PI . Things to do
- 70. 55 Part I: Chapter 2 Fundamentals of C++ Using the ‘# define’ preprocessor directive declare a constant PI = 3.1426 Step 2: Enter the radius of the circle from the user. Things to do Declare a variable ‘radius’ of type int. Using the ‘cin’ statement to input ‘radius’ from the user. Step 3: Calculate the area of the circle and display it on the screen. Things to do Declare a float variable ‘area’ Assign the value obtained by the formula (PI * (radius)2 ) to the variable ‘area’. Display the variable ‘area’ on the screen with the help of the ‘cout’ statement. Program # include<iostream.h> # include<conio.h> # define PI 3.1426 //main function void main() { int radius=0; float area=0; cout<<" Enter the radius of the circlen"; cin>>radius; area= PI * (radius * radius); cout<<"n The area of the circle is"<<area; getch(); } Example 2.19.3 Write a program to calculate the perimeter of a triangle. The length of all the three sides of the triangle will be fed by the user through the keyboard. FUNDAMENTALSOFC++
- 71. 56 Solution of 2.19.3 Again, let us begin with the pseudo code of the program. Pseudo code Step 1: Obtain the length of the three sides from the user Things to do Declare three integer variables ‘length’, ‘breadth’ and ‘width’. Using the ‘cin’ statement obtain values for the ‘length’, ‘breadth’ and ’width’ of the triangle. Step 2: Calculate the perimeter of the triangle and display it on the screen Things to do Declare a variable ‘perimeter’ of type float to store the perimeter of the triangle. Assign the value obtained by the expression (length + breadth + height) to the variable ‘perimeter’ Using the ‘cout’ statement display the ‘perimeter’ on the screen. Program // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { clrscr(); int length, breadth, height; int perimeter=0; cout<<" Enter the three sides of the triangle"; cin>>length>>breadth>>height; perimeter = length + breadth + height; cout<<" The perimeter of the triangle is "<<perimeter<<" Units"; }
- 72. 57 Part I: Chapter 2 Fundamentals of C++ If you notice that our pseudo codes are shrinking in size as we move on. Just as while learning how to ride a bicycle we take help of the support wheels and leave them when we know how to ride one. Similar logic applies to pseudo codes. Now that we can solve simple programs let us try a tricky one. Example 2.19.4 Suppose we have three fractions (a/b), (c/d) and (e/f). Their sum can be obtained by using the formula given below. Formula to add three fractions Write a program which asks the user to enter three fractions. Using the three fractions inputted by the user and the formula given above, find the resultant sum in fractional form and display it on the screen as shown in the example given above. Hint: Use cascading of operators to simplify program. Solution of 2.19.4 This problem is slightly more complicated than the previous ones that we have encountered. So how does one approach such a problem? We again begin with writing the pseudo code for the problem. Pseudo code Step 1: Obtain the three fractions from the user. Things to do Declare six integer variables ‘numerator1’, ‘numerator2’, ‘numerator3’,‘denominator1’,‘denominator2’ ,’denominator3’. Using the ‘cin’ statement obtain the values for the these six variables from the user. FUNDAMENTALSOFC++
- 73. 58 Step 2: Obtain the sum of the three fraction using the given formula.. Things to do Declare two integer variables ‘numerator_sum’ and ‘denominator_sum’. Assign the variable ‘numerator_sum’ the value obtained by Assign the variable ‘denominator_sum’ the value obtained by Step 3: Display the fraction ((a/b) form) obtained after adding the three fractions. Things to do Display the variables numerator1 and denominator1 on the screen using the ‘cout’ statement. Program // Header Files # include<iostream.h> # include<conio.h> void main() { clrscr(); // Declaring Numerators (Step I: Pseudo code) int numerator1, numerator2, numerator3, numerator_sum=0; // Declaring Denominators int denominator1, denominator2, denominator3,denominator_sum=0; // entering the fractions from the user cout<<" Enter the fractions:"<<endl; // entering the numerator ((numerator1)*(denominator2)*(denominator3) + ((numerator2)*(denominator1)*(denominator3) + ((numerator3)*(denominator1)*(denominator2)) + (denominator1*denominator2*denominator3)
- 74. 59 Part I: Chapter 2 Fundamentals of C++ cout<<" Enter the three numerators"; cin>>numerator1>>numerator2>>numerator3; // entering the denominator cout<<" Enter the three denominators"; cin>>denominator1>>denominator2>>denominator3; /* Finding sum using the given formula Calculating the resultant numerator (Step 2: Pseudo code) */ numerator_sum = ((numerator1*denominator2*denominator3)+ (numerator2*denominator1*denominator3)+ (numerator3*denominator1*denominator2)); // Finding the resultant denominator denominator_sum = (denominator1 * denominator2 * denominator3); // Displaying the resultant fraction (Step 3: Pseudo code) cout<<" So the resultant fraction is"<<numerator_sum<<"/"<<denominator_sum; } Example 2.19.5 Find the error(s) in the following piece of code: // Program to calculate the product of three integers entered by the user // Header Files # include <"iostream.h"> Void main { clrscr(); int a; int b; int c; cout>>" Enter the variables whose product has to be calculated /n"; cin <<a<<b<<c; d = a * b * c; cout>>" The sum of three integers entered is ">> d; getch(); } FUNDAMENTALSOFC++
- 75. 60 Solution of 2.19.5 When ever we program, errors will take place. So it becomes important that a programmer can understand and be able to remove errors. The common programming errors described in the chapter should help you in developing skills to be able to create efficient programs. This example is a specially designed to include some of the most common errors. As we know that errors are mainly of three types, i.e., ‘syntax’, ‘logical’ and ‘runtime’ errors. In these errors, syntax and logical are the most common errors. Let us learn how to identify errors using the piece of code given above. The first error in the piece of code occurs in the statement ‘# include <”iostream.h”>’. This is a syntax error as it is not allowed to use codes (“ “) as well as (< >) together when calling header files. The next error occurs in the function header of the main function ‘Void main’. This has two errors both being syntax errors. The first error that will occur is that C++ is a ‘case sensitive’ language and there is a difference between ‘void’ and ‘Void’. Hence the compiler doesn’t recognize ‘Void’ as a keyword. The second error in the statement is that the function header are always followed by parentheses () which are missing in this case. The next few errors occur in function ‘main’. They are shown in the table given below. Statement Type of Error Error clrscr(); Syntax The function is defined under the header file ‘conio.h’ but this header file is not included in this program as a result it would result in an error. Void main Syntax C++ is a case sensitive language and the compiler doesn’t recognize ‘Void’ as a valid return type. Also the function header is always followed by parentheses ‘( )’ which define the number of parameters that can be passed to the function. cout>>" Enter the variables whose product has to be calculated /n"; Syntax Use of extraction operators (>>) is not allowed with the ‘cout’
- 76. 61 Part I: Chapter 2 Fundamentals of C++ statement. Instead we use the insertion operators (<< ) Cin <<a<<b<<c; Syntax Insertion operators (<<) are not used with ‘cin’ statements. Instead we use the extraction operators (>>) D = a * b * c; Syntax Variable‘d’ has not been declared in function main. cout>>" The sum of three integers entered is ">> d; Syntax Usage of extraction operators (>>). getch(); Syntax The function is defined under the header file ‘conio.h’ but this header file is not included in this program as a result it would result in an error These are some of the most typical errors one observes while programming. Using a step wise approach is a viable solution to reduce errors while programming. 2.20 Review Exercise 1 Mark/2 Mark Questions Q1) Header files usually have the following extension a) ‘.h’ b) ‘.head’ c) ‘.dat’ d) ‘.mpeg’ e) None of the above Q2) Preprocessor directives are: a) instructions to the compiler. b) instructions to the preprocessor. c) instructions to the operating system. d) Both ‘a’ and ‘b’ are correct. e) None of the above. FUNDAMENTALSOFC++
- 77. 62 Q3) A key word is a) an example of an identifier b) a reserved word that has a predefined meaning and purpose in the language. c) a preprocessor directive. d) both ‘a’ and ‘b’ are correct e) None of the above Q4) _____ is/are (an) example(s) of a ternary operator. a) The conditional operator (?:) b) The increment/decrement operators c) All relational operators d) All relational and logical operators e) None of the above. Q5) The logical not operator (!) is an example of a) an unary operator. b) a binary operator c) a ternary operator d) a relational operator e) both a and d are correct Q6) State whether the following statements are true or false: i) Preprocessor is a part of the compiler and it is from here that all preprocessor directives are assessed. ii) Header files always have ‘.h’ extension. iii) A variable of type ‘double’ occupies ‘8’ bytes and ‘63’ bits of the computers memory iv) It is not possible to use the ‘modulus’ operator (%) with float and double variables. v) ASCII which stands for ‘American Standard for Internet Information’ Q7) Find the error(s) in the following piece of code? // Program to calculate the sum of two integers entered by the user Header Files # include <"iostream.h"> Void main { clrscr(); // Variable Declrations
- 78. 63 Part I: Chapter 2 Fundamentals of C++ int a; int b; cout>>" Enter the variables whose sum has to be calculated /n"; cin <<a<<b; c = a + b; cout>>" The sum of two integers entered is ">> c; getch(); } Q8) Determine the output of the following piece of code: (a) // Header Files # include <iostream.h> # include <conio.h> void main() { clrscr(); int var1 = 10.5; float var2 = 10.7; float var3; var3 = (float)var1 + var2; cout<<(int)var3<<"t"<<var3; getch(); } (b) // Header Files # include<iostream.h> # include<conio.h> void main() { clrscr(); int a =2, b=2, c=4; c = a % b + c; cout<<" a "<<a%3; cout<<"n a "<<++a<<"t"<<a++; cout<<"n a "<<(c++ + (a-b)/4); } FUNDAMENTALSOFC++
- 79. 64 Q9) Fill the missing blanks ( _ _ _ ) to complete the following C++ program: # include <iostream.h> # include <_ _ _ _ _ > /* A program to calculate the length of each side of triangle given the perimeter of an equilateral triangle */ _ _ _ main() { clrscr(); float perimeter = 0; _ _ _ length = 0; cout<<" Enter the perimeter of the triangle"; cin>> _ _ _ ; length = perimeter/3; cout<<" The length of each side is"<<length; cout<<" The length of each side is"<<_ _ _ ; getch(); } 3/4 Mark Questions Q10) Write a program to calculate the area of the circle using the circumference of the circle provided by the user. Hint: Area of a circle = ^ R2 , Circumferance of the circle = 2 ^ R. Q11) Write a program to enter the currency of five countries from the user. The program should then display the ‘currency’ and the respective country in the format given below:
- 80. 65 Part I: Chapter 2 Fundamentals of C++ Country Currency India Rupee Canada Dollar Nepal Rupee England Pound Singapore Dollar Q12) Write a program to display the following format given below using only ‘cout’ and output manipulators: 2.21 Programming Project Hero’s formula is used to calculate the area of triangle provided all the sides of a triangle are given. Our programming project is to use the ‘Hero’s’ formula to calculate the area of a triangle. The sides of the triangle ‘a’, ‘b’ and ‘c’ will be inputted by the user. The Hero’s formula for calculating the area of a triangle is given below: Hero's formula: A formula for calculating the area of a triangle: _________________ A = s(s a formula for the area of a triangle: _________________ A = (s(s - a)(s - b)(s - c)) where a, b, and c are the lengths of the sides of the triangle, ‘s’ is the half perimeter (s = (Perimeter of triangle)/2). City Population New Delhi 10,00,220 London 90,000 California 100,000 Hamilton 9,200 FUNDAMENTALSOFC++
- 81. 66 2.22 Let us revise! There are three basic types of data recognized by C++: ‘integer’, ‘floating-point’ and ‘character’ data types. Every variable in a C++ program must be declared to inform compiler to allocate suitable space based on the type of variable. A simple C++ program containing declaration statements has the following format: A preprocessor directive is an instruction to the compiler. A part of the compiler known as the ‘preprocessor’ deals with these directives before the actual compilation process of the program begins. The type of file included by ‘#include’ is defined as a header file. Header files usually have the ‘.h’ extension. They are predefined in the compiler but we can also create our own header files. Comments help in increasing the program readability and increase program’s maintainability. In C++, comments can be given in two ways: (i) Single Line Comments: They start with // (double slash) symbol and terminate at the end of the line. (ii) Multi Line Comments: Start with a /* symbol and terminate with a */ symbol. // Header Files # include <iostream.h> // Main function void main() { …. declaration statements; …. …. other program statements; }
- 82. 67 Part I: Chapter 2 Fundamentals of C++ A Token is a group of characters that logically belong together. It is the smallest individual unit in a program A variable is a location in the computer memory which can store data and is given a symbolic name for easy reference. Its value can change during the program execution. There are three fundamental data types recognized by C++. There are the ‘integer’, ‘floating-point’ and the character data types. Integer variables represent integers like 1, 2 and -100. Character variables are used to store integers from the range of (-128 to 127) they are also used to store ASCII characters. ASCII which stands for ‘American Standard for Information Interchange’ is the numerical representation of a character such as 'a' or '@' or an action of some sort. Floating points represent numbers with decimals or they represent those numbers which have a fractional part attached to them for example, 10.2, -10.2 or 2.10. A keyword is a reserved word that has a predefined meaning and purpose in the language. A constant is an expression whose value doesn’t change during program execution. It can be declared using any of the following two methods: (i) The ‘#define’ preprocessor directive: One can define a constant identifier by using the ‘ #define’ preprocessor directive it has the following syntax: Syntax: #define identifier value (ii) The ‘const’ access modifier: The ‘const’ access modifier can also be used to declare constant identifiers. It is done using the following syntax. Syntax: const <data type of the constant > < name of the constant identifier >; Escape sequences are a combination of characters preceded by a code-extension character (also called as escape character) Output manipulators are operators used with the insertion operators (<<) to control or manipulate how data is displayed FUNDAMENTALSOFC++
- 83. 68 When the compiler confronts mixed expressions in same statement it converts the variable belonging to the lower data type to the variable belonging to the higer data type. This process is known as automatic type conversion. It is also possible to type cast a variable manually for special program needs. Notes FUNDAMENTALSOFC++
- 84. Part I CHAPTER 3 CONTROL STRUCTURES AIM OUTLINE To introduce the concept of control structures. To be able to use the conditional structure: if then else. To understand and use looping constructs for, while loop. Learn about ‘multiple selection’ using the switch selection statement. 3.1 Introduction 3.2 Conditional structure: if then else 3.3 Repetitive structures or loops 3.4 Bifurcation of control loops 3.5 The Selective Structure: switch 3.6 Review Examples 3.7 Review Exercise
- 85. 70 3.1 Introduction A program control flow is usually not limited to execute a linear sequence of instructions. During execution of a program the programmer might want to use a set of code repeatedly or may want to jump or branch to another set of instructions based on certain conditions. Let us take up an example, suppose you want to display your name 5 times on the screen then based upon your current knowledge you would generate the following piece of code. // Header Files # include <iostream.h> # include <conio.h> void main () { clrscr(); cout<< "My name is Ankit Asthana"<<endl; cout<< "My name is Ankit Asthana"<<endl; cout<< "My name is Ankit Asthana"<<endl; cout<< "My name is Ankit Asthana"<<endl; cout<< "My name is Ankit Asthana"<<endl; getch(); } It might acceptable if we have to do this 5 times but what happens if we want to display out name a hundred times on the screen. Then in order to achieve this we would have to repeat this statement ‘cout<< "My name is Ankit Asthana"<<endl;’ a hundred times. This would result in wastage of a lot of time and resources. To overcome this problem C++ provides us with control structures which not only allow us to repeat such block of instructions but also allow the programmer to respond to changing circumstances during execution of a program. C++ allows the following types of control structures: (a) ‘If’ then else (b) The ‘while’ loop and do while loop (c) The ‘for’ loop (d) The ‘switch’ case construct 3.2 Conditional structure: if then else The ‘if’ statement allows you to execute an instruction or block of instructions only if the specified condition is true. It has the following syntax:
- 86. 71 Part I: Chapter 3 Control Structures Syntax: if (condition) statement If the statements to be executed in the conditional structure are more than one then the set of statements also referred to as block of instructions to be executed are enclosed in curly braces ({}). if (condition) { block of instructions } Where ‘condition’ is the expression that is being evaluated. If this expression evaluates to ‘true’, then the statement(s) are executed. On the other hand if the ‘condition’ provided in the ‘if’ statement evaluates to ‘false’, then the statement(s) are ignored and the program jumps to the next instruction after the conditional structure. Let us take up an example, The following pieces of code are executed if x has an even value assigned to it. Case 1 Case 2 As one might observe in the second case we have used curly brackets to execute the block of instructions. Now let me ask you a very interesting question. What happens if we want to execute a set of statements if the condition is true and another set of statements if the condition is false? This is where the ‘if…else’ structure comes in. It consists of the ‘if’ statement followed by a set of instructions, followed by the keyword ‘else’, followed by another set of instructions which is executed if the specified condition evaluates to false. The syntax for the ‘if…else’ structure is given below: Syntax: if (condition) if (x %2==0) cout << "x is even"; if (x %2==0) { cout << "x is even"; cout << x; } CONTROLSTRUCTURES
- 87. 72 { // statements executed if specified condition is true statement(s); } else { // statements executed if the specified condition is false statement(s); } For example, When the piece of code given above is executed then the program evaluates the expression ‘x%2==0’. If the expression evaluates to true, the program prints ‘x is even’ on the screen. On the other hand if this expression evaluates to false the message ‘x is not even’ is printed on the screen. The figure 3.1 explains the same concept with the help of a flow chart. Figure 3.1: The 'if...else' statement if (x %2== 0) { cout << "x is even"; } else { cout << "x is odd"; }
- 88. 73 Part I: Chapter 3 Control Structures The if…else structure can also be nested or concatenated in order to verify a range of values. For example, list of grades achieved by a group of students in their final exams (program 3.1). Program 3.1: Nested if…else structure // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); int grade; cout<<" Enter the grade achieved by the student (0 to 100)"; cin>>grade; if (grade<=50) { cout<<" Grade achieved by the student F:Fail "; } else if (grade>50 && grade <=70) { cout<<" Grade achieved by the student is B"; } else if (grade>70 && grade <=100) { cout<<" Grade achieved by the student is A"; } } 3.3 Repetitive structures or loops Loops allow us to repeat instructions or a block of instructions while the specified condition to initiate the loop remains true. Loops have an objective to repeat a statement (or a set of statements) a certain number of times or till the moment the required condition is fulfilled. CONTROLSTRUCTURES
- 89. 74 3.3.1 The while loop. The while loop, is one of the most commonly used loops in most programming language. This loop is generally used when we are not sure how many number of times do we want to execute a loop. It has the following syntax: Its function is to simply repeat statement(s) while expression provided is true. Program 3.2 shows the use a while loop. Program 3.2: The ‘while’ loop // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); // Variable declaration int i; cout << " Enter the starting number > "; cin >> i; // Entering the while loop while (i>0) { cout << i << " "; i--; } cout << " BANG! BANG!"; getch(); // End of program } Syntax: while (expression) { statement(s) }
- 90. 75 Part I: Chapter 3 Control Structures When the program is executed the user is prompted to insert a starting number for the countdown. Assuming that the user enters the starting number as greater than zero the program then enters the ‘while’ loop. As a result of which the instructions provided in the loop are executed. This process is repeated till the point ‘i’ is no longer greater than zero and the condition provided evaluates to false. A brief step by step description of what happens in program 3.2 is given below. Program 3.2 1. User enter the starting number ‘i’. 2. The while instruction checks for the expression ‘( i > 0)’ resulting in two possibilities. True: The program follows to step 3. False: The program jumps the while loop and follows in to step 5.. 3. Executes the block of instructions provided in the while loop. Print the value stored in ‘i’. Decrement ‘i’ by 1. 4. End of block. Return to ‘step 2’. 5. Print ‘BANG! BANG!’ on the screen. Figure 3.2 also illustrates how the while loop operates. Figure 3.2: The ‘while’ loop CONTROLSTRUCTURES
- 91. 76 3.3.2 The do-while loop. The do-while loop as the name suggests is similar to the while loop. As we have already studied in the while loop the condition or the text expression is evaluated at the starting of the loop. If this expression evaluates to false then the program statements are not executed at all but in some cases we actually might want to execute a set of instructions at least once. This is made possible with the help of the do-while loop. It has the following syntax: Syntax: The do-while loop functions is the same manner as the ‘while’ loop except that the condition containing the test expression in the do-while loop is evaluated after the execution of statement(s) hence allowing the statements present in the block to execute at least once. Program 3.3 illustrates how the do-while loop is used in C++ do { statement } while (condition); COMMON PROGRAMMING ERROR Missing curly brackets {} for opening and closing the loops will lead to a compilation/logical error. When using the do while loop not using semicolon (;) at the end of the while condition will lead to a compilation error. Remember that in case we want more than a single instruction to be executed, we must group them in a block of instructions by using curly brackets { }.
- 92. 77 Part I: Chapter 3 Control Structures Program 3.3: Do while loop // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); int i; do { cout << "Enter number (0 to end) (sentinel Value): "; cin >> i; cout << "Number entered: " << i << "n"; } while (i != 0); getch(); } The working of the ‘do while’ is also shown in figure 3.3. Figure 3.3: ‘do while’ loop CONTROLSTRUCTURES
- 93. 78 3.3.3 The for loop. The for loop executes a given section of code a given number of times. A for loop is generally used when it is previously known from before how many times a loop has to be executed. To be truthful the ‘for’ loop works exactly like the ‘while’ loop having space for variable initialization and incrementation in its expression. It has the following syntax: Syntax: As one might observe the syntax for the ‘for’ loop provides space to specify an initialization and an increment instruction. The various fields in the ‘for’ looping construct are explained in table given below. 1. Initialization: ‘Initialization’ allow us to assign a certain value to the counter variable. This step is exercised only once at the beginning of the loop. 2. Condition: If the condition evaluates to ‘true’ the program enters into the body of the loop (‘statement(s)’). Where as if the specified ‘condition’ evaluates to ‘false’ the program exits the loop and jumps to the statement immediately following the loop. 3. Statement(s): The statements(s) or the body of the loop are repeated till the ‘condition’ remains true. 4. Increment: This expression is used at the end of the loop before checking for the ‘condition’ specified in the loop. Let us take up an example to understand how the ‘for’ loop actually works. PROGRAMING TIPS The while loop is an entry controlled loop that is used when the condition given is clearly defined. The do while loop is an exit controlled loop. That is a do while loop is executed once even if the specified condition is not true. for (initialization; condition; increment) { statement(s); }
- 94. 79 Part I: Chapter 3 Control Structures for (int i = 1; i <3; i++) { cout<<"n Value of i:"<<i; } Let us begin with the ‘for’ statement. ‘for(int i=1; i<3; i++)’. Now when the compiler reads this statement. It firsts declares a variable ‘i’ and assigns it a value of ‘1’. Once this is done it then checks for the condition ‘i<3’ provided in the for expression. As ‘i=0’ at this stage the program enters the body of the ‘for’ loop displaying the value of ‘i’ on the screen. As now we have reached the end of the loop the incrementation step is exercised and the counter variable ‘i’ is incremented by a single value. After this the ‘condition’ for loop continuation is again checked for and as the condition is satisfied ‘2<3’ the loop is again repeated. This process is repeated till this condition is satisfied. The working of the ‘for’ loop is also explained in figure 3.4. Figure 3.4: The for loop Now that we understand how the ‘for’ loop works we are now at a stage where we can compare the ‘for’ and the ‘while’ looping constructs. So let us now create program 3.1 again using for loop. CONTROLSTRUCTURES
- 95. 80 Program 3.4: The ‘for’ loop // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); for (int i=10; i>0; i--) { cout << i << " "; } cout << " BANG! BANG!"; getch(); } The program given above compares the ‘for’ and the ‘while’ looping construct. The ‘initialization’ and ‘increase’ fields provided in the ‘for’ statement are optional hence these fields can be left blank but we still have to use the semi colons(;) provided at the end of these fields. For example, If we observe carefully, when the ‘initialization’ and the ‘increase’ fields are left blank the for loop behaves exactly like the ‘while’ loop. The ‘for’ loop also allows us to specify more than one instruction in any of the fields included in the for statement. This is made possible by the use of the ‘comma (,)’ operator which acts as an instruction separator, it separates and allows more than one instruction where only one instruction is generally expected. For example, for ( e=0, i=80 ; e!=i ; e++, i-- ) { // Body of the loop. for ( ; i<5 ; ) { Body of the loop }
- 96. 81 Part I: Chapter 3 Control Structures } This loop will execute 40 times provided neither ‘e’ nor ‘i’ are modified within the for loop. 3.4 Bifurcation of control loops In this section, we will discuss the statements that are used to alter the flow of control. This can be done by the usage of the following statements. In C++ this is termed as bifurcation of control loops. 3.4.1 The break instruction The ‘break’ instruction is used to exit a given loop even if the condition specified evaluates to true. It is generally used to end an infinite loop. Program 3.5 explains how the ‘break’ instruction is used in C++. Program 3.5: The break instruction // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); int i; for (i=10; i>0; i--) { cout << i << " "; if (i==5) { cout << " Stop BANG! BANG!"; CONTROLSTRUCTURES
- 97. 82 break; } } getch(); } In this program as the value stored in ‘i’ approaches to ‘5’ the program exits the loop and the statements following the loop are executed. 3.4.2 The ‘continue’ expression The ‘continue’ expression instructs the program to skip the rest of the loop in the ongoing iteration as if the program has reached the end of statement(s) block. This causes the program to jump to the next iteration. Program 3.6 given below explains how this expression is used in C++. Program 3.6: The ‘continue’ expression // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr(); for (int i=10; i>0; i--) { if (i==3)continue; cout << i << " "; } cout << " BANG! BANG!"; getch(); } In this program as ‘i’ approaches the value ‘3’ the program skips the statements following this instruction to reach the end of the loop and then begins the next iteration. As it is also evident from the output of the program.
- 98. 83 Part I: Chapter 3 Control Structures 3.4.3 The goto instruction The ‘goto’ instruction allows the programmer to make a jump from one point in the program to another point in the program. Program 3.7 explains how this instruction is used in C++. Although the statement might seem very handy it is not advisory to use it. Program 3.7: The goto instruction # include <iostream.h> # include <conio.h> void main () { clrscr(); int i=10; restart: cout << i << " "; i--; if (i>0) goto restart; cout << " BANG! BANG!"; getch(); } 3.5 The Selective Structure: switch. The selective structure: switch is a multiple selection statement structure to handle decision making. The switch-case statement is a multi-way decision statement. Unlike the multiple decision statement that can be created using the ‘if-else’ structure, the switch statement evaluates the conditional expression and tests it against a list of constant values. The branch corresponding to the value that the expression matches is taken during execution. The switch statement along with a series of case labels also contains an optional default case statement. The only draw back of using the ‘switch’ structure is that one cannot use the ‘float’ or the ‘double’ data types. We can also think about switch case statements as a substitute for long if statements that compare a variable to several "integral" values. They have the following syntax: switch (expression) { case constant1: statement(s) 1 break; case constant2: statement(s) 2 break; CONTROLSTRUCTURES
- 99. 84 So how does the ‘switch’ statement work? The switch selection statement begins by evaluating the expression passed to it as a parameter. It then checks for this evaluated value to be equivalent to ‘constant1’. If it comes out to be equal then it executes the block of instructions provided under the case constant until it reaches the optional ‘break’ keyword. This tells the compiler that it has reached the end of the block of instruction as a result of which the compiler jumps to the end of the selective switch structure. On the other hand if the evaluated expression does not equal the 1st case constant then the structure checks the evaluated expression to the second case constant. If it comes out to be equal, it then executes the second block of instructions until it approaches the break keyword. Finally, if the value of the evaluated expression does not match any of the previously specified constants the program executes the block of instructions under the ‘default:’ keyword. Let us take up an example to understand how the switch case operates (Program 3.8). Program 3.8: ‘Switch Case’ statement // Header Files # include<iostream.h> # include <conio.h> void main() { clrscr(); int day; cout<<" Enter the week day (1-5)"; cin>>day; // Switch Case statement switch(day) { case 1: cout<<" The day is Monday"; break; case 2: cout<<" The day is Tuesday"; break; . . . default: statement(s) }
- 100. 85 Part I: Chapter 3 Control Structures case 3: cout<<" The day is Wednesday"; break; case 4: cout<<" The day is Thursday"; break; case 5: cout<<" The day is Friday"; break; default: cout<<" Not a valid Entry"; } getch(); } The following output is observed when this piece of code is executed. Now that we understand how the switch operation works we can now compare the ‘switch’ and the ‘if-else’ conditional structures. switch example if-else switch (z) { case 5: cout << "z is 5"; break; case 10: cout << "z is 10"; break; default: cout << "value of z is not known"; } If (z == 5) { cout << "z is 5"; } else if (z == 10) { cout << "z is 2"; } else { cout << "value of z is not known"; } CONTROLSTRUCTURES
- 101. 86 At this point let me ask you a very interesting question. What happens if we do not include the optional ‘break’ keyword? If we do not include the optional ‘break’ keyword at the end of every block of instructions then the program would not jump to the end of the block of instructions instead it would enter the blocks of other case constants and keep on executing them until it approaches the ‘break’ instruction or reaches the end of the switch selective block. An example of such a case is given below. // Header Files # include<iostream.h> # include <conio.h> void main() { clrscr(); int day; cout<<" Enter the week day (1-5)"; cin>>day; // Switch Case statement switch(day) { case 1: cout<<"n The day is Monday"; case 2: cout<<"n The day is Tuesday"; case 3: cout<<"n The day is Wednesday"; case 4: cout<<"n The day is Thursday"; case 5: cout<<"n The day is Friday"; break; default: cout<<" Not a valid Entry"; } getch(); }
- 102. 87 Part I: Chapter 3 Control Structures As you might observe the piece of code given above is quite similar to program 3.7 the only difference being that in this program we are missing the optional break statements in this ‘switch’ structure. As a result of this the following output is observed which is quite different from the one we observed with program 3.7. Output Screen 3.6 Review Examples Example 3.6.1 Write a program to find the sum of the series 1 + 3 + 5 + 7……n terms. The value of ‘n’ is given by the user during program execution? Solution of 3.6.1 // Header Files # include <iostream.h> # include <conio.h> COMMON PROGRAMMING ERROR Missing the break statements may result in a logical error. Switch case statements can only be used for integral values and usage of switch case with decimal index values with result in a compilation error. Switch case can be only be used to compare an expression with different constants therefore using the following expressions will result in a compilation error. ‘case (i*2)’:, ‘case (a+b*c)’: … CONTROLSTRUCTURES
- 103. 88 // Main Function void main() { int i=1,sum=0,n; cout<<"Enter the value of n n"; cin>>n; while (i<= n*2) { cout<<i<<endl; sum = sum + i; i = i + 2; } cout<<"Sum="<<sum<<endl; } Output Screen Example 3.6.2 Write a program to find the sum of the series 13 + 33 + 53 + 73 + … n terms. Assume that the value of n is given by the user during program execution? Solution of 3.6.2 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int i =1, sum=0,n; Enter the value of n 5 1 3 5 7 9 Sum = 25
- 104. 89 Part I: Chapter 3 Control Structures cout<<" Enter the value of n n"; cin>>n; while ( i<=n+2) { sum = sum + (i * i * i ) ; i = i + 2 ; } cout<<" The sum of the series = "<<sum<<endl; } Output Screen Example 3.6.3 Write a program to find the sum of the following series: 1 + ½! + 1/3! + … n terms. Assume that the value of n is given by the user? Solution of 3.6.3 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int i,j,n; float fact; float sum=0; cout<<" Enter the value of n n"; cin>>n; Enter the value of n 3 The sum of the series = 153 CONTROLSTRUCTURES
- 105. 90 for (i=1;i<=n;i++) { fact=1; for (j=1;j<=i;j++) { fact = fact * j; } sum += (float) 1/fact; } cout<< " The sum of the series "<< sum; getch(); } Output Screen Example 3.6.4 Write a program to print the following pattern: 1 1 12 21 123 321 1234321 Solution of 3.6.4 // Header Files # include<iostream.h> # include<conio.h> void main () { Enter the value of n 2 The sum of the series is 1.5
- 106. 91 Part I: Chapter 3 Control Structures int a,b,c,d,e,f=1,r=2; //Variable Declarations for ( a=1;a<=4;a++) { for ( b=1;b<=a;b++) cout<<b; for (d=5; d>=f ; d=d-1) cout<<" "; for (e=1;e<r;e++) cout<<a; for ( c=1;c<a;c++) cout<<a-c; cout<<endl; f=f +2; if ( a==3) { r=0; } else r =2; } getch(); } Example 3.6.5 Write a program to ask the user to input two integers and an operator. Based on the operator (+,-,/,*,%) perform the desired operation and print the result (use switch case) ? Solution of 3.6.5 # include <iostream.h> # include <conio.h> void main() { clrscr(); CONTROLSTRUCTURES
- 107. 92 int n1,n2,res; char ch; cout<<” Enter 2 numbers n”; cin>>n1>>n2; cout<<” Enter an operator (+,-,/,*,%) n”; cin>>ch; switch(ch) { case ‘+’ : res = n1+n2; cout<<” The result is “<<res; break; case ‘-’ : res = n1- n2; cout<<” The result is “<<res; break; case ‘*’ : res = n1* n2; cout<<” The result is “<<res; break; case ‘/’ : res = n1/n2; cout<<” The result is “<<res; break; case ‘%’ : res = n1%n2; cout<<” The result is “<<res; break; default: cout<<” Invalid choice”; } getch(); } Output Screen Enter 2 numbers 81 9 Enter an operator (+,-,/,*,%) % The result is 0
- 108. 93 Part I: Chapter 3 Control Structures Example 3.6.6 Write a program to ask the user to input a number between 1 to 12 and print the corresponding month (Use switch case) Solution of 3.6.6 # include <iostream.h> # include <conio.h> void main() { clrscr(); int month; cout<<” Enter any number between 1-12n”; cin>>month; switch(month) { case 1: cout <<”Jan”; break; case 2: cout <<”Feb”; break; case 3: cout<<”Mar”; break; case 4: cout<<”Apr”; break; case 5: cout<<”May”; break; case 6: cout<<”June”; break; case 7: cout<<”July”; break; case 8: cout<<”Aug”; break; case 9: cout<<”Sept”; break; case 10: cout<<”Sept”; break; case 11: cout<<”Nov”; break; case 12: cout<<”Dec”; break; default: cout<<”Invalid choice”; } getch(); CONTROLSTRUCTURES
- 109. 94 } Output Screen Example 3.6.7 Write a program to ask the user to input three numbers. The program should then print the largest number on the screen. Solution of 3.6.7 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int n1,n2,n3; cout<<"Enter three numbers n"; cin>>n1>>n2>>n3; if (n1>n2&&n1>n3) { cout<<"The largest number is "<<n1; } else if (n2>n1&&n2>n3) { cout<<"The largest number is "<<n2; } else if (n3>n1&&n3>n2) { cout<<"The largest number is "<<n3; } getch(); } Enter any number between 1-12 4 Apr
- 110. 95 Part I: Chapter 3 Control Structures Output Screen Example 3.6.8 Write a program to ask the user to input a number and find out whether it is a palindrome or not? Solution of 3.6.8 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { int rem,num,temp,reverse=0; cout<<" Enter a number n"; cin>>num; temp=num; while (temp !=0 ) { rem = temp % 10; reverse = reverse *10 +rem; temp = temp /10; } if (reverse == num) { cout<<" Palindrome number "; } else { cout<<" Not a palindrome number"; } Enter three numbers 3 5 6 The largest number is 6 CONTROLSTRUCTURES
- 111. 96 getch(); } Output Screen Example 3.6.9 Write a program to ask the user to input a number and print its table up to the 10th multiple. Solution of 3.6.9 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int n,i=1,p; // Variable Declrations cout<<"Enter a number n"; cin>>n; while ( i<=10) { p = i*n; cout<<n<<"*"<<i<<"="<<p<<endl; i = i +1; } getch(); } Enter a number 31313 palindrome number
- 112. 97 Part I: Chapter 3 Control Structures Output Screen Example 3.6.10 Write a program to ask the user to input the marks in 5 subjects. Calculate the student’s percentage and print the grade? Solution of 3.6.10 // Header Files # include <iostream.h> # include <conio.h> void main() { int m1,m2,m3,m4,m5; float tot; cout<<" Enter the marks in five subjects n"; cin>>m1>>m2>>m3>>m4>>m5; tot = m1 + m2 + m3 + m4 + m5; cout<<" Your total is n"<<tot; float avg=tot/5.0; cout<<"n Your average is n"<<avg; if (avg>=90) { cout<<"n Your grade is A n"; Enter a number 5 5 * 1=5 5 * 2=10 5 * 3=15 5 * 4=20 5 * 5=25 5 * 6=30 5 * 7=35 5 * 8=40 5 * 9=45 5 * 10=50 CONTROLSTRUCTURES
- 113. 98 } else if (avg>=80&&avg<=89) { cout<<"nYour grade is B"; } else if (avg >=70 && avg<=79) { cout<<"nYour grade is C"; } else if (avg>=60 && avg<=69) { cout<<"nYour grade is D"; } else if ( avg>=50 && avg <=59) { cout<<"nYour grade is E"; } else { cout<<"n Your grade is F"; } } Output Screen Example 3.6.11 Write a program to check whether a number entered by the user is an angstrom number or not. The program should carry on executing as long as the user wants? Enter the marks in five subjects 80 88 56 65 69 Your total is 358 Your aggregate is 71.59998 Your grade is C
- 114. 99 Part I: Chapter 3 Control Structures Solution of 3.6.11 // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { int sum=0,num,temp,rem; char ch='y'; again: cout<<" Enter a number n"; cin>>num; temp=num; while(temp != 0 ) { rem=temp%10; sum = rem *rem * rem +sum; temp = temp /10; } if (sum==num) cout<<" Angstrom number"; else cout<<" Not an angstrom number"; cout<<"n Do you want to continue (y/n)"; cin>>ch; if (ch=='y') goto again; else cout<<" Thank you "; }
- 115. 100 Output Screen Example 3.6.12 Write a program to accept a number from the user and display the following menu: 1. First five multiples 2. Squares of first five multiples 3. Cubes of first five multiples 4. Exit Depending on the users choice, the appropriate action should be carried out Solution of 3.6.12 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int n1,opt,i=1,p=1; cout<<" Enter a number n"; cin>>n1; cout<<"1. First five multiplesn"; cout<<"2. Squares of first five multiplesn"; cout<<"3. Cubes of first five multiplesn"; cout<<"4. Exit n"; cout<<"Enter option (1,2,3,4)n"; cin>>opt; if (opt==1) { while (i<=5) { Enter a number 153 Angstrom number Do you want to continue (y/n)y Enter a number 321 Not an angstrom number Do you want to continue(y/n)n Thank you
- 116. 101 Part I: Chapter 3 Control Structures p = i * n1; cout<<n1<<"*"<<i<<"="<<p<<endl; i = i + 1; } } else if (opt==2) { while (i<=5) { p = i * n1; p = p * p; cout<<p<<endl; i = i + 1; } } else if (opt==3) { while (i<=5) { p = i * n1; p = p * p * p; cout<<p<<endl; i = i + 1; } } else cout<<"Thank you"; getch(); } Output Screen Enter a number 1 1.First five multiples 2.Squares of first five multiples 3.Cubes of first five multiples 4.Exit Enter option (1,2,3,4) 3 1 8 27 64 125 CONTROLSTRUCTURES
- 117. 102 Example 3.6.15 Write a program to find the sum of the following series: 3 + (3+6) + (3+6+9) … n terms. Assume that the value of n is given by the user during run time. Solution of 3.6.15 // Program to find sum of the series // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { int a,b,c=0,d=3,e,sum=0; cout<<"Enter the required number n"; cin>>a; for (b=1;b<=a;b++) { c = c + d * b; sum = sum + c; } cout<< " The sum of the series "<<sum; getch(); } Output Screen Enter the required number 2 The sum of the series 12
- 118. 103 Part I: Chapter 3 Control Structures Example 3.6.16 Write a program to find the sum of the series 1-x3 /3! + x5 /5!+x7 /7! ... n terms. The value of n is given by the user. Solution of 3.6.16 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { // Variable Declarations int a,b,d=3,g=1,x=1; // Counter Variables int sign=1; // Determines the sign of each term in the series float sum=0.0,c=1,f=1; // Sum is used to store the sum of the series cout<<"Enter the value of x and no of times n n"; cin>>x>>a; for ( b=1;b<2 * a;b=b+2) { f=1; c=1; for (int t = 1;t<=b;t++) { c = c * t; } for (int y=1;y<=b;y++) { f = f * x; } sum = sum + (float) (f/c * sign); // Type Casting sign = sign * -1; CONTROLSTRUCTURES
- 119. 104 g = g+1; } cout<<" The sum of the series is "<<sum-(x-(x-1)); getch(); } Output Screen 3.7 Review Exercise 2/3 Mark questions Q1) What do you understand by the term ‘Exit controlled loop’? Q2) Differentiate between a while and do-while loop? Q3) What happens if the break statement in the switch case selection statement is not used? Q4) What are jump statements? Q5) Under which header file is the exit() function defined? Q6) What is the syntax for a while loop explain giving suitable example? Q7) How is the ‘switch case selection statement’ different from ‘if else conditional statement’? Q8) Find the errors in the following piece of code: Enter the value of x and no of times n 2 1 The sum of the series is 1
- 120. 105 Part I: Chapter 3 Control Structures # include <iostream.h> void main() { clrscr(); int sum=0, n; i=1; cout<<” Enter the value of n n”; cin>>n; while (i<=n+2) cout<<i<<endl; sum = sum + i cout<<“ Sum = “<<sum<<endl; getch(); } Q9) What will be the output of the following program: // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int month=4; switch(month) { case 1: cout <<"Jan"; break; case 2: cout <<"Feb"; break; case 3: cout<<"Mar"; break; case 4: cout<<"Apr"; case 5: cout<<"May"; case 6: cout<<"June"; case 7: cout<<"July"; CONTROLSTRUCTURES
- 121. 106 case 8: cout<<"Aug"; break; case 9: cout<<"Sept"; break; case 10: cout<<"Sept"; break; case 11: cout<<"Nov"; break; case 12: cout<<"Dec"; break; default: cout<<"Invalid choice"; } getch(); } Q10) Write the output of the following piece of code: // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { int n=7,i=1,p; while ( i<=10) { p = i*n; cout<<n<<"*"<<i<<"="<<p<<endl; i = i +1; } } 4/5 Mark Questions Q1) Write a program to find the sum of the following series: 1-y5 /3! + y7 /5!+y9 /7!..................n terms.
- 122. 107 Part I: Chapter 3 Control Structures The values of variables ‘y’ and ‘n’ are given by the user during program execution. Q2) Write a program to find the sum of the following series 4+ (4+8) + (4+8+12)…………..n terms. The value of n is given by the user during program execution. Q3) Write a program to print the following pattern: 0 0 00 00 000 000 0000000 Note: The program should use for and while loops to achieve the process Q4) Write a program to input 5 numbers and print the largest number out of the given 5 numbers on the screen? Q5) Write a program to accept a number from the user and display the following menu: First Six Multiples Lowest Common Factor (L.C.F.) Highest Common Divisor (H.C.D.) Exit Depending upon the user’s choice, the appropriate action should be carried out? Q6) Encryption and Decryption Write a program to encrypt a text message by replacing each letter in the message by another one in the following way: If the original letter is uppercase, and its integer value (according to ASCII) is value, replace it by the letter whose integer value is (value + ID)% 26 + 65. If the original letter is lowercase, and its integer value (according to ASCII) is value, replace it by the letter whose integer value is (value + ID)% 26 + 97. Any other characters are left unchanged. Assume ASCII character set is used. ID is student number of your university. (Default ID: 0441673) CONTROLSTRUCTURES
- 123. 108 3.8 Let us revise! C++ allows the following types of control structures i) If then else control structure ii) The while and do-while loop iii) The for loop iv) Switch multiple-selection statement If then statement: if (condition) { statement(s) } Loops have an objective to repeat a statement a certain number of times or till the moment the required condition is fulfilled. The while loop: while (condition) { statement(s) } The do-while loop: do { statement(s) } while (expression); The statements that facilitate the unconditional transfer of program control are called jump statements. C++ provides four jump statements: return, goto, break and continue. A goto statement transfers the program control anywhere in the program.
- 124. 109 Part I: Chapter 3 Control Structures The continue statement abandons the current iteration of the loop by skipping over the rest of the statements in the loop body. It immediately transfers control to the evaluation of the test-expression of the loop for the next iteration of the loop. The exit() function is a library function and it breaks out of the program, abandoning the rest of the execution of the program. CONTROLSTRUCTURES
- 125. 110 Notes
- 126. Part I CHAPTER 4 FUNCTIONS AIM OUTLINE Introduce the concept of ‘functions’. To learn declaring and using ‘functions’. To study method of passing parameters and their relative advantages. Defining Global and Local variables. Enlist ‘Character’ and ‘Mathematical’ functions in C++. 4.1 Introduction 4.2 Function Definition 4.3 Accessing a function 4.4 Default Arguments 4.5 Constant Arguments 4.6 Arguments as Reference Parameters 4.7 Inline Functions 4.8 Scope of variables 4.9 Character functions in C++ 4.10 Mathematical functions in C++ 4.11 Recursive Function 4.12 Review Examples 4.13 Review Exercise 4.14 Programming Project 4.15 Let us revise!
- 127. 112 4.1 Introduction Large programs comprising of only a single list of instructions are cumbersome to develop as it is difficult to manage, understand and maintain them. A function in C++ can perform a particular task, and supports the concept of modular programming design techniques. To understand why one needs to use functions let us take up an example. Suppose that you have created a program which uses the sum of two integers inputted by the user to find the average of the two numbers. How will you approach this piece of problem? To solve the problem we would use the following piece of code: // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { clrscr(); float sum=0; int a,b; float average=0; cout<<" Enter two numbers"; cin>>a>>b; sum = a+b; average = sum/2; cout<<" n The average of the two numbers is "<<average; getch(); } Now, what happens if we again want to find the sum of two integers to find the percentage of each number in the sum? To solve this problem on basis of our current knowledge, we would generate the following code: // Header Files # include<iostream.h> # include<conio.h> // Main function void main() { clrscr(); int ch; int a,b; float average=0; float percentage=0;
- 128. 113Part I: Chapter 4 Functions float sum=0; do { cout<<"nn ====== MENU =====n"; cout<<" 1. Average of two numbersn"; cout<<" 2. Percentage of each number in the sumn"; cout<<" 3. Exitn"; cout<<" !. Choicen"; cin>>ch; switch(ch) { case 1: cout<<" Enter two numbers"; cin>>a>>b; sum= a + b; average = sum/2; cout<<"n The average of two numbers is "<<average; break; case 2: cout<<" Enter two numbers"; cin>>a>>b; sum = a + b; percentage= (a/(sum)) * 100; cout<<" n The percentage of "<<a<<" in the sum is: "<<percentage<<"%"; cout<<" n The percentage of "<<b<<" in the sum is: "<<(100-percentage)<<"%"; break; } } while (ch!=3); getch(); } What have we done in this program? All we have done in this program is that we have copied the same piece of code (highlighted part in the code) given above to find the sum of two integers and then use it to find the percentage of each in the sum. At this point, we would wish that we could store this repeated piece of code in a block and call it again when required, instead of typing the same piece of code again and again. This is where functions are useful. A function is named units of a set of program statements which can be invoked from other parts of the program. Much like the copy and paste feature commonly used with Microsoft word. For our example, we can generate a function ‘sum’ which adds the two integers and returns the sum of the two integers when ever required as shown in program 4.1. FUNCTIONS
- 129. 114 Program 4.1: Function ‘sum’ // Header Files # include<iostream.h> # include<conio.h> // Function Header: Tells the compiler that the function is declared later in the program. float sum( int c, int d); // Main function void main() { clrscr(); int ch; int a,b; float average=0; float percentage=0; do { cout<<"nn ====== MENU =====n"; cout<<" 1. Average of two numbersn"; cout<<" 2. Percentage of each number in the sumn"; cout<<" 3. Exitn"; cout<<" !. Choicen"; cin>>ch; switch(ch) { case 1: cout<<" Enter two numbers"; cin>>a>>b; average = sum(a,b)/2; cout<<"n The average of two numbers is "<<average; break; case 2: cout<<" Enter two numbers"; cin>>a>>b; percentage= (a/(sum(a,b))) * 100; cout<<" n The percentage of "<<a<<" in the sum is: "<<percentage<<"%"; cout<<" n The percentage of "<<b<<" in the sum is: "<<(100-percentage)<<"%"; break; } } while (ch!=3); getch(); }
- 130. 115Part I: Chapter 4 Functions /*************************** Name: 'sum' Parameters: 'c' and 'd' Return Type: float Purpose: Returns the sum of two integers ******************************************/ float sum( int c, int d) { int sum_of_integers=0; sum_of_integers= c + d; return sum_of_integers; } This piece of code generates a function ‘sum’ which takes in two integers and returns the sum of the two integers to the main program when ever called by the program. Functions are one of the most fundamental concepts of programming and will be used thoroughly in programs you create later on. We will begin this chapter by teaching you how to declare functions in C++. 4.2 Function Definition The definition of a function consists of the function header and its body. The syntax of a function definition is as under: A function definition comprises of the following parts: ‘type’: It is the type of data returned by the function. The fundamental types of data that can be returned by a function are given in table 4.1. ‘function_name’: It is the name by function is called. Figure 4.1: Return types in C++ type function_name ( argument1, argument2, ...) statement { Body of the function } Function header FUNCTIONS
- 131. 116 Table 4.1 ‘Arguments’: Each argument consists of a type of data followed by its identifier as per procedure for declaring variables in chapter 1. Once declared, these variables act within the function like any other variable. They allow us to pass parameters to the functions when a function is called. To enter more than one argument ‘commas’ are used. A function can have ‘none’ or multiple arguments depending upon the purpose of the function. ‘Body of the function’: Body of a function could be a single instruction or a block of instructions. If the function contains a block of instructions then the function body should be enclosed in curly brackets {}. We are giving a few solved examples to understand how functions are defined and used in the following section. 4.2.1 Solved Examples Example 4.2.1.1 Declare a function ‘multiply’ which takes in two integers as arguments and returns their product as an integer. Solution of 4.2.1.1 So what do we know about the function we are going to declare? In this problem, we have already been given the function name ‘multiply’ and its ‘return type’. So let us put the information already given to us about the function in the general form of a function declaration: Return Type Type of data returned ‘void’ None ‘int’ Integer ‘float’ Floating points ‘char’ Characters and strings
- 132. 117Part I: Chapter 4 Functions Once we have done this then we are only left with the list of arguments and the body of the function. As per the problem statement, the function takes in only two arguments of type ‘integer’ - we can call these two arguments by any name of our choice but it’s a good practice to give names which are context sensitive. We have used ‘integer1’ and ‘integer2’ as these arguments in the piece of code given below: These arguments ‘integer1’ and ‘integer2’ can now be used for any valid purpose in the function just as we can use any variable after it has been declared in function ‘main()’. As per the need of the program, we now have to calculate and return the product of these two arguments. This can be done by declaring and then returning a variable ‘product’ which stores the product of these two arguments as shown below: int multiply( int integer1, int integer2) { int product; product = integer1 * integer2; return product; } One might notice that both ‘product’ and the return type of the function ‘multiply’ belong to the same data type. This is not just a mere coincidence but both the return type of the function and the identifier being returned in the ‘return’ statement always have to be of the same type as they tell the compiler what type of data is being returned by the function. If a function does not return any type of data then we do not need a return statement and we use the keyword ‘void’ before the function name which means null or empty. int multiply ( argument1, argument2, ...) { Body of the function } int multiply ( int integer1, int integer2) { Body of the function } FUNCTIONS
- 133. 118 4.2.2 Function prototype To create and use a function in another function, it becomes necessary that the function being called must be declared before the function in which it is called. Function prototypes allow us to overcome this problem. . The function prototype is a statement, which means it ends with a semicolon. It consists of the function's ‘return type’, ‘name’, and the ‘parameter list’. In syntax, they are very similar to a function header. The function prototype tells the compiler that the function is defined somewhere later in the program and when this function is called the compiler shifts control to that part of the program. For example, the function header for function ‘multiply’ in example 4.2.1.1 would look like this. int multiply( int integer1, int integer2); As already mentioned, the function prototype tells us about the ‘return type: int’, ‘name: multiply’ and the ‘list of parameters and their return types: ‘int integer1’, ‘int integer2’’.As a formatting style all function headers are declared after the preprocessor directives, constants and global variables. Many of the built-in functions, we use in C++ have function prototypes which are already defined in files which we include in our programs by using ‘#include’ preprocessor directive. For functions that we create on our own, we must include function prototypes. 4.3 Accessing a function A function is called (or invoked, or accessed) by providing the function name, followed by its parameters. The parameters are enclosed in parentheses. Program 5.2 declares and uses function ‘addition’ to calculate the sum of the parameters passed to it as parameters. Program 4.2: Accessing a function // Header Files #include <iostream.h> # include<conio.h> int addition (int a, int b) { int add; add=a+b; return (add); } void main () { int sum; sum = addition (1,2); cout << " The sum of the two numbers is " << sum; The sum of the two numbers is 3
- 134. 119Part I: Chapter 4 Functions getch(); } Let us study how this program works. The program execution begins from the main function by declaring the variable ‘sum’ of type integer which is followed by the function call to ‘addition’ using the following statement. sum = addition (1,2); This statement calls the function addition(), the values ‘1’ and ‘2’ present in this function call are known as ‘parameters’ and these parameters have a clear correspondence to the variables ‘a’ and ‘b’ present in the function declaration. At this point the control of the program is passed from function main to function ‘addition’. The values ‘1’ and ‘2’ passed with the function call are assigned to the corresponding variables in the function definition. As a result, the value ‘1’ gets assigned to variable ‘a’ and the value ‘2’ gets assigned to variable ‘b’. Both these variables ‘a’ and ‘b’ are termed as local variables as they are defined and can be used only in function ‘addition’. In this function, we have declared a variable ‘add’ which stores the sum of the two parameters passed to it. This variable ‘add’ is then returned by the function. In this case the function will return the value ‘3’ to the main function which is also known as the return value of the function. The return statement tells the compiler that the role of the function is over as a result of which the control passes back to function ‘main’ after which the compiler replaces the function call by the value returned by the function. In other words, the statement ‘sum = addition (1, 2);’ is simplified to ‘sum = 3’ as the function addition returns the integer value ‘3’. The important thing that one needs to notice is that once the control has been passed back to the function main and the execution begins from the same point from where the control was passed to the function. FUNCTIONS
- 135. 120 A function can be accessed or called as many times as the user wants. As it might be obvious the result of the function depends upon the parameters passed to it. Program 4.3 calls the function ‘product’ multiple times with different parameters every time. Program 4.3: Multiple calls to function ‘product()’ // Header Files # include<iostream.h> # include<conio.h> void product( int c, int d) { int product_operands=0; product_operands= c * d; cout<<"n The product of operands is "<<product_operands; } // Main Function void main() { clrscr(); int a=3,b=4; product(a,b); product(5,6); product(a+3,b+2); getch(); } COMMON PROGRAMMING ERRORS A function can only have one return type. Using more than one data type will result in a compilation error. A function declaration can skip the argument names but a function definition cannot. A function prototype always precedes the function definition. When the function does not return anything then the return type is void. 4.4 Default Arguments When declaring variables, C++ also allows us to assign default values for arguments. These parameters are used if this particular parameter is left blank when the function is called. Although if a value is passed with the function call then the parameters default value is
- 136. 121Part I: Chapter 4 Functions overwritten by the value passed in function call. Program 4.4 illustrates how we use default arguments in C++. Program 4.4: Default arguments in C++ # include<iostream.h> # include<conio.h> int product (int x, int y=3) { int multiply; multiply = x* y; return multiply; } void main() { clrscr(); // First call to function ‘product’ cout<<product(10)<<endl; // Second call to function ‘product’ cout<<product(20,12); } 30 240 In this program, function ‘product’ has been called twice by function ‘main’. In the first call to function ‘product’, we have only specified one argument ‘10’, so the argument ‘x’ in this function obtains the value ‘3’ where as argument ‘y’ takes its default value ‘3’. As a result of which we obtain the result ‘30’ on the screen. In the second call to the function ‘product’ two parameters have been passed to the function as a result of which the parameters value ‘13’ overwrites the default value of arguments ‘y’ resulting in returning a value ‘240’ from the function. 4.5 Constant Arguments Just like we can declare constant variables, C++ also allows us to declare constant arguments which cannot be modified in the function. For example, In this declaration the argument ‘year’ is declared as a constant variable. The keyword const preceding this variable is an access specifier and it tells the compiler that the function should not modify the argument. float example (int day, int month, const int year = 2005); FUNCTIONS
- 137. 122 4.6 Arguments as Reference Parameters So far, we have studied only to pass parameters to functions by their value. Let us take up an example, Now in this call to the function ‘add’ parameters ‘m’ and ‘n’ are not passed to arguments ‘a’ and ‘b’. Only values stored in these variables ‘7’ and ‘10’ are passed to arguments ‘a’ and ‘b’. As illustrated in the figure 4.1 given below. Figure 4.1: Passed by value Any changes made to these variables ‘a’ and ‘b’ in function ‘add’ are not reflected back to variables ‘m’ and ‘n’ declared in the main function. In some cases it might be useful if changes made to arguments could be reflected back to the parameters passed in the function call. C++ allows this by declaring arguments as reference parameters. To understand how this is done look at the following program: Program 4.5: Declaring arguments as reference parameters // Header Files # include<iostream.h> # include<conio.h>
- 138. 123Part I: Chapter 4 Functions void reference( int &a, int b) { a=a+10; b=b-5; } void main() { clrscr(); int operand1, operand2; cout<<" Enter the two operands "; cin>>operand1>>operand2; reference(operand1, operand2); cout<<" Operand1:"<<operand1<< "t Operand2:"<<operand2; getch(); } In this program the function definition for ‘reference’ might seem a bit peculiar. Here we have declared argument ‘a’ as a reference parameter this is done by preceding an ampersand ‘&’ sign before the argument identifier. As a result of this, when function ‘reference’ is called from function ‘main’ then the parameter ‘operand1’ will be passed to it by ‘reference’ where as ‘operand2’ will be passed to the function by value. The difference between them might become more evident from the figure 4.2 given below. Figure 4.2: Arguments as reference parameters FUNCTIONS
- 139. 124 As parameter ‘b’ is passed by value to function ‘reference’, ‘operand2’ is assigned the value ‘5’ but as ‘operand1’ has been declared as a reference argument, so variable ‘a’ is passed to ‘operand1’ and not just the value stored in variable or parameter ‘a’. During this function call variable ‘a’ is also known as the actual parameter and ‘operand1’ is also referred to as the formal parameter. As any changes made to the formal parameter ‘operand1’ are also reflected back to the actual parameter ‘a’ passed to it in the function call. So when the following operations take place in function ‘reference’ then variables ‘a’ and ‘operand1’ get effected in the following manner (see figure 4.3). Figure 4.3: Arguments as reference parameters 4.6.1 Solved Examples Example 4.6.1.1 Determine the output for the following piece of code: # include<iostream.h> # include<conio.h> void function1(int &x, int &y) { x++; y--; } void function2(int &c, int &d, int e=0) { c = c + e;
- 140. 125Part I: Chapter 4 Functions d = d - 2; e = e + c; } void main() { clrscr(); int var1=2, var2=5, var3=12; function1(var1, var2); function2(var1, var3); cout<<"n Var1:"<<var1<<"t Var2:"<<var2<<"Var3:"<<var3; getch(); } Solution of 4.6.1.1 Output Screen Programming Tips When passing the parameter as reference, the parameter is passed to the argument and any changes then made to this argument are reflected back to the passed parameter also known as the actual variable. Where as when passing parameters by value only the value stored in the variable is passed to the argument and changes made to this argument are not reflected back to the actual variable or the actual parameter. Parameters are passed to arguments exactly in the order as specified in the call to the function. Changing the order will result in a different output. When declaring arguments in the function header, default arguments are always declared at the end of the argument list. The following declaration will result in a compilation error: FUNCTIONS
- 141. 126 void xyz( int x = 2, int y, int z) { …………….. } Calling a function before its function prototype or definition is declared will result in a compilation error. 4.7 Inline functions When a function is declared inline, the function is expanded at the calling block. The inline specifier forces the compiler to substitute the body of code where function invocation occurs. As a result of which the control of a program is not shifted to the function being called. Inline functions are used to increase program’s speed and performance. They are declared using the following syntax: Syntax: inline return_type function_name (list of arguments) { function_body } For example, # include<iostream.h> # include<conio.h> inline void inline_functions() { cout<<" Inline functions help in increasing programs performance"; } void main() { inline_functions(); }
- 142. 127Part I: Chapter 4 Functions 4.8 Scope of variables ‘Scope’ of variables refers to the visibility of variables. In other words, it tells us where can we use or access variables. For example, the variables declared within a function or any other closed block of instructions can only be used or accessed in that very function in other words their scope lies only within the function they are declared in. Let us take up a program to explain this concept. Program 4.6: Scope of variables # include <iostream.h> # include <conio.h> int multiply (int op1, int op2) { int product=0; product = op1 * op2; return product; } void main() { int a, b; cout<<" Enter two numbers"; cin>>a>>b; cout<<" The product is "<<multiply(a,b); getch(); } Now if in this program we use variable ‘a’ or ‘b’ in function ‘multiply’ it will give a compilation error. This is so because these variables ‘a’ and ‘b’ are local variables and as they have been declared in function main their scope lies only in this function. Similarly, ‘op1’, ‘op2’ and ‘product’ of type integer are also local variables and their scope lies only within function ‘multiply’. C++ also allows us to declare ‘global variables’. A global variable is a variable that does not belong to any subroutine in particular and can therefore can be accessed from any context in a program. They are declared using the same syntax the only difference being that they have to be declared outside any function or block of instructions. Program 4.7 illustrates how global variables are declared and used in C++. Program 4.7: Global variables /* To illustrate how to declare and use global variables */ # include <iostream.h> # include <conio.h> FUNCTIONS
- 143. 128 // Declration of a global variables 'PI' and 'radius' float PI = 3.14; int radius; float area() { float evaluate_area=0; cout<<"n Enter the radius of the circle"; cin>>radius; evaluate_area= PI * (radius) * (radius); return evaluate_area; } float circumference() { float evaluate_circumference=0; cout<<"n Enter the radius of the circle"; cin>>radius; evaluate_circumference= PI * 2 * radius; return evaluate_circumference; } void main() { clrscr(); int ch; do { cout<<"n 1. Calculate the area of the circle"; cout<<"n 2. Calculate the circumference of the circle"; cout<<"n 3. Exit the program"; cout<<"n ! Enter choice"; cin>>ch; switch(ch) { case 1: cout<<" n The area of the circle is: "<<area()<<" square units n"; break; case 2: cout<<" n The circumference of the circle is: "<<circumference()<<" units n"; break; } } while (ch!=3); getch(); }
- 144. 129Part I: Chapter 4 Functions In this program, we have used ‘PI’ and ‘radius’ as global variables and used them in functions ‘area()’ and ‘circumference()’. You may observe that declaring these variables as global variables not only saves us the trouble to separately declare these variables in these functions but also saves the computers memory space. Although global variables do allow or increase the chances of accidental changes of data a concept that you will learn in later chapters. 4.8.1 Solved Examples Example 4.8.1.1 Determine the output of the following piece of code: // Header Files # include<iostream.h> # include<conio.h> void function1(int a, int b) { a++; b = b + 5; cout<<"n Function1 a:"<<a<<"t Function1 b:"<<b; } void function2(int a, int c) { a--; c=c+5; cout<<"n Function2 a:"<<a<<"t Function2 c:"<<c; } // Main Function void main() { clrscr(); int a=7, b=0,c=2; function1(a,b); function2(a,c); cout<<"n Main a: "<<a<<"t Main b: "<<b<<"t Main c:"<<c; } FUNCTIONS
- 145. 130 Solution of 4.8.1.1 We would observe the following output when this piece of code is executed. We start by declaring a set of variables ‘a = 7’, ‘b =0’ and ‘c=2’ of type integer in function ‘main’. Then, we pass the values stored in variables ‘a’ and ‘b’ to ‘function1’. The important thing to remember here is that just because ‘function1’ uses the same named variables ‘a’ and ‘b’ it does not mean that they point to the same memory blocks. The scope of variables ‘a’ and ‘b’ in function ‘main’ is limited to that function and hence they are not accessible to ‘function1’. When ‘function1’ is called by the ‘main’ function, the compiler again declares variables ‘a’ and ‘b’ assigning them values ‘7’ and ‘0’ but pointing to different memory blocks as shown in figure 4.4. Figure 4.4: Example 4.7.1.1 As a result of this, any changes made to variables ‘a’ and ‘b’ of ‘function1’ do not alter values stored in variables ‘a’ and ‘b’ with memory locations ‘100’ and ‘102’ of function ‘main’ and we obtain the first line in the output screen as . This is also illustrated in the figure given below. Using the same logic now try to find the rest of your output on your own.
- 146. 131Part I: Chapter 4 Functions Figure 4.5: Example 4.7.1.1 4.9 Character functions in C++ The table 4.2 summarizes some character functions. To use these, you need to include the header file <ctype.h> in your programs. Table 4.2: Character functions in C++ Function Description Example isalnum() Returns a non-zero if the argument is a letter or a digit. char ch; cin>>ch; if (isalnum(ch)) cout<<”Alphanumeric”; isalpha() Returns a non-zero if the argument is an alphabet. char ch; cin>>ch; if (isalpha(ch)) cout<<”Alphabet”; isdigit() Returns a non-zero if the argument is a digit. char ch; cin>>ch; if (isdigit(ch)) cout<<”Digit”; islower() Returns a non-zero if the argument is a lowercase alphabet. char ch; cin>>ch; if (islower(ch)) FUNCTIONS
- 147. 132 cout<<”Lower case Alphabet”; isupper() Returns a non-zero if the argument is a uppercase alphabet. char ch; cin>>ch; if (isupper(ch)) cout<<”Upper case Alphabet”; tolower() Returns the lowercase equivalent of the argument. char ch; cin>>ch; switch(tolower(ch)) { case ‘a’: case ‘e’: case ‘i’: case ‘o’: case ‘u’: cout<<”Vowel”; break; default; cout<<”Consonant”; } toupper() Returns the upper case equivalent of the argument. char ch; cin>>ch; if(toupper(ch)==’A’) cout<<”You entered A” 4.10 Mathematical functions in C++ Table 4.3 summarizes the mathematical functions available in C++. They are defined under the header file <math.h>. Table 4.3: Mathematical functions in C++ Function description Example fabs(x) Calculates the absolute value of floating point number x. float a=-98.987 cout<<fabs(a); returns 98.987
- 148. 133Part I: Chapter 4 Functions abs(x) Calculates the absolute value of an integer x. int a=-76; cout<<abs(a); returns 76 log(y) Calculate natural algorithm of y. cout<<log(2); returns 0.693147 log10(y) Calculates natural logarithm of y. cout<<log(2); returns 0.30103 pow(x,y) Calculates the power x to y cout<<pow(3,2); returns 9 4.11 Recursive functions Recursion is a technique where a function calls itself. It is generally used to implement a recursive algorithm. It solves a problem by representing it as an instance of the same problem with a smaller input size in each iteration of the function. Each recursive function should have the following properties: Base case: Must have at least one case which can be solved without recursion. Making progress: At each step there must be some progress toward the base case so that the base case condition is eventually met. Design rule: during the design of the recursive algorithm, assume that all simpler recursive calls work. For gaining in depth knowledge on recursive functions please see Appendix F (pp. 589-595). 4.12 Review Examples Example 4.12.1 Write a program to take two numbers say, ‘x’ and ‘y’ from the user and add these numbers using a function and then display the sum of these two numbers on the screen. FUNCTIONS
- 149. 134 Solution of 4.12.1 // Header Files # include<iostream.h> # include<conio.h> // Function prototype int add(int x, int y); // Main function void main() { int a,b; cout<<" Enter two variables"; cin>>a>>b; cout<<" Sum:"<<add (a,b); getch(); } /***************************** Name:add Arguments: 'x' and 'y' Purpose: Calculates the sum of two integers ********************/ int add(int x,int y) { int sum=0; sum = x + y; return sum; } Example 4.12.2 Write a program which declares a function ‘int accept(int a, int b, int c);’ which accepts three parameters ‘a’, ‘b’ and ‘c’ inputted by the user and returns the value obtained by computing value of the expression ‘a * b +c’ to the main program which displays the returned value. Solution of 4.12.2 // Header Files # include<iostream.h> # include<conio.h> // Function accept int accept(int a, int b, int c)
- 150. 135Part I: Chapter 4 Functions { return (a*b+c); } // main function void main() { int x, y, z; cout<<" Enter the three integers"; cin>>x>>y>>z; cout<<" Result after evaluation: "<<accept(x,y,z); getch(); } Example 4.12.3 Write a program to find the roots of a quadratic equation. Assume that the quadratic equation only has unequal or equal roots. The program should use separate set of functions to determine whether the equation has equal or unequal roots. Solution of 4.12.3 // Header Files # include <iostream.h> # include <conio.h> # include <math.h> // Function Prototypes void exitprogram(); void equalroots(float, float); void unequalroots(float, float, float); // Main Function void main() { /******** Variable Declrations Quadratic Equation: ax^2 + bx + c Discriminant:D ********/ float a, b, c, D; cout<<"Enter the Coefficients of the quadratic equation:"; cin>>a>>b>>c; FUNCTIONS
- 151. 136 /************* Calculating D= b^2 - 4ac case 1: if D = 0 the equation has real and equal roots case 2: if D > 0 the equation has real and unequal roots ********/ D = (b*b) - 4*a*c; if (D<0) { cout<<"n The quadratic equation has unreal roots."; exitprogram(); } else if (D>0) { unequalroots(a,b,D); exitprogram(); } else if(D==0) { equalroots(a,b); exitprogram(); } } /******** Function: exitprogram Purpose: To exit the program *****************************/ void exitprogram() { cout<<"n Thank you... For using the program"; } /******** Function: equalroots Purpose: To determine the unique equal root
- 152. 137Part I: Chapter 4 Functions ********************************************/ void equalroots(float a, float b) { float root; root = (-b/(2*a)); cout<<"n The quadratic equation has equal roots:"; cout<<"n The only root is "<<root<<" with multiplicity 2"; return; } /******** Function: unequalroots Purpose: To determine the unequal roots ********************************************/ void unequalroots(float a, float b, float D) { float root1,root2; root1 = (-b + sqrt(D))/(2*a); root2 = (-b - sqrt(D))/(2*a); cout<<" The quadratic equation has unequal roots: "<<root1<<" and "<<root2; return; } Example 4.12.4 Write a program to display the following type of pattern using functions (The value of n entered by the user)? * * * * * * * * * * * * * * * Solution of 4.12.4 // Header Files # include <iostream.h> # include <conio.h> FUNCTIONS
- 153. 138 // Function Prototype void pattern ( int g); // Main function void main() { int a; cout<<" Enter for how many do the numbers have to be entered "; cin>>a; pattern (a); } /************************************************************** // Function Name: Pattern // Parameters: g (type:int) // Purpose: To create the required pattern **************************************************************/ void pattern (int g) { int t=g; for (int b=1;b<=g;b++) { for (int d = 0 ; d<t;d++) cout<<" "; for (int c=1;c<=b;c++) cout<<"*"; cout<<"n"; t=t-1; } getch(); } 4.13 Review Exercise 2/3 Mark questions Q1. What is functions single most important role? Q2. What do you understand by the term ‘function argument’? Q3. Can preprocessor directives be passed as ‘parameters’ as well? Q4. What is the difference between arguments passed by value or passed as reference?
- 154. 139Part I: Chapter 4 Functions Q5. How many values can be returned by a single function? Q6. What does the function header tell us? Q7. Overloaded functions are i. same named functions ii. functions having the same return type. iii. functions performing the same process. iv. examples of data redundancy. Q8. A ‘function calling statement’ must supply arguments that i. match the function header in number ii. match the function header in number and order iii. match the function header in number, order, and data type iv. match the function header in number, order, data type, and names v. match the function header in number, names, and data type Q9. How does a function report its answer back to its calling (or boss) function? i. by altering or storing the result into one of the variables passed as an argument ii. by altering or storing the result directly into the caller’s variable iii. by cout-ing the result to the screen iv. by asking the user via an input function v. by executing a return statement with the answer. Q10. Which of the following appear in a function header? i. the function name, return type, and argument types ii. the function name, return type, argument types and argument names iii. the function name, return type, and argument names iv. the function name and the supplied arguments Q11. What is a default argument? Q12. Write the main function to call the following function: int sum(int a, int b=2) { return ( a + b); } Q13. Assuming that the main program has already been compiled successfully will the following function also compile: FUNCTIONS
- 155. 140 float sum ( int b=2, int a) { return ( a+b); } 4/5 Mark questions Q14. Determine the values of y after each of the following statements a) y = isalnum(‘a’); b) y = isalpha(‘b’); c) y= fabs(99.12); Q15. For each of the following set of integers, write separate functions to find the sum of these series where n (number of terms to be included) is passed as a parameter. a) 2..4..6..8..10…(2n). b) 3..5..7..9..11...(2n+1). c) 6..10..14..18…(2n+4). Q16. Write a function that displays at the left side of the screen a solid square of dollar signs ($) where the side length (s) is passed as a parameter by the user. Q17. Write a function which returns the greatest of the three floating point numbers passed to it as parameters. Q18. Write a function that takes an integer as a parameter and returns 1 if the number with the digits reversed is same as the initial number passed to it as a parameter. (Palindrome Number), For example: 121 is such a number. Q19. Identify the errors in the following program. a) int main ( void ) { int sum( void ) { cout<<” The sum is “>>a+b; } }
- 156. 141Part I: Chapter 4 Functions b) int difference ( int x, int y) { int diffrential; diffrential = x – y; } c) void remainder( float y) { float result; result = y % 4; } Q20. Write the function headers for each of the following: a) Function ‘area’ that takes two floating point variables viz. height and side and returns the calculated area as a float parameter. b) Function ‘even’ which takes in an integer type and returns the value 1 if the number is even or 0 if the number is odd. c) Function ‘smallest’ which takes in 5 parameters and returns the smallest parameter divisible by 3. 4.14 Programming Project In the 12th century, Leonardo Fibonacci discovered a simple numerical series that is the foundation for an incredible mathematical relationship behind phi. Starting with 0 and 1, each new number in the series is simply the sum of the two before it. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 . . . The programming project is to create a program which declares a function after the man who discovered the series to print the terms of the Fibonacci series on the screen, where the number of terms to be displayed is passed as a parameter by the user. FUNCTIONS
- 157. 142 4.15 Let us revise! Function is a named unit of group of program statements. Function call is the statement invoking the function. Function definition is the function declaration plus the body of the function. Actual parameters are the variables passed to a function. Formal parameters are the variables which receive the incoming parameters in the function. Global variable is a variable available to all the functions in the program. Local variable is a variable whose scope lies within the function in which it is declared. Functions help in data organization, increase program efficiency and reduce data redundancy. The general format for a function definition is: return_type function_name ( parameters…) { function definition } Arguments can be passed by value where the functions make a copy of the passed argument and changes made to this copied formal parameters do not affect the actual parameters. A function prototype declares the return type of the function and declares the number, type and the order of parameters. The function which returns no value is a void function. A function can only return a single value. C++ allows function names to be overloaded, i.e. the same identifier name can be used in the same scope of both functions. The compiler decides the function to be used based on the number and types of parameters that are used while invoking the function. Function overloading is an example of ‘polymo- rphism’ a concept is explained in Part II, ‘Chapter 1’ of this book (Object Oriented Programming).
- 158. Part I CHAPTER 5 ARRAYS AIM OUTLINE • Introduce the concept of ‘Arrays’ • To understand the use of arrays to store, sort, and search lists of data in One/Multi dimensional arrays and strings. • To be able to pass One/Multi dimension arrays to functions. • To understand basic sorting techniques and memory allocation of arrays. 5.1 Introduction 5.2 Declaring Arrays 5.3 Passing arrays as parameters to functions 5.4 Traversing of Arrays 5.5 Searching element in an array 5.6 Merging of arrays 5.7 Sorting with arrays 5.8 Arrays as strings 5.9 Two-dimensional arrays 5.10 Solved Examples 5.11 Review Exercise 5.12 Programming Project 5.13 Let us revise!
- 159. 144 5.1 Introduction Till now, we have studied on writing, printing and inputting single values. Each data type defined so far is stored in a single unique location and referred to as a single variable having its own name like ‘sum’, ‘count’, etc. Suppose, we have to make a program to store the names of ten items, we could start creating variables ‘item1’, ‘item2’, and ‘item3’ and so on but what if we have to create a program to store names for say 100 or say a 1000 items then what? We could use the same approach but then the resulting program will be too long and cumbersome. Now suppose that instead of treating them as 100 similar but separate data items we could set aside a space large enough for storing all 100 values and assign a name to the storage area. Data items stored in this way are termed as ‘arrays’. 5.2 Declaring Arrays Arrays are data structures which hold multiple variables of the same data type. An array is a sequence of objects all of which have the same type. The objects are called elements of the array and are numbered consecutively 0, 1, 2, and so on. These numbers are called the index values or subscripts of the array (see Fig 5.1). The term ‘subscript’ is used because as a mathematical sequence, an array would be written with subscripts n0, n1, n2, and so on. These numbers locate the elements position within the array, thereby giving direct access into the array. Figure 5.1: Array description The elements of a one-dimensional array store all the elements in a consecutive order of entry into an array name. The subscripts of array are increased sequentially one-by-one to input the values into the array. A one-dimensional array is a list of variables that are all of the same type and referenced through a common name. An individual variable in the array is called an array element. Arrays form a convenient way to handle groups of related data. Let’s take an example to further understand the concept of arrays. Suppose you want to store 5 different integer values into the memory, and then find the sum of the values. Then you would use the following piece of code:
- 160. 145Part I: Chapter 5 Arrays void main() { clrscr(); int num1, num2, num3, num4, num5; int sum=0; cout<<” Enter the five values”; sum=num1+ num2+ num3+ num4+ num5; cout<<” The sum of the five values “<<sum; } The above piece of code declares five different locations in the computer memory for processing five integer values. By using arrays, we can allocate these five memory locations by a single variable name called array with a subscript value ‘5’. A subscript is a number, within brackets, that differentiates one element of an array from another. Elements are the individual variables in an array. To declare an array the following syntax is used: Syntax: Where ‘data_type’ is any C++ data type, ‘array_name’ is the name of the array and ‘subscripted range’ is the length of the array. For example, int employees[5]; This statement declares an array ‘employees’ of type integer and it can store ‘5’ integer values. Fig 5.2 shows how this array ‘employees’ is represented in the computer’s memory. data_type array_name[subscripted range]; ARRAYS
- 161. 146 Figure 5.2: Array 'employees' in the computer’s memory The indexing of an array always starts from ‘zero till ‘size of the array -1’. The array elements are always stored in contagious memory locations (‘100’,’102’...). Data entries in an array are accessed by using square brackets with an index value. For example, to access the first index of array ‘employees’ we would write ‘employees [0];’. Similarly to access the second index of the array we would write ‘employees [1];’ and so on. 5.2.1 Solved Examples Example 5.2.1.1 Write a program to enter five integer values from the user and print their sum on the screen. Solution of 5.2.1.1 // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { // Declaring an array 'num' int num[5]; int i, sum=0; // Inputting the values in the array for ( i=0; i<5;i++ ) { cout<<" Enter value "<<i<<"->"; cin>>num[i]; } // Calculating the sum of the values for ( i=0; i<5;i++) { sum = sum + num[i]; }
- 162. 147Part I: Chapter 5 Arrays cout<<" The sum of all the five elements is"<<sum; } Output Screen Let us try to understand how this program actually works. We begin by declaring an array by the statement: ‘int num[5];’. This tells the compiler to set aside the memory space for 5 integer variables grouped together under the name ‘num’. As one might notice in the figure given below these ‘5’ integer variables which are grouped together are stored in contagious memory locations and can be accessed by changing the subscripts assigned with the array. Figure 5.3: Array declaration Once this is done next comes entering elements in our array ‘num’. In example 5.2.1.1 this is done using the following piece of code. // Inputting the values in the array for ( i=0; i<5;i++ ) { cout<<” Enter value “<<i<<”->”; Enter value 0 -> 100 Enter value 1 -> 50 Enter value 2 -> 400 Enter value 3 -> 300 Enter value 4 -> 250 The sum of all the five elements is 1100 ARRAYS
- 163. 148 cin>>num[i]; } The index variable ‘i’ takes values from ‘0’ to ‘4’ within the ‘for’ loop which are used to enter the array elements of array ‘num’. Let us assume that the user enters the values ‘100’, ‘50’, ‘400’, ‘300’ and ‘250’. These values are then assigned to num[0], num[1], … num[4], i.e., in the order they were entered by the user (see fig. 5.4) which are finally used to calculate the sum of the array elements displayed on the screen. Figure 5.4: Array 'num’ Example 5.2.1.2 Suppose that in your computer science class, each student secures a certain percentage of marks and the teacher decides to give grace of 10 % to each student. If the class contains 20 students, write a C++ program to do the needful? Solution of 5.2.1.2 // Header Files # include <iostream.h> # include <conio.h> // Main Function void main() { // Variable Declrations float per[20]; int i; cout<<" Enter the students percentages for the 20 students";
- 164. 149Part I: Chapter 5 Arrays // Entering student information from the student for( i=0;i<20;i++) { cin>>per[i]; } // Display the increased percentage of the 20 students for (i=0;i<20;i++) { per[i] = per[i] + 10; cout<<per[i]<<endl; } getch(); } 5.3 Passing arrays as parameters to functions Arrays are passed to functions as reference parameters. To pass an array argument to a function the array name is specified without any brackets. The syntax for passing arrays to functions is given below: The subscripts of all the arrays should always begin with 0. For example in the array int a [5], the subscripts should begin with 0 as explained above. When using loops to initialize values into the array one should make sure that the index variable always contains a positive value. If the number of values assigned exceeds the size of the array or the program tries to access an index outside the bounds of the array such a statement is an example of a logical error. As the concurrent C++ compilers are very user friendly the compiler may not give an error but it will although produce unexpected results. For example, int A[3]={9,5,2,3} is an incorrect statement as we are assigning an integer value ‘3’ to unallocated set of memory Common Programming Errors ARRAYS
- 165. 150 Syntax: return_type function_name( data_type array_name) For example, if array ‘weeklyupdate’ is defined as: int weeklyupdate[10]; then the function call for a function ‘modifyarray’ which takes in an array as a parameter would look like ‘modifyarray(weeklyupdate);’. Let us take up a program to further understand the concept of passing arrays to a function. Program 5.1 // Header Files # include<iostream.h> # include<conio.h> void calculate( int arr[5]) { for (int i=0 ; i<5 ;i++) { arr[i] = arr[i] + 10; } } void main() { clrscr(); int array_example[5]; cout<<" Enter the elements of the array"; for (int a=0; a<5; a++) { cin>>array_example[a]; } // Modifying array calculate(array_example); cout<<" The modified array is as followsn";
- 166. 151Part I: Chapter 5 Arrays for (a=0; a<5;a++) { cout<<array_example[a]<<"n"; } getch(); } Now in this program, we have passed the array ‘array_example’ as a parameter to function ‘calculate’ which increments each array entry by 10 using the concept that arrays are passed to functions as reference parameters. That is the array ‘arr’ defined in the function call for functions ‘calculate’ acts as a formal parameter to ‘array_example’. Hence any changes made to this formal parameter are also reflected back to the actual parameter ‘array_example’. The output of this program is given below (Fig. 5.5): Output Screen Figure 5.5: Output Screen 5.4 Traversing of Arrays It is an operation in which each element of an array (i.e. from first element of an array to the last element of the array) is visited. Processing of the visited element can be done according to the requirements of the problem. Program 5.2 shows how traversal of arrays takes place. Program 5.2: Traversal in an array // Preprocessor Directives # include <iostream.h> # include <conio.h> void main() ARRAYS
- 167. 152 { clrscr(); int a[15],i,n; do { cout<<” Enter the number of elements in array less than 16”; cin>>n; } while(n>15); // Traversing // Reading the array for(i=0;i<n;i++) { cout<<” Enter “<<i+1<<” element : “; cin>>a[i]; // Displaying original array cout<<” n Original Array “; for (i=0;i<n;i++) cout<<a[i]<<” “; // Traversing (a[i]=a[i]*10) for(i=0;i<n;i++) a[i]=a[i]*10; // Displaying the transformed array cout<<” n The transformed array :- n“; for (i=0;i<n;i++) cout<<a[i]<<” “; getch(); } 5.5 Searching element in an array The process of finding a particular element in an array is called as searching. This process can be carried out in two ways. 5.5.1 Linear Search In linear search, each element of the array is compared with a search ‘key’. The process of carrying out linear search is given below in program 5.3.
- 168. 153Part I: Chapter 5 Arrays Program 5.3: Linear Search // Header Files # include<iostream.h> # include<conio.h> // Function Prototypes void display (int [], int); void input (int [], int &); int lsearch (int a[],int n, int data); // Main Function void main() { // Variable and array declrations int a[10], n, data; input (a, n); cout<<" n The Array ->"<<endl; display (a,n); cout<<" n Enter element to be searched for: "; cin>>data; int x= lsearch(a,n,data); if (x!=-1) { cout<<data<<" Present at location "<<x<<endl; } else { cout<<" Not present"; } getch(); } /***************** Function:lsearch Parameters: a[], n, data Purpose: To search for the desired data element entered by the user in the array ARRAYS
- 169. 154 ************************************************/ int lsearch (int a[],int n,int data) { int c=0,found=0; while ((c<n) && !(found)) { if (a[c]==data) { found++; } else c++; } if (found) { return (c+1); } else { return (-1); } } /***************** Function:input Parameters: a[], n Purpose: To enter the elements of the array *********************************************/ void input (int a[],int &n) { int i; do { cout<<" Enter the number of elements <= 10:"; cin>>n; } for(i=0;i<n;i++) { cin>>a[i]; } }
- 170. 155Part I: Chapter 5 Arrays /***************** Function:display Parameters: a[], n Purpose: To display the elements of the array **********************************************/ void display (int a[], int n) { int i; cout<<" The elements of the array are: n"; for (i=0;i<n;i++) { cout<<a[i]<<endl; } } 5.5.2 Binary search To understand the concept of binary search lets take up the following example. Let’s say one is looking for the name ‘Paul’ in a phone book. Open up the phone book approximately half way and look for the name on the opened page. What does it say? Probably one sees a name that begins from ‘M’. One would then think does ‘Paul’ come before or after ‘M’ in the phone book and realize that it comes in the latter portion of the phone book As a result of this we can ignore the entire first half of the phone book (Part before ‘M’). Now open the remaining portion of the book half way. One is probably some where near the data entries starting with the alphabet ‘T’ in the phone book. We can then again ask our selves the same question does ‘Paul’ come before or after data entries starting from alphabet ‘T’. As the data entry ‘Paul’ comes before data entries starting from alphabet ‘T’, we can ignore the latter half of the phone book and then keep searching in the volume between letters ‘M’ and ‘T’. Repeating this process, we would eventually find the data entry ‘Paul’ we are looking for. This is how most of us search for data in huge volumes like a dictionary or a phone book. This process of searching data through a sorted list is also known as binary search. In brief the process is to search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty. ARRAYS
- 171. 156 The algorithm for binary search is given below: Algorithm: o The element to be searched for is taken from the user say the ‘Key’. o The given data array is taken. Any arbitrary lower limit or the lower bound for the data values is taken. Similarly one defines the upper limit or the upper bound. Then we obtain the middle value using. Middle = (lower bound + upper bound)/2; o The search ‘key’ is compared to the Middle value if ‘Key=Middle’ then the search is complete. If Key<Middle, the program continues and the search is confined to the first half only. On the other hand if Key>Middle then also the program continues and the search is confined to the latter half. This process is continued till the time ‘Key = Middle’. Using binary search in a phone book containing one million entries, one can locate any specific data entry in at most 20 comparisons. The program to carry out binary search is given below: Problem 5.4: Binary Search /************* Name:bsearch Parameters: ‘a’, ‘n’, ‘data’ Purpose: To perform binary search to sorted array ‘a’ *******************************************/ int bsearch ( int a[], int n, int data) { int mid,lbound=0,ubound=n-1,found=0; while ((lbound<=ubound) && !(found)) { mid = (lbound + ubound)/2; if (data > a[mid] ) lbound = mid +1; if (data <a [mid] ) ubound = mid -1; else found++; } if (found) return (mid + 1); // Returning location, if present else return (-1); // Returning -1, if not present
- 172. 157Part I: Chapter 5 Arrays } Common Programming Errors Binary search only works when the array to which it is applied is sorted. 5.6 Merging of arrays This operation merges two sorted arrays into a single sorted array. Suppose ‘a’ is a sorted array with ‘r’ elements and ‘b’ is the sorted array with ‘s’ elements, then the operation that combines the elements of ‘a’ and ‘b’ into a single array ‘c’ is called merging. Program 5.4 provides you with the code to merge two sorted arrays ‘a’ and ‘b’. Program 5.4: To merge two arrays (sorted in ascending order)? // Header Files # include<iostream.h> # include<conio.h> // Main Function void main() { const int r=5; // size of array A const int c=5; // size of array B const int m=10; // size of the merged array A and array B int a[r],b[c],res[m],i,j,k; // Enter Details of the first sorted array cout<<" Enter the first sorted array ==> "<<endl; for(i=0;i<r;i++) { cout<<" Element "<<i+1<<":"; cin>>a[i]; } // Enter the details of the second array cout<<" Enter the second sorted array ===>"<<endl; for(i=0;i<c;i++) { cout<<" Element "<<i+1<<":"; cin>>b[i]; } ARRAYS
- 173. 158 // Initializing all index values to zero i=0, j=0, k=0; // Merging Arrays 'A' and 'B' while ((i<r) && (j<c)) { if(a[i]<b[j]) { res[k++]=a[i++]; } else if (a[i]>b[j]) { res[k++]=b[j++]; } else { res[k++]=a[i++]; j++; } } int t; for (t=i;t<r;t++) { res[k++]=a[t]; } for (t=j;t<c;t++) { res[k++]=b[t]; } // Displaying Merged Array cout<<" Merged array ==>"; for (i=0;i<k;i++) { cout<<endl<<res[i]; } getch(); }
- 174. 159Part I: Chapter 5 Arrays 5.7 Sorting with arrays Sorting is one of the most important operations performed by computers. The history of sorting goes back to the days when magnetic tapes were used for data storage .Before the emergence of modern databases, sorting was certainly the most common operation performed by computers as most database updating was done by sorting transactions. Sorting still plays an important part for presentation of data extracted from databases as most people prefer to get reports sorted into some relevant order before wading through series of data. We will discuss two of the simplest method of sorting ‘bubble’ and ‘selection’ which will give you an idea of how sorting takes place. Bubble Sort Bubble sort is the simplest but unfortunately also the slowest method of sorting arrays. The basic idea when using bubble sort is to compare two neighboring objects and to swap them if they are present in the incorrect order. Given an integer array ‘X’, of size ‘N’ the piece of code to employ bubble sort is given below: // Bubble Sort in Ascending Order for (i=0; i<N; i++) { for (j=0; j<N-1-i;j++) { if (X[j+1]<X[j]) /* Compare two neighboring elements */ { temp = X[j]; /* Swap X[j] and X[j+1] */ X[j] = X[j+1]; X[j+1]=temp; } } } As one might observe the piece of code given above consists of two nested for loops. The inner loop having ‘j’ as the index variable is used to compare adjacent entries in the array ( ‘j’ and ‘j+1’). While the outer loop causes the inner loop to make repeated passes through the array in the process sorting the array. If the data elements are being sorted in the ascending order then after the first pass the largest element is shifted to the last position in the array. Similarly, after the second pass the second largest data element is placed next to the largest element. Which is the reason that in the inner loop the upper ARRAYS
- 175. 160 bound for ‘j’ is stated as ‘N-1-I’ as one does not need to revisit the end of the array. Let us take an example to understand how bubble sort works better. Suppose the array to be sorted ‘X’ consists of the following data entries: Pass 1: Bubble sort begins by comparing the first two data elements in the array. As in this case the data element at 1st index of the array ‘60’ is greater than the data element at the zeroth index ‘21’ no swapping takes place. Once this is done the piece of code then compares the data elements at the first and the second index of the array. As ‘60’ is greater than ‘29’ the two data elements are swapped with the help of a temporary variable ‘temp’. The result of which is that the largest of the three elements is now at the second index of the array. This process is repeated until all the indexes of the array have been compared with their adjacent indexes as a result of which largest data element bubbles to the end of the array. This marks the end of the 1st pass. After which the value of ‘i’ becomes ‘1’ and the upper bound of nested loop with index variable ‘j’ is decreased by 1 (‘ N – 1 – i’) as now we only have to compare the first 4 data elements in the array. Similarly in the second pass through the array the second largest element is placed next to the largest. The process again involves the same steps as observed in the 1st pass the only difference being that the effective array size has now been decreased to ‘4’. This entire process is repeated till the entire array is sorted. Selection Sort Selection sort is also one of the simpler methods of sorting. It is almost twice as faster then bubble sort and works best for smaller files. Selection sort begins by first selecting 21 60 29 52 17 21 29 60 52 17 21 29 52 60 17 21 29 52 17 60
- 176. 161Part I: Chapter 5 Arrays the smallest data element in the unsorted list and then exchanging it with the data element in first position of the array. Once this is done, it then selects the second smallest element and exchanges it with the data element at the second position of the array. This process is carried on till the entire array has been sorted. Given an integer array ‘X’, of size ‘N’ the piece of code to employ selection sort is given below: for (i = 0; i < N-1; i++) { min = i; for (j = i+1; j < N; j++) { if (X[j] < X[min]) min = j; } temp = X[i]; X[i] = X[min]; X[min] = temp; } Figure 5.6 explains the working of selection sort with the help of an example. Array X Pass 1 Pass 2 Pass 3 Pass 4 Figure 5.6: Selection Sort 5.8 Arrays as strings C++ does not have a string data type rather it implements strings as single-dimension character arrays. A string is defined as a character array that is terminated by a null character ‘/0’. For this reason, the character arrays are declared one character longer than the largest string they can hold. This makes room for the null character at the end of the string. Individual characters of a string can be easily accessed as they make the elements of the character array. The index – refers to the first character, the index 1 to the second, 2 to the third, and so forth. The end of a string is determined by checking for a null character. For example, 21 60 29 52 17 17 60 29 52 21 17 21 29 52 60 17 21 29 52 60 17 21 29 52 60 ARRAYS
- 177. 162 A character array can be initialized using a string literal. For example, char myfirststring[] = “Hello”; This statement declares a character string ‘myfirststring’. The size of the string is automatically determined by the compiler based upon the length of the string. In this example, the compiler would determine the size of the string as ‘6’. As this string consists of ‘5’ characters and a special null character (‘0’) included at the end of each string. Strings can also be declared using the following syntax: char myfirststring[] = {‘H’, ‘e’, ‘l’, ‘l’, ‘o’, ‘0’}; When assigning the null character ‘0’ you might think that we are assigning two characters ’’ and ‘0’ but this is not true the ‘’ symbol just tells the compiler that you are dealing with a special character which happens to be the null character ‘0’. As strings also are array of characters, they can also be accessed as any integer array. For example, myfirststring[0] is the character ‘H’ and myfirststring[1] is the character ‘e’. Figure 5.7 given below shows how strings are stored in the computer’s memory. Figure 5.7: 'myfirststring' The functions to display and print the string on the screen are defined under the <stdio.h> header file and are given in table 5.1. Table 5.1: The <stdio.h> header file gets(string_name); The function assigns the string inputted by the user to the parameter passed to it. For example, char string_name[15]; Function Usage
- 178. 5.8.1 String functions The <string.h> header file enlists a number of useful functions for dealing with null terminated strings. A brief description of the functions is given in table 5.2. Table 5.2: The <string.h> header file gets(string_name); Returns the string inputted by the user to the character array ‘string_name’. It works exactly like the ‘cin’ function defined under the <conio.h> header file which is used to enter integers, floats and characters. puts(string_name); The function prints the character array passed to it as a parameter on the screen. It works exactly like the ‘cout’ function which is used to display integers, floating points and characters on the screen. For example, char string_name[]=“This is my first string”; puts(string_name); Output Screen Function Description strlen(string_name); ‘string_name’ is a string The function calculates the length of the string passed to it as a parameter (‘string_name’) by counting the number of characters from the 0th index of the array to the index containing the null character marking the end of the string. ARRAYS
- 179. 164 For example, char string_name[20]; cout<<" Please enter the string: "; gets(string_name); cout<<" The length of the string is:"<<strlen(string_name); Output Screen strcpy(string1,string2) Where ‘string1’ is the destination string and ‘string2’ is the source string. C++ does not allow the programmer to directly assign one string to the other. For example the statement ‘string1 = string2’ will result in a syntax error. Hence for copying or assigning strings to a character array the ‘strcpy’ function is used. The function copies the string stored in ‘string2’ to the first string ‘string1’ which are passed to it as parameters. Let us look at a few examples to understand how this function works: Examples, Output Screen char string1[]="My first string"; char string2[]="My last string"; cout<<strcmp(string1,string2);
- 180. 165Part I: Chapter 5 Arrays Output Screen Strcmp(string1,string2 The function is used to compare two strings passed to it as parameters by the programmer. The function starts comparing from the first character in each string and continues with subsequent characters until the corresponding characters differ or until the end of the strings is reached. Based upon this comparison the function returns the following values. Return Value Comparison ‘0’ It returns a value ‘0’ if both the strings passed to it as parameters (‘string1’, ‘string2) are equivalent. Example, ‘>0’ It returns a value ‘>0’ if ‘string1’ is greater than ‘strings2’. The magnitude of the value returned is the difference between the ASCII values of the first dissimilar characters observed during the char string1[20], string2[20]; cout<<" Enter a string:"; gets(string2); strcpy(string1,string2); cout<<" The entered string is:"; puts(string1); char string1[]="My first string"; char string2[]="My last string"; cout<<strcmp(string1,string2);
- 181. 166 Strcat(string1,strings2); Where ‘string1’ and ‘string2’ are both strings. String comparison For example, The function will return the value ‘6’ as a result of the comparison between the strings passed to it. This is so because the first difference between the two strings occurs at the 4th index. As ‘1’ and ‘f’ are not equivalent the difference in their ASCII values comes out to be ‘+6’. The functions returns a value (<0) if ‘string2’ is greater than ‘string1’. For example, When this piece of code is executed then a value of ‘-6 is observed on the screen instead of ‘+6’ as in the previous example when ‘string1’ was greater than ‘string2’. The functions strcat appends a copy of ‘string2’ at the end of ‘string1’. As a result of which the length of ‘string1’ becomes (strlen(string1) + strlen (string2)). For example, char string1[]="My first string"; char string2[]="My last string"; cout<<strcmp(string1,string2); char string1[]="My first string"; char string2[]="My last string"; cout<<strcmp(string1,string2); Char string1 []=”My first”; char string1[20]; strcpy(string1,"My first string"); puts(string1);
- 182. 167Part I: Chapter 5 Arrays 5.8.2 Solved Examples Example 5.8.2.1 Write a program to check whether the string entered by the user is a palindrome or not? Solution 5.8.2.1 // Preprocessor Directives #include<iostream.h> #include<conio.h> #include<stdio.h> #include<string.h> // Main Function void main() { clrscr(); char string[40],i; int j=0,flag=0; cout<<" Enter string"; gets(string); j=strlen(string)-1; for (i=0;i<strlen(string)-1;i++) { if (string[i] != string[j]) { flag++; } j--; } if (flag==0) Output Screen ARRAYS My first string
- 183. 168 { cout<<" n |Entered string is palindrome|"; } else cout<<" Entered string is not a palindrome"; getch(); } OUTPUT SCREEN Example 5.8.2.2 Write a program to enter a string from the user and then display the inputted string with all the spaces converted into ‘#’s? Solution of 5.8.2.2 // Preprocessor Directives # include<iostream.h> # include<conio.h> # include<string.h> # include<stdio.h> // Main Function void main() { clrscr(); char string[15]; cout<<" Enter the string"; gets(string); for (int i=0;i<strlen(string);i++) { if (string[i]==' ') { string[i]='#'; } } cout<<" n Displaying the converted string n"; puts(string); Enter the string MaM |Entered string is a palindrome|
- 184. 169Part I: Chapter 5 Arrays getch(); } Output Screen Example 5.8.2.3 Write a program to count the no. of vowels in the given string? Solution of 5.8.2.3 // Header Files # include<iostream.h> # include<conio.h> # include<string.h> # include<stdio.h> void main() { clrscr(); char string[15]; int vowels=0; cout<<" Enter the string"; gets(string); for (int i=0;i<strlen(string);i++) { if (string[i]=='a' || string[i]=='e' || string[i]=='o' ||string[i]=='i' || string[i]=='u') { vowels++; } } cout<<" n The no. of vowels are"<<vowels; getch(); } Enter the string My sisters name is Deepti Displaying the converted string My#sisters#name#is#Deepti ARRAYS
- 185. 170 OUTPUT SCREEN 5.9 Two-dimensional arrays A two-dimensional array is an array in which each element is itself an array. For instance, an array A[M][N] is an M by N table with M rows and N columns containing M * N elements figure 5.8. Figure 5.8: Two Dimensional Arrays The number of elements in a 2-D array can be determined by multiplying number of rows with number of columns. For example, the number of elements in an array A[5][3] is calculated as 15. This two-dimensional array may also be visualized as a one-dimensional array of arrays. An alternative view of this array is given in the figure 5.9. Enter the string Ankit The no. of vowels are 2
- 186. 171Part I: Chapter 5 Arrays Figure 5.9: Two dimensional Arrays The simplest form of a multi-dimensional array, the two-dimensional array, is an array having single-dimension arrays as its elements. The general form of a two-dimensional array declaration in C++ as follows: Type array-name[rows][columns] Where type is the base data type of the array having name array-name; rows, the first index, refers to the number of rows in the array and columns, the second index, refers to the number of columns in the array. The following declaration declares a character array ‘name’ having ‘5’ rows and ‘25’ columns. Just like one dimensional arrays two dimensional arrays can also be initialized at the time of declaration. The following declaration assigns values for three grades for each of the first two students in the array. int grades[Student_Number][Grades_achieved] = { {55, 60, 65},{95, 90, 85}}; Note: Data entries for each row are enclosed in individual curly braces. 5.9.1 Accessing Two Dimensional Arrays Two dimensional arrays are accessed/ initialized with the help of two square brackets. The following piece of code given below is used to access an array ‘A’ of size ‘M*N’ (‘M rows, N columns) and initialize all its members to zero. char name[5][25]; ARRAYS
- 187. 172 … { … // Accessing two dimensional arrays int i, j; for(i = 0; i < M; i++) { for(j = 0; j < N; j++) { A[i][j] = 0; } } } When accessing two dimensional arrays it is important to put each subscript in its own, correct pair of brackets. Trying to initialize or access elements of a two dimensional array would produce syntax errors. For example, /* Incorrect Statements */ int A[M, N]; A[i, j] = 0; A[j][i] = 0; 5.9.2 Passing Two Dimensional Arrays as a Parameter to Functions As one dimensional array, two dimension arrays are also passed to functions by reference automatically. Program 5.5 declares a two dimensional array ‘actual_2darray’ which is initialized by the user and then passed to the function ‘display’ where it is displayed in matrix form. Program 5.5: Two dimensional arrays as function arguments // Header Files # include<iostream.h> # include<conio.h> // Function prototype void display(int formal_2darray[3][3]); // Main Function void main() { clrscr(); int actual_2darray[3][3]; // Enter 2d Array
- 188. 173Part I: Chapter 5 Arrays cout<<" Enter the elements of the 2 dimensional array:"; for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { cin>>actual_2darray[i][j]; } } //Passing 2d array to function display(actual_2darray); } /****************************** Name: display Purpose: To display the 2d array in Matrix form **************************************/ void display(int formal_2darray[3][3]) { cout<<endl<<" Displaying the 2d array in Matrix Form... n"<<endl; for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { cout<<"t"<<formal_2darray[i][j]<<" "; } cout<<endl; } } The following output is observed when the program 5.5 is executed: OUTPUT SCREEN ARRAYS
- 189. 174 5.9.3 Mapping of Two Dimensional Arrays Although, we visualize two dimensional arrays by rectangular tables this is not the way two/multi dimension arrays are stored in the computers memory. The computer’s memory only supports a linear placement of data items akin to a one-dimensional array as a result of which the memory simulates the two/multi dimensional array as a one dimensional array. This process is called as Mapping of an array. For example, for a two dimensional array ‘Two_dimension_array[M][N];’ having ‘M*N’ as the maximum number of elements that can be stored into it. The computer maps the two dimensional array into one dimension using the following formula: Two_dimension_array[ i * N + j ] Where ‘i’ represents the corresponding row, ‘j’ represents the corresponding column and ‘N’ represents the maximum number of columns allowed in the two dimensional array ‘Two_dimension_array’. Assuming ‘N’ to be ‘10’ the data entry stored at ‘Two_dimension_array[2][3]’ is mapped into a one dimensional array say, Single_dimension_array[2*10 +3]. Hence, we can also use this for ‘Single_dimension_array[i * N+j]’ in place of ‘Two_dimension_array[i][j]’ as both point or reflect to the same memory block. The entire process of mapping is explained in the part III of the book. 5.9.4 Solved Examples Example 5.9.4.1 Write a program to accept a sequence of strings, storing them in a two dimensional array, and then print them on the screen? Solution of 5.9.4.1 // Header Files # include<iostream.h> # include<string.h> # include<stdio.h> // Main Function void main() { clrscr(); char name[5][25];
- 190. 175Part I: Chapter 5 Arrays cout<<"Enter the names with a maximum of 24 characters"; for(int i=0;i<5;i++) gets(name[i]); cout<<" n The names stored are"; for (i=0;i<5;i++) puts(name[i]); getch(); } OUTPUT SCREEN 5.10 Solved Examples Example 5.10.1 Write a program to create a matrix of order 4*4 containing integer elements using a function swap which swaps the value of two integers print the transpose of this matrix (without using a 2nd matrix)? Solution of 5.10.1 // Header Files # include <iostream.h> # include <conio.h> # include <iomanip.h> // Function Prototype void swap(int e, int f); Enter the names with the maximum of 24 characters Ankit Asthana Kartik Gupta Deepti Asthana Karan Chopra Neha Maheshwari The names stored are Ankit Asthana Kartik Gupta Deepti Asthana Karan Chopra Neha Maheshwari ARRAYS
- 191. 176 // Main Function void main() { int a,b,c,d,e,f; int trans[5][5]; cout<<"Enter the size of the 2-dimensional array A(Rows(<5) and Columns(<5))"; cin>>a>>b; swap(a,b); cout<<"Thank you for using the program "; getch(); } /******************************** Name: swap Parameters: int e, int f Purpose: Transpose the matrix ********************************/ void swap( int e, int f) { int arr1[5][5]; int arr2[5][5]; int a,b,c,d; // Transposing the array cout<<" Enter array A"; for (int p=0;p<e;p++) for (int q=0;q<f;q++) cin>>arr1[p][q]; for (int i=0;i<e;i++) for (int j=0;j<f;j++) arr2[j][i]=arr1[i][j]; cout<<" Displaying"; for (int t=0;t<10;t++) { cout<<"."; } cout<<" Elements of A"<<endl; for (i=0;i<e;i++) { for (int j=0;j<f;j++) cout<<setw(8)<<arr1[i][j]; cout<<endl; }
- 192. 177Part I: Chapter 5 Arrays cout<<" Transposed array A as follows nn"; for (int j=0;j<f;j++) { for (i=0;i<e;i++) cout<<setw(8)<<arr2[j][i]; cout<<endl; } } OUTPUT SCREEN Example 5.10.2 Write a program to create two matrices A and B and to store the product of these two matrices into another matrix C. The program should use a separate function to calculate the product of the two matrices. Solution of 5.10.2 // Header Files # include <iostream.h> # include <conio.h> # include <dos.h> // Function Prototype ARRAYS
- 193. 178 void multiplication (float arr1[5][5], float arr2[5][5], int s, int t, int u, int v); // Main Function void main() { float A[5][5],B[5][5]; int a,b,c,d; cout<<"n Enter the size of array 1 "; cin>>a>>b; cout<<" Enter the elements of array 1"; for (int e=0;e<a;e++) { for (int f=0;f<b;f++) { cin>>A[e][f]; } } cout<<"n Enter the size of the array 2"; cin>>c>>d; cout<<" Enter the elements of the array 2"; for (e=0;e<c;e++) { for (int f=0 ; f<d;f++) { cin>>B[e][f]; } } multiplication (A,B,a,b,c,d); } /********************* Name: multiplication Purpose: To calculate the product of two matrices passed to it as parameters **************************************************************/ void multiplication (float arr1[5][5], float arr2[5][5], int s, int t, int u, int v) { int C[5][5];
- 194. 179Part I: Chapter 5 Arrays if (t==u) { for (int i=0;i<s;i++) { for (int j=0;j<v;j++) { C[i][j]=0; for (int k=0;k<t;k++) C[i][j]= C[i][j] + arr1[i][k] * arr2[k][j]; } } // Displaying array 1 cout<<" nArray 1 in matrix form is as follows nn"; for (i=0;i<s;i++) { for (int j=0;j<t;j++) { cout<<"t"<<arr1[i][j]<<" "; } cout<<endl; } //Displaying array 2 cout<< "nArray 2 in matrix form is as follows nn"; for (i=0;i<u;i++) { for (int j=0;j<v;j++) { cout<<"t"<<arr2[i][j]<<" "; } cout<<endl; } // Displaying array 3 cout<<"nArray C = A * B in matrix form is as follows n"<<endl; for (i=0;i<s;i++) { for (int j=0;j<v;j++) { cout<<"t"<<C[i][j]<<" "; } cout<<endl; } } else ARRAYS
- 195. 180 { cout<<"n Matrices not Compatiable"; } } Example 5.10.3 Write a program to create a matrix of the order of ‘5 * 5’ containing integer elements. Write user defined functions to do the following: (1) Find the sum of the elements lying on the left diagonal (2) Find the sum of the elements lying on the right diagonal (3) Find the sum of the elements lying on both the diagonals Solution of 5.10.3 // Header Files # include <iostream.h> # include <conio.h> // Function Prototypes void suml ( int arr1[ ][5], int c, int d) ; void sumr ( int arr1[ ][5], int c, int d) ; void sumb ( int arr1[ ][5], int c, int d) ; void display (int arr1[ ][5],int c ,int d); // Main Function void main() { int a,b,e,f; int arr[5][5]; do { cout<<"nEnter the dimensions of the square matrix(rows and columns):"; cin>>a>>b; if (a!=b) { cout<<"nInvalid Dimensions.n"; } }while (a!=b); cout<<"nEnter the elements of the array:"; for (e=0;e<a;e++) for (f=0;f<b;f++) cin>>arr[e][f]; display(arr,a,b);
- 196. 181Part I: Chapter 5 Arrays suml(arr,a,b); sumr(arr,a,b); sumb(arr,a,b); getch(); } /********* Name:suml Parameters: arr1[][], c, d Purpose: To calculate the sum of the left diagonal ***************************************************/ void suml( int arr1[5][5], int c,int d ) { int lsum=0; int i; int j; // sum of left diagonal for (i=0; i<c; i++) { for (j=0; j<d; j++) if (i==j) { lsum+= arr1[i][j]; } } cout<<"nSum of the elements on the left diagonal "<<lsum<<endl; } /********* Name:sumr Parameters: arr1[][], c, d Purpose: To calculate the sum of the right diagonal ***************************************************/ void sumr(int arr1[5][5],int c,int d) { int rsum=0; int j=d-1; // Sum of the right diagonal for (int i=0;i<c;i++) ARRAYS
- 197. 182 { rsum+=arr1[i][j]; j--; } cout<<endl<<"The sum of the elements on the right diagonal "<<rsum<<endl; } /********* Name:sumb Parameters: arr1[][], c, d Purpose: To calculate the sum of both diagonal ***************************************************/ void sumb(int arr1[5][5], int c,int d) { int lsum=0; // Sum of left diagonal for (int i=0;i<c;i++) { for (int j=0;j<d;j++) if (i==j) { lsum+=arr1[i][j]; } } int lrsum=0; int rsum=0; int j=d-1; // Sum of the right diagonal for (i=0;i<c;i++) { rsum+=arr1[i][j]; j--; } // Sum of both the diagonals if (c%2==0) { lrsum= lsum+rsum ; }
- 198. 183Part I: Chapter 5 Arrays else { int b=c/2; lrsum= lsum + rsum - arr1[b][b]; } cout<<endl<<"The sum of both the diagonals "<<lrsum<<endl; } /********* Name:display Parameters: arr1[][], m, n Purpose: To display the matrix ***************************************************/ void display (int arr1[5][5], int m,int n ) { // To clear the screen cout<<"nMatrix -> n"<<endl; for (int i=0;i<m;i++) { for (int j=0;j<n;j++) { cout<<arr1[i][j]<<" t "; } cout<<endl<<endl; } } 5.11 Review Exercise 1/2 mark questions Q1) Is the following declaration valid? ‘unsigned int arr(7);’ Q2) State whether the following statements are true or false: (i) An array element is accessed using a dot operator. (ii) An array can store many different types of values. (iii) This is an syntactically correct statement int arr[2.3]; (iv)An individual array element is passed to a function and not the entire array. Q3) Arrays are passed to functions as: (i) reference parameters. (ii) formal parameters. ARRAYS
- 199. 184 (iii) actual parameters (iv) (i) and (iii) are correct (v) None of the above Q4) The number of bytes occupied by the two dimensional array A[10][10] are: a) 100 bytes b) 10 bytes c) 20 bytes d) 40 bytes e) None of the above Q5) State whether the following are true or false: a) Arrays can have more than two dimensions b) There is no difference between a character array and a string. c) Arrays are a collection of elements belonging to the same data type. d) An array element is accessed using the dot operator. Q6) Write C++ statements for the following questions: a) Array birds having 50 elements of type int. b) A float array sum of index value 23. c) A statement that checks for the occurrence of string “Dog” in the string s1. Q7) Are the following statements syntactically correct: a) int x[2][2] = { {0,1},{1,2}}; x[1,1]=5; b) int name1[15]; char name2[15]; strcat(name1,name2); c) int x[4][0] = { (0,0)}; Q8) Write statements to accomplish the following: i) Display the value of the fifth element of array ‘A’. ii) Print a value into element 5 of single-subscripted long int array A. iii) Initialize each of the eight members of single subscripted array A to 0. iv) Print the grand total of integer array ‘num’ containing 100 elements. v) Copy all the elements of integer array ‘A’ into floating-point array ‘B’. vi) Determine the smallest and the largest integer values from a character array ‘A’ containing 15 elements. Hint: Use type casting (ASCII Values) Q9) Consider a 5 by 5 integer array ‘A’. a) Write C++ statement to define this array? b) How much space (in bytes) does this declaration involve using the declaration? c) How many columns does it have? d) How many rows does it have? e) How many elements does it have? f) Write the names of all the elements in the second row in array A.
- 200. 185Part I: Chapter 5 Arrays g) Using loops initialize each member of array A to a value of 0. h) Using loops input values for each member of Array A. i) Determine the sum of elements on the main diagonal of this array A inputted in the above statements. j) Print this array A in matrix (tabular) form. Q10) Using binary search how many comparisons will the array of following size take i) 1023 elements ii) 1048576 elements (Hint: Binary search eliminates one half of the elements in a sorted array after each comparison) 4/5 mark Questions Q11) Write a program to initialize an integer array numbers containing 100 elements. Initialize these elements values from 1 to 100 in ascending order. Create another array even which stores those elements of array numbers that are even. Also display all the elements of array even? Q12) Write a program to create two matrices a of the order m*n and matrix b of the order p *q. Write a user defined functions to do the following: a. Create another matrix ‘c’ where each element of ‘c’ is the sum of the corresponding elements of the array ‘a’ and ‘b’. b. Create another matrix‘d’ where each element‘d’ is the difference of the corresponding elements of the arrays ‘a’ and ‘b’. c. Write a function display to display the following matrices ‘a’, ‘b’, ‘c’ and ‘d’. d. Find the transpose of matrix a and display it Q13) Write user defined functions to do the following a. Create 2 matrices ‘x’ & ‘y’ of the order ‘a*b’, ‘c*d’ respectively containing integer values b. Create a 3rd matrices which has the sum of the array X and array Y 5.12 Programming Project In section 5.8.4, we studied how the computer memory stores two dimension arrays by simulating them in a single dimension. The programming project for this chapter is to state how C++ maps a three dimensional array (i.e. array_name[row][col][depth]) to one dimensional memory. Write a fully commented and structured ANSI C++ program that uses a single looping structure (of your choice) and your 3-D to 1-D mapping to print out the following array: int map_3d[2][2][2] = {5,10,15,20,25,30,35,40}. ARRAYS
- 201. 186 5.13 Let us revise! An array is a collection of variables of the same type that are referenced by a common name. Arrays can be one dimensional, two dimensional or multi dimensional. An array is declared by specifying its base type, name and size. An array of type char can be used to store a character string. In C++, arrays are passed to functions as reference that is any changes made to the formal array will be reflected back to the actual array. To receive an array argument the formal array should be of the same type as the actual array. The data type of array elements is known as the base type of the array. In C++, array indices start from 0, and are up to size 1. C++ considers strings as single dimension arrays. The total number of bytes stored in a two dimensional array is given by the formulae: number of rows (1st index) * number of columns (2nd index) * sizeof(base type). Linear search compares each element of the array with a specific element (Search key). Works very well for smaller arrays or unsorted arrays. Binary search also a method for searching elements in an array. The algorithm for the binary search eliminates one half of the array after each comparison. If and when element is found the elements position is stored. Binary search works great for larger arrays but the only limitation is that the array should be sorted. The data type of array elements is known as the base type of the array. In C++, array indices start from 0, and are up to size 1. The total number of bytes stored in a two dimensional array is given by the formulae: number of rows(1st index) * number of columns(2nd index) * sizeof(base type). Sorting is to arrange the various elements of an array in either ascending or descending order. It is of three types bubble, insertion, selection C++ considers strings as single dimension arrays. The total number of bytes stored in a two dimensional array is given by the formulae: number of rows(1st index) * number of columns(2nd index) * sizeof(base type). Sorting is to arrange the various elements of an array in either ascending or descending order. It is of three types bubble, insertion, selection
- 202. Part I CHAPTER 6 STRUCTURES AIM OUTLINE Introduce the concept of ‘Structures’ How to declare and use Structures, Arrays of Structures and Nested Structures. Close comparison between Structures and arrays How to create Pointer to a Structure Pass and return structures to functions Appreciate the concept of Structures 6.1 Introduction 6.2 Structure Definitions 6.3 Structure Variables 6.4 Accessing Members of a Structure 6.5 Structure Arrays 6.6 Structures with Functions 6.7 Nested Structures 6.8 Solved Examples 6.9 Review Exercise 6.10 Programming Project 6.11 Let us revise
- 203. 188 188 6.1 Introduction Suppose one has lots of different but related data, one would prefer to store this data together in a single unit. Let me take a very common example, which we have often observed while using computers. Think about random folders and icons lying on your desktop like ‘Recycle Bin’ which stores all the deleted files together, ‘My Documents’ which stores all documents and spread sheets and ‘My Network Places’ which stores all files related to ‘Networking’. All in all, these icons consist of similar files which are put together making up these single units to help organize data and files. This process in C++ is achieved by the help of structures. A structure is a heterogeneous collection of data. It is a collection of data of different data types. A structure empowers us to create our own data types by using the collection of fundamental and derived data types available in C++ which serves as an efficient way of grouping several pieces of related information together. Just the same way as random lying papers are grouped in a file and words are put together to form sentences. 6.2 Structure Definition Till now, we have learnt about simple data types like int, char, and float. Structure allows us to declare user defined data types thus increasing the number of data types. A structure is declared using the following C++ statements: Syntax: The structure definition tells us about how the structure is organized. It informs the compiler about the various data members present inside the structure. Figure 6.1 helps in understanding structure declaration. struct < tag name> { data_type1 var1; data_type2 var2; data_type3 var3; };
- 204. 189 Part I: Chapter 6 Structures Figure 6.1: Structure Definition To declare a structure one has to start with the keyword ‘struct’. The keyword ‘struct’ informs the compiler that a structure is being declared. The identifier ‘student’ is the structure name. The variables within the structure ‘student’, ‘roll_no’ of type integer, a character array ‘student_name’ and a floating variable ‘student_avg are known as structure’s members which are enclosed between a pair of curly braces ({ }). The semicolon (;) is used to terminate the structure definition. The important thing to understand at this point is that this structure definition alone does not actually create any variables in other words no memory has been assigned to these variables. With this declaration one has declared a type of structure which we will use further to create structure variables. They are used to declare new data types. 6.3 Structure Variables As we already know that the structure definition does not declare any variables. To do this we need to declare a structure variable for the defined structure. This is done by again starting with the structure name already defined with the structure variable name. For example, if one wishes to create structure variables for the structure ‘student’ defined above one would just write the following c++ statement. Keyword ‘struct’ Structure Tag/Name (;) Terminates Structure definition Structure Members
- 205. 190 190 student s1; This statement would declare a variable ‘s1’ of type student. We could easily understand this by comparing it to a normal declaration that is declaring a variable ‘s1’ of say, type integer. int s1; The only difference being in the two declarations is that in the 1st case s1 is a variable of type ‘student’ and in the second case s1 is a variable of type int. Now let me ask you a very interesting question what is the size of this structure variable ‘s1’ defined above? The answer to this question is very simple, as we know that a structure is nothing but a heterogeneous collection of data hence the size of a structure variable can be calculated easily by adding up individually the size of all the structure members present in the structure. Using this logic we can easily calculate the size of the structure variable ‘s1’ which comes out to be 21 bytes. Fig (6.2) gives an idea of how memory is allocated by the computer to the structure variable ‘s1’. Figure 6.2: Memory allocation for structure variables (s1 occupies memory locations say, from 99 to 122, i.e., 21 bytes) In addition to this one can also define multiple data structures, also all the information stored in one structure can also be copied into another structure of the same type which we will study in more detail in the next section. 6.3.1 Solved Examples Example 6.3.1.1 (a) Declare a structure ‘date’ having the following members: day -- A string of size[3] Represents the day part of the date month -- A string of size[20] Represents month name of a date
- 206. 191 Part I: Chapter 6 Structures year -- A string of size [4] Represents the year part of the date (b) What is size of a variable of type date hence created? Solution of 6.3.1.1 a) struct date { char day[3]; char month[20]; char year[4]; }; b) The size of a structure variable of type ‘date’ will occupy 27 bytes of computers memory. Note: The size of structure might differ from one machine to another also the size of the structure is also determined by a concept known as ‘word alignment’ although this concept is out of the scope of the book. Example 6.3.1.2 (a) Declare a structure ‘Employee’ having the following members: Emp_Name -- A string of size[15] Represents the name of the employee Emp_IdNo. -- A Long int variable Represents the employees Id number Emp_Designation -- A string of size[20] Represents the designation of the employee Emp_salary -- A float variable Represents the salary of the employee (b) Declare C++ statement to declare two structure variables ‘Emp1’, ‘Emp2’ of type ‘Employee’ Solution of 6.3.1.2 (a) struct Employee {
- 207. 192 192 char Emp_Name[15]; long int Emp_IdNo; char Emp_Designation[20]; float Emp_salary; }; (b) Employee Emp1,Emp2; We can also define the structure variables after the structure, in such a case the tag name becomes optional but then one cannot declare any more structure variables in the main function. For example, consider the structure ‘Employee’, for this structure we can also declare the structure variables ‘Emp1’, ‘Emp2‘without giving a name to the structure as shown below. struct <optional tag/name> { char Emp_Name[15]; long int Emp_IdNo; char Emp_Designation; float Emp_salary; }Emp1,Emp2; The only disadvantage using this feature is that now we cannot create any more structure variables of this type. 6.4 Accessing members of a structure We can access members of a structure with the help of two operators in C++. The operator (.) and the structure pointer operator (->) also known as the arrow operator. The structure member is accessed with the help of the structure name. Lets take an example, to access the structure member ‘Emp_salary’ of structure variable Emp1 of type Employee defined in the example above one would write the following C++ statement. Emp1.Emp_salary = 2000.20 ; The important thing to note here is that the part of the statement before the (.) operator is the structure variable name ‘Emp1’ and the part after the dot operator ‘Emp_salary’ is the
- 208. 193 Part I: Chapter 6 Structures name of the structure member to be accessed. The result of this statement is that the structure member ‘Emp_salary’ is given a value of ‘2000.20’. Structures members can be treated just like any other ordinary variables available in C++. That is all the various operations which can be done on any ordinary variables are also valid with structure members. Consider the following figure 6.3 given below. This figure(6.3(a)) contains the structure definition for the structure ‘Hospital’. The members of this structure can be accessed using the (.) operator figure (6.3(b)). (a) (b) Figure 6.3(a): Structure Definition for ‘Hospital’ (b): Structure Variable ‘Patient1’ As we already know that one can create several structure variables using the same structure. Suppose we create two structure variables for structure ‘Hospital’ ‘Patient1’ and ‘Patient2’ and we assign some particular ‘Patient_Name’, ‘Patient_No’ and ‘Patient_Details’ to ‘Patient1’ then if we say: Patient2 = Patient1;
- 209. 194 194 Then this statement will assign the structure variable ‘Patient1’ to the other structure variable ‘Patient2’ figure(6.4) and all the following conditions will hold. Figure 6.4: Copying structure variables We can now look at program 6.1 which implements the same concept. Program 6.1 // Header Files // Program 6.1: Structure Variables # include<iostream.h> # include<conio.h> # include<stdio.h> /******************************************** Structure definition Name: Student Patient2.Patient_Name = Patient1.Patient_Name Patient2.Patient_No = Patient1.Patient_No Patient2.Patient_Details = Patient1.Patient_Details
- 210. 195 Part I: Chapter 6 Structures Structure Members and Type: roll_no: int student_name[15]: char avg: float ********************************************/ struct student { //Data_type1 Variable1 int roll_no; //Data_type2 Variable2 char student_name[15]; //Data_type3 Variable3 float avg; }; //Main Program void main() { // Declaring structure variable s1 of type student student s1; cout<<" ENTER DETAILS"; // Entering Data for student cout<<endl<<" Enter Roll No:"; cin>>s1.roll_no; // Accessing the Structure member 'roll_no'
- 211. 196 196 cout<<endl<<" Enter Student Name:"; gets(s1.student_name); // Accessing the structure member 'student_name' cout<<endl<<" Enter Student Average"; cin>>s1.avg; // Accessing the structure member 'avg' cout<<endl<< " PRINTING DERAILS"; cout<<endl<< " Roll No:"<<s1.roll_no; cout<<"t"<< " Student Name:"<<s1.student_name; cout<< " Student Avg"<<s1.avg; } When this program is executed at first the compiler creates a structure ‘student’ with the above mentioned structure members. Then the compiler declares a structure variable ‘s1’ of type student. In this program, the user enters the details of a student which include students roll number, name and average and then the program displays these details on the screen. The program could be easily understood by reading the comment entries provided with the program. PROGRAMING TIPS Tag Names for structures should always be initialized to increase the scope of creating structure variables at time of structure declaration. All structure members should have different names to prevent logical errors.
- 212. 197 Part I: Chapter 6 Structures 6.4.1 Solved Examples Example 6.4.1.1 Using the structure definition for structure ‘Library’ having the following members write a c++ program which declares a structure variable L1: Book_Name -- A string of size[15] Represents the name of the book Book_IdNo -- A int variable Represents the books Id number Book_Shelf -- A string of size[5] Represents the Book shelf number where the book is stored. The user should then be asked to enter the details for the structure variable. The program should also check whether the book name is “Cleopatra” if so, the program should print a suitable message on the screen. Solution of 6.4.1.1 // Header Files # include <iostream.h> # include <conio.h> # include <string.h> # include <stdio.h> // Structure Library struct Library { char Book_Name[15]; int Book_IdNo; char Book_Shelf[5]; }; // Main Function void main() { Library L1; char str[]="Cleopatra"; cout<<"Enter Respective Details:"; cout<<endl<<"Book Name";
- 213. 198 198 gets(L1.Book_Name); cout<<endl<<"Book_IdNo"; cin>>L1.Book_IdNo; cout<<endl<<"Book Shelf"; gets(L1.Book_Shelf); if (strcmp(str,L1.Book_Name)==0) { cout<<endl<<"Condition Satisfied"; } else { cout<<endl<<"Condition not Satisfied"; } } COMMON PROGRAMMING ERRORS • Not ending the structure definition with a (;) semicolon is a compilation error. • Not using the dot operator (.) to reference to structure members is a compilation error. 6.5 Structure Arrays Before we learn about arrays of structures lets compare arrays and structures. Both arrays and structures are collection of data types. The major difference in them being that an array is a collection of data of the same type, where as a structure helps to group together data that may vary in type. Let us look at the following comparison between array and structures. Structures Arrays struct Employee char Emp_Name [15]; { char Emp_Name[15]; char Emp_Designation[15]; int Emp_IdNo.; char Emp_Designation[15]; };
- 214. 199 Part I: Chapter 6 Structures Now as we can see the structure ‘employee’ groups data of three different data types: char, int and float where as the array ‘name[15]’ is a collection only of character type of data. Structures allow us to build data bases, especially if one considers the possibility of building arrays of them but is it possible to have arrays of structures? The answer to this question is ‘Yes’. Arrays of structure are declared by declaring the structure variables as arrays. Let us look at program 6.2 given below: Program 6.2: // Header Files # include <iostream.h> # include <conio.h> # include <stdio.h> // Structure Employee struct Employee { char Emp_Name[15]; int Emp_IdNo; char Emp_Designation[15]; }; // Main Program void main() { Employee E1[10]; // Array of structures for (int i = 0; i<10;i++) { cout<<" Enter Respective Details for Employee"<<i+1; cout<<endl<<"1. Name"; gets(E1[i].Emp_Name); cout<<endl<<"2. Employee Id No."; cin>>E1[i].Emp_IdNo; cout<<endl<<"3. Employee Designation"; cin>>E1[i].Emp_Designation; } cout<<endl<< " Printing Employee database";
- 215. 200 200 for (i=0;i<10;i++) { cout<<endl<<"1. Name"<<E1[i].Emp_Name; cout<<endl<<"2. Employee Id No."<<E1[i].Emp_IdNo; cout<<endl<<"3. Employee Designation"<<E1[i].Emp_Designation; } } Arrays of structure have the properties same as a normal array the only difference being is that this time they are collection of a ‘struct’ data type. All operations valid on ordinary Arrays are also valid with arrays of structure. For example, in the program above we have declared an array ‘E1’ with index value 10 of type ‘Employee’. This array of structure helps us to create a database for a set of 10 employees with the name, designation and Id No. for each employee. Each index value in this array of structure is accessed in a similar manner with the help of the (.) operator. Let us look at the following C++ statements from program 6.2 above. One uses for loop to enter the details for employee name into each index of array ‘E1’. Each index entry of this array is accessed using ‘E1[i].Emp_Name’. The variable ‘i’ refers to each index of the array ‘E1’ and the part after the (.) operator refers to the ‘Emp_Name’ of the structure variable ‘E1’. Similarly, the other for loop is used to print the details of this database on the screen. An array of structure is very similar to a two-dimensional array, it can be thought as if it consists of columns, each of which may consist of distinct data types. Fig 6.3 given below represents the array ‘E1’ in such a manner. for (int i = 0; i<10;i++) { cout<<" Enter Respective Details for Employee"<<i+1; cout<<endl<<"1. Name"; cin>>E1[i].Emp_Name; . . . }
- 216. 201 Part I: Chapter 6 Structures Figure 6.3: Memory Allocation for Structure variable E1 6.5.1 Solved Examples Example 6.5.1.1 (a) Declare a structure ‘Team’ having the following members: Player_Name -- A string of size[15] Represents the name of the employee Player_No. -- An integer variable Represents the employees Id number Player_Type -- A string of size[20] Represents the designation of the employee (b) Write C++ statements to declare an array of structure variables ‘Players’ for a team of 11 players and create a menu driven program which takes in details from the user for creating a team, displays the created team and also searches for a specific player in the created team using the Player_No. as the search key? Solution of 6.5.1.1 // Header Files # include <iostream.h> # include <conio.h> # include <stdio.h> // Structure Definition struct team { char Player_Name[15];
- 217. 202 202 int Player_No; char Player_Type[20]; }; // Main Function void main() { team Players[11]; int flag=0; int search=0; int ch; do { cout<<endl<<"1. Create Team"; cout<<endl<<"2. View Team"; cout<<endl<<"3. Search for a player"; cout<<endl<<"4. Exit"; cin>>ch; switch(ch) { case 1: { for (int i=0; i<11;i++) { cout<<" Enter Details for player "<<i+1; cout<<endl<<" Enter Name:"; gets(Players[i].Player_Name); cout<<endl<<" Enter Player No."; cin>>Players[i].Player_No; cout<<endl<<" Enter Player Type"; gets(Players[i].Player_Type); flag++; } break;
- 218. 203 Part I: Chapter 6 Structures } case 2: if (flag!=0) { cout<<"=== TEAM ==="; for (int i=0;i<11;i++) { cout<<endl<<" Player No:"<<Players[i].Player_No; cout<<endl<<" Player Name:"<<Players[i].Player_Name; cout<<endl<<" Player Type:"<<Players[i].Player_Type; } } break; case 3: cout<<" Enter the Player No. to be searched"; cin>>search; for (int i=0;i<11;i++) { if (search == Players[i].Player_No) { cout<<" Player Found"; cout<<endl<<" Name"<<Players[i].Player_Name; cout<<endl<<" Type"<<Players[i].Player_Type; cout<<endl<<" Player No."<<Players[i].Player_No; break; } } } } while (ch!=4); } 6.6 Structures with functions As variables of any other data type, structure variables can also be passed to functions. Structure variables can be passed to functions in the following ways: Passing the entire structure variable as a parameter Passing individual structure members to the function Passing a pointer to a specific structure* * Topic of pointers is introduced and explained in Chapter 7.
- 219. 204 204 In addition to all this, we can also have functions which return structures. The important thing to note is that whenever structure variables or individual structure members are passed to the functions they are passed by value. So any changes made to the formal parameters are not reflected back to the actual parameters. Structures can be passed by reference by preceding the structure variable with an ‘&’ sign when the function is being called. Arrays of structures are automatically called by reference. 6.6.1 Passing structure variables as parameters Program 6.3 shows structure variables being passed as parameters. In this example, the structure variable ‘point_actual’ is passed by value. The result of the program is that it plots ‘**’on the screen at a specified location by the user. Program 6.3: // Header Files # include<iostream.h> # include<conio.h> // structure graph struct graph { int x; int y; }; // Function Header for function plot void plot (graph point_formal); // Main Function void main() { // Declaring structure variable point_actual graph point_actual; // Passing structure variable 'point_actual' to function 'plot', by value. plot(point_actual); getch();
- 220. 205 Part I: Chapter 6 Structures } /*************************************** Name: plot Purpose: Prints '**' at the specified location ***********************************************/ void plot (graph point_formal) { cin>>point_formal.x; cin>>point_formal.y; /* The ‘gotoxy’ function General Syntax: gotoxy(xcoordinate,ycoordinate): Purpose: Sets the cursor to the specified x and y coordinate specified in its call. It is defined under the <conio.h> header file. It works exactly like the locate command in some other programming languages. */ gotoxy(point_formal.x,point_formal.y); cout<<"**"; } COMMON PROGRAMMING ERRORS • Equating structures variables belonging to different structures is an compilation error • Assuming that structure variables are passed by reference when being passed to functions is a logical error. 6.6.2 Passing structure members as parameters Members of structure variables declared can also be passed as parameters to a function. Again, these parameters can be passed by value or passed by reference as per need of the program. Let us take an example, Program 6.4: // Header Files # include<iostream.h> # include<conio.h>
- 221. 206 206 // Example of a structure variable being passed to a function struct graph { int x; int y; }; void plot (int x, int y); void main() { clrscr(); // Declaring structure variable point_actual graph point_actual; cout<<" Enter required points to plot the graph"; cin>>point_actual.x; cin>>point_actual.y; // Passing structure variable 'point_actual' to function 'plot' plot(point_actual.x,point_actual.y); getch(); } /************************************** Function Name: plot Purpose: Plot the point at the specified location **************************************/ void plot (int x,int y) { /* gotoxy: function defined under the <conio.h> header file. It allows the user to display text at any specified location on the screen */ gotoxy(x,y); cout<<"**"; } You might have already noticed that Program 6.4 is very similar to Program 6.3. Both these programs perform the same function. The only difference being that in Program 6.3 the entire structure ‘point_actual’ is not passed as a parameter where as in this program we have instead passed only the members of this structure ‘x’ and ‘y’ to the function.
- 222. 207 Part I: Chapter 6 Structures 6.6.3 Passing a pointer to the structure as a parameter Pointers pointing to structures can also be passed as parameters to the function. An example of such a case is shown in Program 6.5. Program 6.5 // Header Files # include<iostream.h> # include<conio.h> // Structure ‘graph’ struct graph { int x; int y; }; void plot (graph *ptr1); void main() { clrscr(); // Declaring structure variable point_actual graph point_actual; graph *ptr; ptr = &point_actual; cout<<" Enter required points to plot the graph"; cin>>ptr->x; cin>>ptr->y; // Passing structure variable 'point_actual' to function 'plot' // Passing the structure by reference plot(ptr); getch(); } void plot (graph *ptr1) { gotoxy(ptr1->x,ptr1->y); cout<<"**";
- 223. 208 208 } 6.7 Nested Structures A nested structure is a structure which contains other structures variables as its members. Once we have created a structure we can use the struct keyword to create nested structures. Consider the two structure definitions given below. We can observe that in the structure declaration for structure ‘Telephone’ we have declared a structure variable ‘Date_Bill’ for another structure ‘Date’ also shown in figure 6.5. Figure 6.5: Nested Structures We can initialize the members of the structure variable ‘Mobile’ with the extended use of again the dot operator(.). This operator is also used to access members of structures that are themselves members of a larger structures in other words in our example, we use this operator to access the structure members (day, month, year) of the structure ‘Date’. To be more clear say; we want to access the member ‘day’ of structure ‘date’ from a structure variable of structure ‘telephone’ and intialize it to a value of ‘4’ then one would write the following C++ statements. struct Date { int day; int month; int year; }; struct Telephone { int Telephone_Number; int Telephone_Bill; struct Date Date_Bill; } Mobile; Telephone Bill1; Bill1.Date_Bill.day=4;
- 224. 209 Part I: Chapter 6 Structures Now let me ask you a very interesting question can a structure contain an instance of its own type? The answer to this question is ‘no’ we cannot. We can only declare structures as structure members to a structure provided the structure whose variable is a member to the nested structure is already defined. 6.7.1 Solved Examples Example 6.7.1.1 Write a structure definition for structure ‘Race’ having the following attributes. A character array ‘Player_Name’ which represents the name of the player playing in the race, an integer variable ‘Player_No’ representing the player number and a structure variable ‘Player_Time’ of structure ‘Time’ which represents the time taken by each player in the race. The definition for structure ‘Time’ is given below: struct Time { int hrs; int mins; int secs; }; Solution of 6.7.1.1 struct Time { int hrs; int mins; int secs; }; struct Race { char Player_name[20]; int Player_No; struct Time Player_Time; };
- 225. 210 210 6.8 Solved Examples Example 6.8.1 Write a program to create an array of the given structure ‘Emp’ struct Emp { int Emp_No; char Emp_Name[15]; float Emp_Basic; }; Create user defined functions to (a) Sort the array of structures declared in ascending order of Emp_Basic (b) Calculate the net salary of each employee using : Net Salary = Basic + H.R.A.+ D.A. – P.F. H.R.A. = 25 % of Basic D.A = 10 % of Basic P.F. = 10% of Basic Solution of 6.8.1 /* Header Files */ # include <iostream.h> # include <conio.h> # include <stdio.h> // Structure 'Employee' // Structure Members: Employee_No: Represents Employee Number // : Employee_Name: Represents Employee Name // : Employee_Basic: Represents Employee Basic struct Employee { int Employee_No; char Employee_Name[15]; float Employee_Basic; }; // Function Headers void accept (Employee E1[], int index); void display (Employee E1[], int index); void sort (Employee E1[], int index);
- 226. 211 Part I: Chapter 6 Structures void Netsalary (Employee E1[], int index); // Main Function void main() { clrscr(); Employee E[50]; int ch; int flag=0; int Array_index; do { clrscr(); cout<<"n =========== DATABASE: EMPLOYEE =============="<<endl; cout<<" 1. Create Employee database"<<endl; cout<<" 2. Print Enteries form database"<<endl; cout<<" 3. Sort the Employees in Order of Basic Salaries"<<endl; cout<<" 4. Calculate the Net salary of each Employee"<<endl; cout<<" 5. Exit Database"<<endl; cout<<" !. Press Choice"; cin>>ch; switch (ch) { case 1: if (flag==0) { clrscr(); cout<<" For how many employees should the database be created"; cin>>Array_index; accept(E,Array_index); flag++; } else { cout<<" Option Not Available"; getch(); } break; case 2: clrscr(); display(E,Array_index); break; case 3:
- 227. 212 212 sort(E,Array_index); break; case 4: Netsalary(E,Array_index); break; } } while (ch!=5); getch(); } /******************************************* Name: accept Formal Parameters: E1[], index Purpose: Enter values into structure array *******************************************/ void accept( Employee E1[], int index) { cout<<" Enter the following Details"; for (int i=0; i<index;i++) { cout<<endl<<" Employee No:"; cin>>E1[i].Employee_No; cout<<" Employee Name:"; gets(E1[i].Employee_Name); cout<<" Employee Basic:"; cin>>E1[i].Employee_Basic; cout<<"--------------------"; } } /******************************************* Name: display Formal Parameters: E1[], index Purpose: Display the structure array *******************************************/ void display( Employee E1[], int index) { for (int i=0; i<index;i++) { cout<<endl<<" Employee No:"<<E1[i].Employee_No; cout<<endl<<" Employee Name:"; puts(E1[i].Employee_Name);
- 228. 213 Part I: Chapter 6 Structures cout<<" Employee Basic:"<<E1[i].Employee_Basic; cout<<endl<<"--------------------"; } getch(); } /********************************************************************** Name: sort Formal Parameters: E1[], index Purpose: Sort the structure array in ascending order of basic salary **********************************************************************/ void sort (Employee E1[], int index) { clrscr(); struct Employee temp; for (int i=0; i<index;i++) for (int j=0; j<index-i-1;j++) if (E1[j].Employee_Basic>E1[j+1].Employee_Basic) { temp = E1[j]; E1[j]=E1[j+1]; E1[j+1]=temp; } cout<<"Sorting Complete"; getch(); } /******************************************* Name: Netsalary Formal Parameters: E1[], index Purpose: Calculate the Net salary of the employees using Net Salary = Basic + 25 % Basic(H.R.A) + 10%Basic(D.A.) -10% Basic(P.F.) *******************************************/ void Netsalary (Employee E1[], int index) { clrscr(); cout<<"==================================="<<endl; cout<<"Net Salaries for all employees"<<endl; float Salary[50];
- 229. 214 214 for ( int i=0; i<index;i++) { Salary[i]= (1.25) * E1[i].Employee_Basic; cout<<"Employee No: "<<E1[i].Employee_No<<" Net Salary: "<<Salary[i]<<endl; } getch(); } Example 6.8.2 Write a program to define a structure called ‘Award’ with the following data members (i) Name of the winner: String of length 30 characters (ii) Prize Money of Award: Long integer variable Declare an array of the structure with n elements. Write user defined functions for: (a) Input data values into structure members; (b) Display the contents in tabular form; (c) Display the details of students who have achieved awards having prize money > $50000; and (d) Display details of the student achieving the highest prize money. Solution of 6.8.2 // Header Files # include <iostream.h> # include <conio.h> # include <stdio.h> // Structure definition for'Award' struct Award { char Name[30]; char type[15]; long prize; }; // Function Headers void input (Award A[], int n); void display (Award A[],int n); void details (Award A[],int n); void largest (Award A[],int n); // Main Function void main()
- 230. 215 Part I: Chapter 6 Structures { Award A[50]; int n; int flag=0; int ch; do { clrscr(); cout<<" ================== Award Database ====================="; cout<<endl<<" 1. Input Award Information "; cout<<endl<<" 2. Display Award Information"; cout<<endl<<" 3. Details: Awards awarded for prize money > 50000$"; cout<<endl<<" 4. largest Prize money award"; cout<<endl<<" 5. Exit"; cout<<endl<<" Enter Choice :?: "; cin>>ch; switch(ch) { case 1: if (flag==0) { clrscr(); cout<<" For How Many Awards"; cin>>n; input (A,n); flag++; } else cout<<" Option not available"; getch(); break; case 2: display (A,n); getch(); break; case 3: details (A,n); break; case 4: largest (A,n); break; } }
- 231. 216 216 while (ch!=5); getch(); } /**************************************************** Name: input Parameters: A[], n Purpose: Input data in structure array 'A[]' ****************************************************/ void input ( Award A[], int n) { clrscr(); for (int i=0; i<n;i++) { cout<<" Enter Name of person"; gets(A[i].Name); cout<<" Enter type"; cin>>A[i].type; cout<<" Enter Prize Money $"; cin>> A[i].prize; cout<<"======================"<<endl; } } /**************************************************** Name: Details Parameters: A[], n Purpose: Print all members of array structure 'A[]' ****************************************************/ void display (Award A[], int n) { clrscr(); for (int i=0; i<n;i++) { cout<<"Name :"; puts(A[i].Name); cout<<"Type :"<<A[i].type<<endl; cout<<"Prize Money :"<<A[i].prize; cout<<endl<<"====================="<<endl; } }
- 232. 217 Part I: Chapter 6 Structures /**************************************************** Name: Details Parameters: A[], n Purpose: Print details of structure members of Array 'A[]' where prize money>50000$ ****************************************************/ void details (Award A[], int n) { clrscr(); for (int i=0;i<n;i++) { if (A[i].prize>50000) { cout<<endl<<"Name:"<< A[i].Name; cout<<endl<<"Type:"<< A[i].type; cout<<endl<<"Prize Money:"<<A[i].prize; } } getch(); } /**************************************************** Name: largest Parameters: A[], n Purpose: Print details of structure array 'A[]' with the largest prize money ****************************************************/ void largest (Award A[], int n) { clrscr(); int max=A[0].prize; int index=0; for (int i=0;i<n;i++) { if (max<A[i].prize) { max = A[i].prize; index = i; } }
- 233. 218 218 cout<<" Name:"<< A[index].Name; cout<<endl<<" Type: "<<A[index].type; cout<<endl<<" Prize: "<<A[index].prize; getch(); } Example 6.8.3 Using the following structure definition for nested structure ‘Ticket’ given below, Write a program to enter the details of 5 tickets and calculate the ticket fares for each of the following depending on distance traveled ( Table Given Below). Ticket Members Type PNR string DOT date Passenger person From string To string Traversed int Solution of 6.8.3 Distance Traveled Fare < =200 Miles >200&<300 Miles >=300 Miles $5/Mile $4/Mile $3/Mile // Header Files # include<iostream.h> # include<conio.h> // Comments: Use Header File <stdio.h> and use functions gets(), puts() to // manipulate with strings for better results struct date { int day; int month; int year ; }; struct person { char Name[15]; int age; };
- 234. 219 Part I: Chapter 6 Structures // Structure Definition for structure 'person' struct person { char Name[15]; int age; }; // Structure Definition for structure 'date' struct date { int day; int month; int year; }; // Structure Definition for structure 'ticket' struct ticket { char PNR[15]; date DOT; person Passenger; char From[15]; char To[15]; int Traversed; }; // Function Headers void accept( ticket T[], int n); void fare (ticket T[], int n); void display(ticket T[], int n); // Main Function void main() { clrscr(); int choice; ticket T[50]; int index; do
- 235. 220 220 { clrscr(); cout<<"######## Ticket Booking ########"<<endl; cout<<"!1! Enter Passenger information"<<endl; cout<<"!2! Ticket fare for passengers"<<endl; cout<<"!3! Print Tickets"<<endl; cout<<"!4! Quit Program"<<endl; cout<<"!?! Enter Choice"<<endl; cin>>choice; switch(choice) { case 1: clrscr(); cout<<" For how many Passenger(s) the ticket(s) to be booked for <50"; cin>>index; accept(T,index); break; case 2: clrscr(); fare(T,index); break; case 3: clrscr(); display(T,index); break; default: if (choice!=4) cout<<"n Please Press Valid Choice"; } } while (choice != 4); getch();
- 236. 221 Part I: Chapter 6 Structures cout<<"Enter the following details:"<<endl; for (int i=0; i<n;i++) { cout<<"Enter the PNR Number"; cin>>T[i].PNR; cout<<endl<<"Enter Date:"; cout<<endl<<" Day:"; cin>>T[i].DOT.day; cout<<endl<<" Month:"; { cout<<"n PNR:"<<T[i].PNR; cout<<"n Fare:"<<4*T[i].Traversed; } if (T[i].Traversed>=300) { cin>>T[i].DOT.month; cout<<endl<<" Year:"; cin>>T[i].DOT.year; cout<<endl<<"Enter Passenger Details:"; cout<<endl<<" Name :"; cin>>T[i].Passenger.Name; cout<<endl<<" Age :"; cin>>T[i].Passenger.age; cout<<endl<<"Source Station:"; cin>>T[i].From; cout<<endl<<"Destination Station:"; cin>>T[i].To; cout<<endl<<"Distance Traversed:"; cin>>T[i].Traversed; cout<<endl<<"^^^^^^^^^^^^^^^^^^^^"<<endl; } } /*************************************** Name: Fare Parameters: T[], n Purpose: To calculate Ticket Fare **************************************/ void fare (ticket T[], int n) { clrscr(); for (int i=0;i<n;i++)
- 237. 222 222 { if (T[i].Traversed <=200) { cout<<"n PNR:"<<T[i].PNR; cout<<"n Fare:"<<5*T[i].Traversed; } if (T[i].Traversed >200 && T[i].Traversed<300) cout<<"n Fare:"<<3*T[i].Traversed; cout<<"n PNR:"<<T[i].PNR; } cout<<"--------------"; } getch(); } /*************************************** Name: display Parameters: T[], n Purpose: To display ticket details **************************************/ void display(ticket T[],int n) { int i; for ( i=0; i<n;i++) { // Entering Ticket Details cout<<endl<<" Entry :"<<i+1; cout<<endl<<" PNR Number: "<<T[i].PNR; cout<<endl<<" Date:"; cout<<endl<<" Day:"<<T[i].DOT.day; cout<<endl<<" Month:"<<T[i].DOT.month; cout<<endl<<" Year:"<<T[i].DOT.year; cout<<endl<<" Passenger Details:"; cout<<endl<<" Name :"<<T[i].Passenger.Name; cout<<endl<<" Age :"<<T[i].Passenger.age; cout<<endl<<"Source Station:"<<T[i].From; cout<<endl<<"Destination Station:"<<T[i].To; cout<<endl<<"Distance Traversed:"<<T[i].Traversed; cout<<endl<<"^^^^^^^^^^^^^^^^^^^^"<<endl; } getch(); }
- 238. 223 Part I: Chapter 6 Structures 6.9 Review Exercise 1 Mark/ 2 Mark Q1) When referencing structure members, the identifier before the (.) operator is the a) Structure Variable b) Structure Tag c) Key word ‘struct’ d) None of the above e) All of the above f) None of the above Q2) A structure is a collection of a) Variables of different data types b) Variables of similar data types c) Variables of same data types d) Smaller nesting structures e) All of the above f) None of the above Q3) Write a C++ statement which sets the ‘secs’ member of time structure to ‘10’. Q4) Write C++ statements to accomplish the following: a) Write structure definition for structure ‘Computer’ having integer variable ‘ComputerNo’, character array ‘ComputerName’ with names as long as 20 characters. b) Declare a structure variable ‘Num1’ for structure ‘Computer’. c) Initialize the structure member ‘ComputerName’ to “I B M” . Q5) Give the output for the following piece of code: struct graph1 { int x; int y; int z; }; graph1 increment (graph1 g1) { g1.x*= g1.y-- + ++g1.z; g1.y++;
- 239. 224 224 g1.z++; return g1; } void main() { clrscr(); graph1 g1; graph1 g2; g1.x=1; g2.x=2; g1.z=2; g1.y=0; g2=increment(g1); cout<<g2.x<<g2.y<<g2.z; getch(); } Q6) Find the error(s) in the following piece of code: # include<'iostream.h'> # include<'conio.h'> structure car { char car_name[15]; int car_number char car_type; }; diesel, petrol; main() { Program to enter details for structure variables cout<<" Enter Car Name"; gets(diesel.car_name); cout<<" Enter Car type"; gets(car_type); cout<<" Car number"; cin>>petrol.car_type; }
- 240. 225 Part I: Chapter 6 Structures Q7) State whether the following statements are true or false. a) Nested structures are collection of structures having same data types. b) Structures are similar to two-dimensional arrays. c) The name of the structure is also known as structure tag. d) (.) operator is used to create structure variables. e) One can create recursive structures. Q8) How are structures more advantageous than arrays explain giving suitable examples? Q9) Create a structure of type ‘time’ that contains three integer members: ‘hour’, ‘minute’ and ‘second’. Then write a program that should take in time from the user in the following format ‘hrs.min.sec’ and display this time entered by the user on the screen. Q10) Using the following structure definition for structure ‘coordinate’ given below, Write a function which takes a structure of type ‘coordinate’ as a parameter it swaps the x and the y coordinate of the structure variable without using any other external variables and returns a structure of type coordinate. struct coordinate { int x; int y; }; Hint: Take 2 variables A and B, we can swap there values without using any other variable using the following method: Use similar trick to solve this problem. 4 Mark/ 5 Mark Q11) Write a program (WAP) to create a structure of type ‘Team’ given below. Write user defined functions to: a) Input values of all the members of the structure for ‘n’ elements of the array. b) Display function to show data of all the cells entered separately in tabular form A = A+B B = A-B A = A-B
- 241. 226 226 struct team { char name[20]; int num; int age; float avg; char category; }; Q12) Write a program to create an array of given structure ‘Train’: struct Train { int Train_No; char Train_Name[10]; int Train_fare; }; Write user defined functions to: a) Sort the array in ascending order of Train_fare, Train_No b) Print the sorted array in tabular form Q13) Write a program to define a structure called ‘Scholarship’ with the following data members. i) Name of the scholarship: string ii) Type of scholarship : char type iii) Number of scholarships : short Int type Declare an array of the structure with ‘N’ elements write user defined functions for the following: a) Input: To read in structure members from the user b) Display: To display the contents in the array in a tabular form c) Display_Selective: To display the details of all the Scholarships whose number is greater than 10. d) Display: To display the name of the scholarship having the highest occurrences. Q14) Declare a structure Air_Canada having the following members: Flight_Name : A character array Flight_Fare : A long int Flight_No: ‘int’ Type:
- 242. 227 Part I: Chapter 6 Structures Write a menu driven program to create an array of the above structure and read in values for these members from the user. Write a function and pass the structure using a pointer to the structure and search for an entry with flight name “AC084’. 6.10 Programming Project Complex numbers are an extension of real numbers, in which all non-constant polynomials have roots. Every complex number consists of a real part and an imaginary part. They have the form a + bi where a and b are real numbers and i satisfies the equation i2 = -1. Multiplication is denoted by ( a+bi )( c+di ) = ( ac-bd ) + ( ad+bc ) i , and addition is denoted by ( a+bi ) + ( c + di ) = ( a+c ) + ( b+d ) i. Given the structure Complex: struct Complex { float real; float img; }; Write a program which consists of user defined functions to perform the following functions with the help of a menu driven program. (i) Accept: To accept values into an array of type ‘Complex’ for a set of 10 complex numbers. (ii) Add: To add any two complex numbers passed to a function the function should return a structure variable (iii) Multiply: To multiply any two complex numbers passed to the function and return the resultant complex number (iv) Sort: To sort the list of complex numbers present in the structure array and print the sorted list on the screen (Sorting Key: Real Part). (v) Largest: To print the largest complex number on the screen (vi) Insert: To insert a complex number (Using Pointers)
- 243. 228 228 6.11 Let us revise! Structures are collection of heterogeneous type of data Structures are very similar to arrays the only difference being that arrays are only collection of same type of data Structures are defined using the following format struct <tag/name> { data type var1; … data type var4; }; Key word ‘struct’ introduces the structure definition which is followed by the structure name. The body or the list of variables making up the structure are enclosed in {} brackets. The structure definition is terminated using the (;) operator. The tag name of the structure is optional. If the structure is defined without a tag name, the structure variables should be defined in the structure definition. It is also possible to create array of structures which are very similar to two dimensional arrays. The major use of arrays of structures is to create databases. Arrays of structures can be created by creating structure variables as arrays. Structure variables and their members can also be passed to functions. They can be passed by value or passed by reference. In addition to this, we can also have functions having structures as there return types. Structure variables are passed by reference by passing their address in the call to the function. Arrays of structures are automatically passed by reference. It is also possible to embed a structure within another structure. To reference variables of the nested structure the (.) operator is used. Structures are a powerful and an efficient way of grouping data from different fields into a single entity.
- 244. PART I CHAPTER 7 POINTERS AIM OUTLINE • To introduce the concept of ‘Memory Management’ • To understand the concept of ‘Pointers’ • Learn to use ‘Pointer’ and ‘Pointer Operators’ • Comparison between ‘Arrays’ and ‘Pointers’ • To introduce and use concept of ‘Dynamic Memory Allocation’ using the ‘new’ and ‘delete’ operators. 7.1 Introduction 7.2 Addresses and Pointers 7.3 Pointer to an Array 7.4 Pointer to Pointers 7.5 Pointers with Functions 7.6 New and Delete operators 7.7 Array of Pointers 7.8 Pointer to a Structure 7.9 Solved Examples 7.10 Programming project 7.11 Let us revise
- 245. 230 7.1 Introduction Pointers are among C++’s most difficult concepts to understand and master. In the earlier chapters, we have already seen how variables are stored in the computer memory in a series of memory cells that can be accessed by the use of identifiers. What is though more important to know is the fact that the computer memory is nothing but also a series of successive 1 byte cells, and each of these cells is assigned a unique address? We can compare the structure of the computer memory to a street in a city. In a street every house has a unique address for example, say 110 Winston Street which can be compared to an identifier in programming terminology. Now if want to find this house we can easily find this house without any trouble since there is only one house by the address of 110 Winston street and it must lie between 109 and 111. Similarly, in a computer each memory location has a unique address. The addresses are in ascending order (in succession of 1 Byte), so if we wish to locate memory location say 110, the operating system knows that there is only one location with that specific address and that it lies between 109 and 111. Pointers are nothing but variables that hold memory addresses in C and C++. They provide much power and utility for a programmer to access and manipulate data in ways not seen in some other languages. Pointers enable us to create data structures, such as, linked lists, stacks, queues which grow and shrink in size during execution of programs. 7.2 Addresses and Pointers Till now we have worked with variable types which store characters, integers and floating point numbers. Addresses are also stored in a similar manner with the use of pointers or pointer variables. A pointer that holds an address of a variable is called a pointer variable. Look at the following statement: In this statement, we are declaring a pointer variable where ‘ptr’ is the name of the variable. The '*' informs the compiler that we want a pointer variable, i.e., to set aside however many bytes is required to store an address in the memory. The int says that we intend to use our pointer variable to store the address of an integer and such a pointer variable is called as a pointer to an integer. We have already learnt that as soon as a variable is declared it is stored in a distinct location in adjacent cells (the memory). In some cases, it might be interesting to know the exact location of where this variable is stored. C++ makes this possible with the use of the ‘address of’ operator. It is a unary int *ptr;
- 246. 231Part I: Chapter 7 POINTERS operator and it is used by preceding a variable identifier by an ampersand sign ( ). Let’s look at the following statement: This statement would assign to the pointer variable ‘address’ the address of the variable ‘unknown’. It is important to know that whenever one sees an ampersand (&) sign preceding the name of a variable, it implies that we are not referring to the content stored in the variable but to the address of the variable at which it is stored in the memory. This is further explained in Figure 7.1 given below: Figure 7.1: Address of operator Now that we know what pointer variables are let’s look at the following program and try to understand its working: int *address; address = &unknown; value = unknown;
- 247. 232 Program 7.1: // Pointer Examples #include <iostream.h> #include <conio.h> void main () { int num1 = 5, num2 = 15; int * ptr; ptr= &num1; *ptr = 0; ptr = &num2; *ptr = 2; cout << " Number 1:>" << num1 << "/ Number 2:>" << num2; } Number 1:>0 / Number 2:>2 When this program is executed, num1 and num2 have the values of 5 and 15 initially. As the program goes on, we assign to the pointer variable the address of num1 using the address of (&) operator. Then, we have put value ‘0’ in the memory location pointed by ptr, which is pointing to the address of num1, so we have modified num1 indirectly. Similarly, we change the value stored in num2 indirectly using the same pointer. To further understand the concept of pointers lets look at a more complicated program. Program 7.2: // Pointers Examples # include <iostream.h> # include <conio.h> void main () { int num1 = -5, num2 = 0; int *ptr1, *ptr2; ptr1 = &num1; // ptr1 = address of num1 ptr2 = &num2; // ptr2 = address of num2 *ptr1 = 0; // value pointed by ptr1 = 0 *ptr2 = *ptr1; // value pointed by ptr2 = value pointed by ptr1 Number 1:>0 / Number 2:>2
- 248. 233Part I: Chapter 7 POINTERS ptr1 = ptr2; // ptr1 = ptr2 (value of pointer copied) *ptr1 = 2; // value pointed by ptr1 = 2 cout << " Number 1:>" << num1 << "/ Number 2:>" << num2; } It is easy to understand the working of the program by reading the comments assigned to the program statements. The important thing to note is that Program 7.1 & 1.2 generate the same output. PROGRAMING TIPS FOR POINTERS While learning pointers, the most important advice that can be offered to any beginner is that draw a lot of Figures. The relationship between the address of a cell and the value it contains is easier to understand if one can see that relationship depicted on a page. The following simple rules will help: Draw boxes for every variable in the part of the program that one is trying to understand. The positioning of these boxes is immaterial but the only thing to keep in mind is that these array elements are laid out contiguously in memory. Thus, if working with an array variable, one should draw that array as a connected series of boxes. All pointer variables occupy 4 bytes of computers memory. Assign arbitrary address numbers to the boxes drawn. It doesn’t matter whether one knows what the real addresses are, just invent some for the purpose of understanding what is going on in the program. As you assign addresses, remember that variables require different amounts of memory depending on their type. That is an integer variable type will occupy two bytes of memory where as a float type will occupy 4 bytes of memory. As an example, let us look at the following piece of code. int x, y, *z, *w; z=&x; w=&y;
- 249. 234 One should be able to create the following diagram using the steps given: Draw boxes for integer variables ‘x’ and ‘y’ and give them any arbitrary addresses (1000, 1002) in our case. Draw similar boxes for pointer variables ‘z’ and ‘w’ having addresses of x and y respectively stored in them. Introduce arrows between the pointer variables and those variables whose addresses are stored by these pointer variables. Now look at the following Figure: Figure 7.2: Pointer Diagrams This Figure also known as ‘Pointer Diagram’ depicts that ‘x’ and ‘y’ are integer variables having addresses ‘1000’ and ‘1002’ respectively. It also shows that ‘z’ and ‘w’ are pointer variables having the addresses of ‘x’ and ‘y’ stored in them. The arrows marked denote the relationship between ‘x’, ‘z’ and ‘y’, ‘w’. Let us now come to a slightly different problem. What happens if we want to access the content stored in the memory block to which a pointer points to? To solve this problem one can take help of the indirection operator (*). For example, Now when we use the indirection operator with the pointer name except when we are declaring the pointer then we are referring to the memory block whose address this int temp; int value; int *ptr; ptr = &value; temp = *ptr;
- 250. 235Part I: Chapter 7 POINTERS pointer stores or in other words to the memory block to which this pointer points to. So by initializing ‘temp=*ptr’, what we are actually doing is pointing to the address stored in the pointer ‘ptr’ which is the integer variable ‘value’. The net result of this statement is that what ever value is stored at this location to which the pointer ‘ptr’ points to is stored in the temporary variable ‘temp’. Let us take-up the following example to under stand this concept better. 7.2.1 Solved Examples Example 7.2.1.1 What will be the output for the following piece of code? int amt = 900; int *p; p=&amt; amt+=100; (*p)-=50; cout<<”amt=”<<amt<<” *ptr=”<<ptr<<endl; cout<<”&amt=”<<&amt<<” p=”<<p<<endl; Solution of 7.2.1.1 The above program will display the following output as ‘p’ holds the address of ‘amt’ and hence any change in ‘amt’ will be same as change in *p. OUTPUT SCREEN Now let’s review the ‘Pointer diagrams’ and learn how to use them with the indirection operator with the help of the following example. PROGRAMING TIP EXAMPLE Draw pointer diagrams and hence determine the output for the following piece of code. # include<iostream.h> # include<conio.h> amt=950 *ptr=950 &amt=0*8ffebab4 p=0*8ffebab4
- 251. 236 void main() { clrscr(); int a=10, b=20; int *ptr1, *ptr2; ptr1=&a; (*ptr1)++; ptr2=&b; (*ptr2)++; *ptr1= *ptr2; cout<<*ptr1<<" "<<*ptr2<<" "<<a<<" "<<b; getch(); } We start with the basic steps, i.e., first make boxes for the integer variables ‘a’ and ‘b’ containing values 10 and 20 having any arbitrary addresses. Step 1: Now make similar boxes for the pointer variables ‘ptr1’ and ‘ptr2’ : Step 2: Now take a look back at the piece of code provided to us and observe the following statements
- 252. 237Part I: Chapter 7 POINTERS ptr1=&a; (*ptr1)++; In the first statement, we assign the address of the integer variable ‘a’ to the pointer variable ‘ptr1’. In other words, the pointer is now pointing to the integer variable ‘a’. In the second statement, we have used the indirection operator, which signifies that we are now referring to the variable to which the pointer variable ‘ptr1’ is pointing to and incrementing it by a value of ‘1’. To make things look even simpler one can also replace ‘*ptr1’ by ‘a’ as both the operations will produce same result. So we will observe the same result even if we replace the statement (*ptr1)++; by a++; as the pointer ‘ptr1’ is currently pointing to the integer variable ‘a’. Let us now make a pointer diagram for these statements: We can now follow the same steps for these statements and arrive to the following pointer diagrams. ptr2=&b; (*ptr2)++;
- 253. 238 Step 3: Now in the next statement: *ptr1= *ptr2; We have assigned the value stored in the integer variable to which the pointer ‘ptr2’ is pointing to the value stored in the variable to which the pointer ‘ptr1’ points to. In simple terms, we can also say that we have assigned the integer variable ‘a’ the value of integer variable ‘b’ indirectly and the statement can also be written as a= b; as long as ‘ptr1’ points to ‘a’ and ‘ptr2’ points to ‘b’ as also clear in our pointer diagram. Step 4: In the last step, we just print the values stored at the specified locations provided in the program on the screen using our pointer diagram, to observe the following output. Output Screen By now it must be clear how useful pointer diagrams are, they make pointers look simple and hence one should always use them to determine output of programs. 21 21 21 21
- 254. 239Part I: Chapter 7 POINTERS 7.2.2 The ‘size of’ operator The ‘sizeof’ operator is used to calculate the size of an object/variable belonging to any data type, structure or class. The ‘sizeof’ operator yields the size of its operand with respect to the size of type char. It has the following syntax: Syntax: sizeof ( object/variable_name ); The result of the sizeof operator is of type size_t, an integral type defined in the include file <STDDEF.H>. For example, … int var; cout<<sizeof(var); When these statements are executed the program will display ‘4’ on the screen. The sizeof operator can also be applied to arrays. In such a case the size of operator will return the size of the entire memory block occupied by the array passed to it as a parameter. For example, … float var[10]; cout<<sizeof(var); As expected when the following statements are executed the integer ‘40’ is displayed on the screen (4 * ‘10’ bytes). 7.2.2.1 Importance of the ‘sizeof’ operator To understand the importance of the ‘sizeof’ operator look at the following program: // Header Files # include <iostream.h> # include<conio.h> / // Main Function void main() { Test test1; cout<<sizeof(test1);
- 255. 240 } / Structure Definition struct Test{ int var1; char var2; }; Now when this program is executed one should expect ‘5’ to be displayed on the screen as the structure definition for structure ‘Test’ has an integer variable (‘4’ bytes) and a character variable (‘1’ byte) but to our surprise we notice that a number ‘8’ is displayed on the screen. Now this is not the compilers fault, but this is because of a concept known as ‘data wrapping’. A concept introduced briefly in the previous chapter. What happens is that the compiler always tries to align all the variables to the word boundary ‘4 or multiples 4 bytes’ for better performance of your programs. So if the amount of space occupied by a variable is say ‘5’ the compiler then aligns it to the nearest multiple of ‘4’ which happens to be ‘8’ in this case. The overall understanding of this concept is however out of the scope of the book but with this example, it should be clear now as why we need the ‘sizeof’ operator. 7.3 Pointer to an Array A pointer can also be used to point to an array. A pointer, which stores the address of an array, is called as a pointer to an array. In fact, there is a close relationship between pointers and arrays. An array name can be thought of as a constant pointer. Actually, the identifier of an array is equivalent to the address of its first element, like a pointer is equivalent to the address of the first element that it points to. Look at the following piece of code: Now the array name ‘num’ is a constant pointer. As we already know that arrays are self reference parameter so while using a pointer to point to an array we do not use ‘The address of (&) operator’ as we have done in the case of variables earlier in this chapter. The following statement sets the pointer ‘ptr’ to point to the array ‘num’: This statement is equivalent to taking the address of the first element of the array ‘num’ as follows: int num[10]; int *ptr; ptr = num;
- 256. 241Part I: Chapter 7 POINTERS Let us now review the comparison between arrays and pointers. At this point the pointer ‘ptr’ and the array ‘num’ are exactly equivalent and they have the same properties, the only difference is that we could assign another value to the pointer ‘ptr’ as it is a pointer variable whereas ‘num’ will always point to the zeroth index of the array ‘num’. COMMON PROGRAMMING ERRORS • Ignoring the ‘*’ variable while declaring a pointer variable will result in a logical error. • Not dereferencing a pointer to obtain a value to which the pointer points to will result in a syntax error . 7.3.1 Pointer expression and pointer arithmetic Pointer arithmetic is quite different than conventional arithmetic. To begin with, only addition and subtraction operations are allowed to be conducted, the others make no sense in the world of pointers. But both addition and subtraction have a different behavior with pointers according to the size of the data type to which they point. The important ones which are commonly used are the increment (++) and the decrement (--) operators. Let’s take up an example to understand the use of these operators with pointer variables. Assume an array say, int x[5] and say that it is stored at a position 100 in the memory. Now as the data type is ‘int’ so each variable of this type would occupy 2 bytes of memory (see Figure 7.3). We can now initialize the pointer ‘ptr’ to point to the array ‘x’ with either of the equivalent statements. int *ptr=num; or int *ptr; ptr=&num[0]; int *ptr=x; ptr = x; ptr = &x[0];
- 257. 242 Figure 7.3(a): Pointer to an array In simple arithmetic 100 + 2 would give 102 but this is not the case when one is operating using pointer arithmetic. When using pointer variables as operands then in such a case they are not just incremented or decremented by the value of an integer but size of object the pointer refers to times the value of the integer. So in the following statement ptr+=2; it would produce final address as 104 (100 + 2*2) and not 102 and hence after this statement the pointer will now point to X[2]. Figure 7.3 (b): Pointer to an array An array pointer can also be alternatively referenced with the following pointer expression. * ( ptr + 2) Note: One should remember to use parenthesis as the precedence of operators that is * is of higher precedence than +. Let us better understand pointer arithmetic with the help of the following example:
- 258. 243Part I: Chapter 7 POINTERS Program 7.3 // Program 7.3: Pointer to an array // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { int x[5]; // Pointer Variable ’ptr’ int * ptr; ptr = x; *ptr = 10; ptr++; *ptr = 20; ptr = &x[2]; *ptr = 30; ptr = x + 3; *ptr = 40; ptr = x; *(ptr+4) = 50; for (int y=0; y<5; y++) { cout << x[y] << ", "; } } 10, 20, 30, 40, 50, The working of this program is explained with the help of Figure 7.4.
- 259. 244 Figure 7.4: Pointer Arithmetic COMMON PROGRAMMING ERRORS • Using Pointer arithmetic with a pointer to refer to an address which does not exist • Using ‘Multiplication’ or ‘Division’ with pointer variables is not valid.
- 260. 245Part I: Chapter 7 POINTERS • Forgetting to use parenthesis when undergoing pointer arithmetic will result in an logical error. For Example: *ptr + 3; instead of *(ptr +3); 7.4 Pointers to Pointers ‘Pointers to Pointers’, the topic name might seem very complicated but in reality it is very simple. In order to create pointers to a pointers all we need to do is add another asterix (*) for each level of reference. This concept is explained through C++ statements in Program 7.4. Program 7.4 We can also explain this concept with the use of memory addresses (see Figure. 7.5). Suppose that a, pa, ppa have the following memory locations 100, 200, 300. Figure 7.5: Pointers to pointers Then, ppa is a variable of type (char **) having value ‘200’ *ppa refers to the integer value stored in variable ‘pa’ i.e. 100 **ppa refers to the character value stored in variable ‘a’ i.e. 'I' char a; char *pa; // Pointer to ‘a’ char **ppa; // Pointer to pointer ‘pa’ a = 'I'; pa = &a; // Pointer ‘pa’ stores address of character variable ‘a’ ppa = &pa; // Pointer ‘ppa’ stores address of character variable ‘pa’
- 261. 246 7.4.1 Solved Examples Example 7.4.1.1 Find the output for the following piece of code: # include <iostream.h> # include <conio.h> # include <ctype.h> # include <string.h> void newtext (char string [], int &position) { char *pointer = string; int length = strlen (string); for ( ; position <length -2; position +=2 , pointer ++) { *(pointer + position ) = toupper (*(pointer + position)); } } void main () { clrscr(); int location = 0; char message [] = “Dynamic Act”; newtext (message, location); cout<<message<<” # “<<location; } Solution of 7.4.1.1 OUTPUT SCREEN DynAmiC ACt# 10
- 262. 247Part I: Chapter 7 POINTERS Example 7.4.1.2 Give the output of the following program segment # include<iostream.h> # include<conio.h> # include<ctype.h> void main() { clrscr(); char *s = "MeTropoliTan"; for ( int x =0; s[x] != '0';x++) { if (x%2!=0 && islower(s[x])) s[x] = toupper(s[x]); else if (isupper(s[x])) s[x] = tolower (s[x]); else s[x]= s[x]--; } cout<<s<<endl; } Solution of 7.4.1.2 In this piece of code‘s’ is declared as a pointer variable pointing to the first character of the string ‘MeTropoliTan’. The program converts some of the characters to upper case and some to lower case depending upon the condition provided in the if else loop construct. The working of this program should be easily understood as we already observed similar problems above. OUTPUT SCREEN mEtRoPoLitaN
- 263. 248 7.5 Pointer with functions As we have already studied in the earlier chapters that there are two ways to pass arguments to a function ‘Call by Value’ and ‘Call by reference’. In C++ ‘Call by reference’ is practiced by using either the address of operator or by the use of the pointers. In part I chapter 3 if we recall we were using the address of operator (&) when we were calling functions by reference. What we were actually doing though was passing the address of the variable to the function and hence reflecting any changes made in the formal parameters to the actual parameters. Call by reference can also be replicated by the use of pointers in C/C++. The programs 7.5 (a), (b) present similar functions being called by reference by using the address of operator (&) in the 1st case and using the indirection operator (*) in the other. The important thing to note is that both these programs are equivalent and give the same result. Program 7.5 (a) (b) Using Function with Pointers Using Functions as Reference // Header Files # include<iostream.h> # include<conio.h> # include<string.h> # include <stdio.h> /*************************** Name: pointer_example Arguments: char *ptr, int length ***************************/ void pointer_example( char *ptr, int length); // Main Function void main() { // Header Files # include<iostream.h> # include<conio.h> # include<string.h> # include <stdio.h> /********************** Name: reference_example Arguments: char string[] **********************/ void reference_example( char string[]); // Main Function void main() { clrscr(); char string_actual[15];
- 264. 249Part I: Chapter 7 POINTERS clrscr(); char string[15]; int length_string; cout<<" Enter String (<15 )"<<endl; gets(string); length_string=strlen(string); // Call to the function pointer_example pointer_example(string, length_string); getch(); } // Function Definition and Body void pointer_example( char *ptr, int length ) { for (int i=0; i<length; i++) { cout<<*ptr; ptr++; } } cout<<" Please enter the string (<15 characters>)"<<endl; gets(string_actual); // Call to the function reference_example reference_example(string_actual); getch(); } // Function Definition and Body void reference_example(char string_formal[]) { puts(string_formal); } When these programs are executed, one would notice that they result in the same outcome. In fact, both these programs are equivalent piece of code. Let us try to understand why is it so? In program 7.4 (a) if we notice we have used a character pointer in the function definition for function ‘pointer_example’. When the character array ‘string’ is passed to it as a parameter. The pointer points to the 1st character of this character array ‘string’. When we enter the function body we fall into a for loop which executes till the length of the string passed to it and the pointer ‘ptr’ is incremented every time in each iteration and hence the whole string is printed on the screen as a result of this loop (Figure III1.2) loop. In program 7.4 (b) the same thing happens the only difference being that in this case we have passed the character array as a reference parameter using the address of operator (&). PROGRAMING TIPS Always use parenthesis to avoid compilation and logical errors when using pointer arithmetic. Pointer arithmetic is different from conventional arithmetic. In pointer arithmetic only ‘addition’ and ‘subtraction’ using operators is allowed. Pointers and arrays are very similar concepts Using the address of (&) operator is same as using pointers.
- 265. 250 7.6 New and Delete Operators To understand the use of the new and delete operator let us look at the following array program 7.6. Program 7.6 All this program does is input a name from the user and then print that name in the next line. What is though more important to observe is that when we declare the character array string with an index value 15. The computer reserves space for 15 characters. Now when we input the string say ‘Computer’. It has 8 characters and should occupy 8 characters but instead it would too use up all the space assigned for 15 characters as at time of declaration for this string we declared it for a set of 15 characters. Hence we are wasting the rest space for 7 characters and hence the computers memory. C++ does not allow us to specify the array size at the time of program execution. The compiler requires the array size to be a constant. Hence, while declaring an array you need to know how big the array is likely to be which may always not be possible as we just saw in the previous program. C++ overcomes this problem by providing the ‘new’ and the ‘delete’ operators which allow us to use the computer memory efficiently. 7.6.1 The New Operator ‘New operator’ in C++ returns the addresses of a block of unallocated bytes (depending on data type a pointer is pointing to). In simple words it helps in obtaining blocks of // Program Prints the Name of a person // Header Files # include <iostream.h> # include <conio.h> # include <stdio.h> // Main Function void main() { clrscr(); char string [15]; cout<<" Enter the Name of the person "; gets (string); puts(string); }
- 266. 251Part I: Chapter 7 POINTERS memory returning a non zero pointer pointing to its starting point.The following program shows how the new operator can be used in C++. Program 7.7: // Header Files # include <iostream.h> # include <conio.h> # include <stdio.h> // Main Function void main() { clrscr();// Clear Screen char *ptr; // Pointer to char int size; cout<<" Enter the number of characters in the persons name"; cin>> size; ptr = new char [size+1]; // Set aside memory: size cout<<endl<<" Enter the Name of the person "; cin>>ptr; cout<<endl<<ptr; delete [] ptr; // Frees the allocated block of memory getch(); } Let us take a look at this expression: ptr = new char [size+1]; returns a pointer to a section of memory equal to size bytes just large enough to hold the name of the person to be inputted. By adding +1 to the size we also make sure that it has enough space to even include the null operator ‘0’ at the end of the string. 7.6.2 The Delete Operator As we know that the computers memory is limited therefore while working with these operators one has to be careful of the amount of memory a program occupies. As if a
- 267. 252 program uses large chunks of memory then all the available memory will be reserved for the program and the program will eventually crash. So for safe and efficient use of memory, the ‘new’ operator has a counter part known as the ‘delete’ operator which returns all the memory blocks allocated to the program using the ‘new’ operator back to the operating system. Syntax: Let us look at following statement in program 7.7: delete [] ptr; This statement returns to the operating system whatever memory pointed to by our pointer ‘ptr’. The square brackets tell the compiler that we are deleting the entire array block. 7.7 Array of pointers An array, who’s each element is a pointer type, is known as an array of pointers. Arrays of pointers are used to create arrays of strings. Each entry in the array is a string and this pointer points to the 1st character of a string. Consider the following definition of a character array ‘flowers’. char * flowers[3] = { “ ROSE”, “LILY”, “LOTUS” }; The highlighted part (char *) indicates that each element of the array flowers is a pointer to char. All entries in the array ‘flowers’ store addresses of these strings Each index entry of this array points to the 1st character of their corresponding string (see Figure 7.6). Figure 7.6: Array of pointers We can now also look at Program 7.8 which also shows the use of array of pointers in C++. delete pointer_name;
- 268. 253Part I: Chapter 7 POINTERS Program 7.8 # include <iostream.h> # include <conio.h> void main() { clrscr(); int *a[3]; // Declaring an array of pointers for (int i=0;i<3;i++) { a[i] = new int; // Allocating block of memory for the character array *a[i]= i * 25; } for (i=2;i>=0;i--) cout<<"*A["<<i<<"]="<<*a[i]<<endl; for (i=0;i<3;i++) delete a[i]; // De allocating block of memory for the character array getch(); } In this program, a[i] is an array of pointers and each element of this array a[0], a[1], a[2] points to a memory block of type int. Again each element of this just stores the pointers which point to their respective memory blocks. ‘Array of pointers’ displays the data structuring capabilities of C++. We will further learn about them in later chapters. 7.8 Pointer to a structure A pointer which stores the address of struct type data is known as pointer to structure. An example of a pointer to a structure is shown in the following program 7.9. Program 7.9 # include <iostream.h> # include <conio.h>
- 269. 254 struct graph OUTPUT SCREEN { int x, y; }; void main() { clrscr(); grpah *g; g = new graph; // *g.x = 24; Not allowed // g.*x = 24; Not allowed // g.x = 24; Not allowed g->x=24; g->y=g->x*2-2; cout<<" G->X="<<g->x<<" G->Y="<<g->y<<endl; delete g; getch (); } When using pointer to a structure the only difference is that we do not use the deference operator instead we use a ‘–‘ and a greater than ‘>’ to refer to the values pointed by the pointer ‘g’. Hence if say we are referring to the data member ‘x’ in the structure ‘graph’ the following c++ statements are invalid. *g.x = 24; // Not Valid g.*x = 24; // Not Valid g.x = 24; // Not Valid One has to use the ‘->’ to refer to data member ‘x’ of the structure as also shown in the program above. G->X=24 G->Y=46
- 270. 255Part I: Chapter 7 POINTERS 7.9 Solved Examples Example 7.9.1 Find the output for the following piece of code: # include <iostream.h> # include <conio.h> void main() { clrscr(); int a = 100, *b=&a; cout<<*b<<a<<endl; (*b)+=10; a-=100; cout<<*b<<a<<endl; int *c; c = new int; *c=100; *c=a; a+=100; cout<<*c<<a<<endl; c=b; (*b)+=90; a+=10; cout<<*c<<*b<<a<<endl; getch(); } Solution of 7.9.1 OUTPUT SCREEN 100100 1010 10110 210210210
- 271. 256 Example 7.9.2 Consider the given declarations and give the outputs. All parts of code are independent of each other. # include <iostream.h> # include <conio.h> void main() { clrscr(); int s=10, t=90, *u, *v; //(i) u=&s; v=&t; cout<<*u<<*v<<t<<s<<endl; //(ii) u = &t; v=u; (*v)++; s=s+t; t=t+s; cout<<s<<t<<*u<<*v<<endl; //(iii) s++; --t; v=&t; cout<<s<<*v; u=v; v=&s; s=t-s; cout<<*u<<*v<<endl; getch(); }
- 272. 257Part I: Chapter 7 POINTERS Solution of 7.9.2 OUTPUT SCREEN Example 7.9.3 What will the user see when the following piece of code is executed. (i) # include <iostream.h> # include <conio.h> struct point { int x,y; }polygon[]={{1,2},{1,4},{2,4},{2,2}}; void main() { point *a; a= polygon; a++; a->x++; cout<<polygon[0].x<<polygon[0].y<<endl; cout<<a->x<<a->y<<endl; getch(); } (ii) # include <iostream.h> # include <conio.h> void main() { clrscr(); int x=20, y=30, *b=&x; (*b)+=15; cout<<x<<*b<<endl; i) 10909010 ii) 101192192192 iii)11898978
- 273. 258 x++; cout<<*b<<x<<endl; b=&y; y+=x; int *c=b; b=&x; y+=12; x*=4; cout<<*b<<*c; getch(); } (iii) # include <iostream.h> # include <conio.h> void main() { clrscr(); int a[]={2,4,6,8,10}; int *k,*p1; p1=a; k = new int; (*p1)+=10; *k=55; (*p1)++; *p1=*k; (*k)*=2; cout<<*p1<<*k<<endl; getch(); } (iv) # include <iostream.h> # include <conio.h> void main() { clrscr(); char *p = "Difficult"; char c; c=++*p++; cout<<c<<*p<<endl; getch(); } (v) # include <iostream.h> # include <conio.h>
- 274. 259Part I: Chapter 7 POINTERS void main() { clrscr(); int a=32, *ptr=&a; char ch='A'; char &f= ch; f+=a; *ptr+=ch; cout<<a<<" "<<ch<<endl; getch(); } Solution of 7.9.3 (i) Output Screen (ii) Output Screen (iii) Output Screen (iv) Output Screen (v) Output Screen 12 24 3535 3636 14478 55110 Ei 129 a
- 275. 260 Example 7.9.4 Write a program which reads in a group of 10 numbers and places them in an array of type int. It then calculates the sum of all the odd numbers in the set of these 10 numbers and prints the odd numbers and their sum on the screen. Use pointer notation where ever possible. Solution of 7.9.4 # include <iostream.h> # include <conio.h> void main() { clrscr(); int *ptr[10]; int sumodd=0; int flag=0; cout<<" Enter the 10 numbers in the array"; for (int i=0; i<10;i++) { ptr[i] = new int; cin>>*ptr[i]; } for (i=0; i<10;i++) { if (*ptr[i]%2!=0) { if (flag==0) cout<<" The odd numbers are"; sumodd = sumodd + *ptr[i]; cout<<*ptr[i]<<endl; flag++; } } cout<<" The sum of odd numbers in the array is"<<sumodd; getch(); }
- 276. 261Part I: Chapter 7 POINTERS Example 7.9.5 Declare an array of pointers to strings representing the 5 week days. Also, declare a function to sort these strings in alphabetical order using bubble sort. Sort the pointers to the string as and not the actual strings? Solution of 7.9.5 // Header files # include <iostream.h> # include <conio.h> # include <stdio.h> # include <string.h> // Function: b_sort // Purpose: To sort the array passed to it as a paramter void b_sort(char *str_pointers[]) { char *temp; for (int j=0;j<5;j++) for (int i=0;i<4-j;i++) { if (strcmp (str_pointers[i],str_pointers[i+1])>0) { temp = str_pointers[i+1]; str_pointers[i+1]=str_pointers[i]; str_pointers[i]=temp; } } } // Main Function void main() { char *weekdays[5]={"Monday","Tuesday","Wednesday","Thursday","Friday"}; puts("nWeekdays: Before Sorting"); for (int i=0; i<5; i++) { puts(weekdays[i]); } b_sort(weekdays); puts("nWeekdays: After Sorting"); for (i=0; i<5; i++)
- 277. 262 { puts(weekdays[i]); } getch(); } Example 7.9.6 Find the output for the following piece of code: # include <iostream.h> # include <conio.h> void main() { clrscr (); int *ptr; int number = 5; ptr = &number; cout<<ptr<<endl; cout<<&number<<endl; cout<<*&ptr<<endl; cout<<*(*&ptr)<<endl; cout<<*&*&*(*&ptr)<<endl; getch(); } Solution of 7.9.6 OUTPUT SCREEN 0x1fa70ffa 0x1fa70ffa 0x1fa70ffa 5 5
- 278. 263Part I: Chapter 7 POINTERS 7.10 Review Exercise 1 Mark/ 2 Mark Questions Q1) A pointer stores a. an address of a variable b. the values of the variable it points to c. is a set of unallocated bytes of memory d. a data type of an address variable Q2) The expression *post; a. is a pointer to the variable post b. holds the address of variable post c. refers to the contents of the post. d. refers to the contents stored to the variable to which post points to. Q3) Will the following piece of code compile without errors: int ptr; int *ptr1; int **ptr2 ptr1=&ptr; ptr2=*ptr1; ptr = ptr2; Q4) Are the following c++ statements equivalent: i) int ptr[2] and *(int ptr +2) ii) int *ptr; and int *ptr; ptr = &add; ptr = *& add; Q5) Is pointer arithmetic the same as conventional arithmetic. If not how is it different from conventional arithmetic? Give suitable examples? Q6) Write c++ statements i. to input a set of two integers say, ‘a’ and ‘b’ from the user and ii. swap their values using pointers. Q7) Assuming the code given below, find the error in each of the program segments a), b) and c).
- 279. 264 Code: int *ptr, number; int num[5] = { 5,10 ,15 , 20, 25, 30}; ptr = num; Statements: a) // Use pointer to get the address of first value of array. number = ptr; b) // Change the value ’10’ at the index position to 2 of array num to ‘5’ *(ptr+1)=5; Q8) Find the output for the following piece of code. # include <iostream.h> # include <conio.h> struct coordinate { int x int y; }; void display ( coordinate n) { cout<<" Row ="<<n.x<<"t Column = "<<n.y<<endl; } void main() { clrscr(); coordinate A = {20,50}, B; coordinate *c; c=&b; c->x += 20; display (b); c = &a; c->y +=20; display (a); getch(); }
- 280. 265Part I: Chapter 7 POINTERS Q9) Write a program which reads in a group of 10 numbers and places them in an array of type int. It then calculates the sum of all the even numbers in the set of these 10 numbers and prints the even numbers and their sum on the screen. Use pointer notation where ever possible. Q10) Write a program to input an array of strings. The program should also contain a function to convert all the characters of the string to upper case and also count the number of lower case characters. Use pointer notation where ever possible. 3 Mark/ 4 Mark Questions Q1) Find the output for the following piece of code: int arr[] = {2,3,4,5}; int *arrptr=array; int val = *arrptr; cout<< val<<endl; val = * arrptr++; cout<<val<<endl; val = *arrptr; cout<<val<<endl; val = *++arrptr; cout<<val<<endl; Q2) Find errors in the following piece of code: /************************** The program swaps the values of two integers with the help of pointers **************************/ // Header Files # include <iostream.h> # include <conio.h> // Main Function void main () { clrscr (); int *ptr; int num1; int num2; num1=&ptr; num2= *ptr;
- 281. 266 Q3) Create a database for 20 employees in a company, the database should include employee name, employee number, employee designation, employee salary, employee department. Create separate functions to sort the database in ascending order with employee name, employee id and employee salary. Also create function Emp_search to search any employee with employee number as the search key. Use structures and pointer notation where ever possible? Q4) Write C++ statements for the following: i) Initialize a pointer to an integer. ii) Initialize a pointer to a string. iii) Initialize a pointer to an array of 10 characters. iv) Initialize a pointer to a pointer to a character, a floating point. v) Initialize a pointer to the following structure: struct graph { int x; int y; }; Also then initialize x = 5, y =10 using the pointer variable. Q5) Read a sequence of words from the user. Keep reading the sequence till the point the user enters the word ‘STOP’. Print these words in the order of sequence do not print any word starting from the character ‘s’. Also write a function Seq_sort which returns a sorted list of words to be printed on the screen. Use pointer notation where ever possible. Q6) State the difference between an array of characters and a string. How is size of an array of characters different from a string? What is the size of the array ‘str’? char str[] = “This is Question 6”; Declare an array of strings in which the strings contain the names of all the days. Ask the user to input a set of 10 random strings. Print those days in order of occurrence. Use pointer notation where ever possible. 7.11 Programming Project A very popular puzzle which most of us have played when we were young is the ‘10- puzzle’. This puzzle’s frame can hold 10 tiles. It consists of blocks number from digits 1 to 9 and an empty space. The purpose of the game is to slide the blocks. The goal is to slide the blocks using the empty space block such that each block is arranged in the ascending order when observed row wise. For example if ‘X’ represents the empty space
- 282. 267Part I: Chapter 7 POINTERS given a valid initial configuration one should achieve the final configuration as shown in Figure 7.6. Our project is to write c++ a program for which the user inputs the set of 9 integers in the order of sequence. The program should print all the moves to reach the desired configuration of blocks. Use pointers and structures where ever possible. Figure 7.6: The '10's Puzzle
- 283. 268 7.12 Let us revise! The address (&) of operator returns the address of its operand. A pointer that holds an address of a variable is called a pointer variable. The pointer definition is: int *ptr; A pointer, which stores the address of an array, is called as a pointer to an array An array name is a constant pointer and points to the first element of the array. Arrays are automatically passed by reference because the value of the array name is the address of the array. The ‘*’ operator, referred to as the indirection operator when used with the pointer variable returns the value of the object to which it points. Pointer arithmetic is different than general arithmetic. Only addition and subtraction is allowed in pointer arithmetic. In pointer arithmetic, the pointer is just not incremented or decremented by the value of the integer instead it is incremented or decremented by the value of the integer times the size of the data type the pointer points to. The ‘new’ operator in c++ returns the address of a block of unallocated bytes. The ‘delete’ operator in c++ reverses the process of the new operator, by releasing the memory location from a pointer. We can also have pointers to pointers. These variables are defined using double or triple asterix (*) depending upon the degree of branching, for example, int ** ptr (for degree of branching 2). An array of pointers is an array who’s each element is a pointer type. A pointer which stores the address of a struct data type is known as a pointer to a structure.
- 284. 269 PPAARRTT IIII OOBBJJEECCTT OORRIIEENNTTEEDD PPRROOGGRRAAMMMMIINNGG
- 286. Part II CHAPTER 1 OBJECT ORIENTED PROGRAMMING AIM OUTLINE • To introduce the concept of Object Oriented Programming (OOP) • To understand the difference between procedural and object oriented programming. • To understand the various features and advantages of OOP • To appreciate the concept of OOP 1.1 Introduction 1.2 Basic features of Object Oriented Programming 1.3 Classes and Objects 1.4 Constructors and Destructors 1.5 Review Examples 1.6 Review Exercise 1.7 Let us revise
- 287. 272 1.1 Introduction As programming languages evolved there occurred many problems with the procedural programming language. In procedural programming language emphasis was on doing things that were functions, it consisted of global variables, and there existed always a possibility of accidental change in data. Hence, a new system of programming was developed which was named as Object Oriented Programming. In short it was termed as ‘OOP’. OOP enables the programmers to group together data and the code that uses data into discrete units. In OOP the stress is given on the object in spite of the procedure. An object is defined by two terms: attributes and behaviors. If you look at a person and visualize him as an object, you will find that a person has attributes, such as, eye color, age, height, and so on. A person also has behaviors, such as walking, talking, breathing, and so on. In its basic definition, an object is an entity that contains both data and behavior. The word both is the key difference between the more traditional programming methodology, procedural programming, and OOP. In procedural programming, code is placed into totally distinct functions or procedures. Ideally, as shown in Figure 1.1, these procedures then become "black boxes," where inputs go in and outputs come out. Data is placed into separate structures, and is manipulated by these functions or procedures. Figure 1.1 Black boxes. 1.1.1 Difference between OO and Procedural Programming In Object Oriented design, the attributes and behavior are contained within a single object, whereas in procedural or structured design, the attributes and behavior are normally separated. As illustrated in Figure 1.2, in procedural programming the data is separated from the procedures, and sometimes the data is global so it is easy to modify data that is outside your scope. This means that access to data is uncontrolled and unpredictable (i.e. several
- 288. 273Part II: Chapter 1 Object Oriented Programming functions may have access to the global data). Further, because you have no control over who has access to the data; testing and debugging are much more difficult. Objects address these problems by combining data and behavior into a nice, complete package. Figure 1.2 Using global data. 1.2 Basic Features of Object Oriented Programming • Emphasis is on data • It follows a bottom-up approach in program design • It has concepts like data hiding which prevents the accidental change of data • It has properties like o Polymorphism o Inheritance o Encapsulation These properties are explained later in this chapter. 1.3 Classes & Objects An object has a similar relationship to a class that a variable has to a data type. In short an object is an instance of a class. A class is a user defined data type used to implement an abstract object. It contains members of three types: ‘private’, ‘public’ or ‘protected’ by default the data members of the class are private. Contents of a class are known as members of the class, data declared inside the class is known are known as data members and functions included inside the class are known as member functions.
- 289. 274 An object is said to be instance of a class, in the same way ‘keyboard’ and ‘mouse’ belong to the class ‘hardware’, or ‘me’ and ‘you’ are instances of a class ‘HUMAN'. Figure 1.3: Classes a collection of data and functions A class with no object is known as an Abstract class and a class having object(s) is known as Concrete class. Every object has its own identity, state and behavior. Identity is the property of the object, which distinguishes it from the other objects. Static and Dynamic properties associated with the object signify the state of the object. The operations/functions associated with the object exhibit the behavior of the object. A class can be declared by using the following syntax: class Class-name { private: // its optional to write private access specifier data-member; data-member; member-function protected: data-member; data-member; member-function public: data-member; data-member; member-function }; : class_name List-of-objects; For example: class HUMAN { int HIEGHT,EYECOLOR,AGE; char TYPE; // M: Man // W: Women
- 290. 275Part II: Chapter 1 Object Oriented Programming public: void input(); void output(); }; : HUMAN Steve, Jordan ; In the above example Steve and Jordan are object declarations. Members can be kept in any of the three accessibility modes private, public and protected. Members in the public visibility mode of the class are accessible to the objects of the class whereas members in the private accessibility mode are not accessible to the objects of the class. They are only accessible inside the class to the member functions of the class. NOTE: Protected members will be explained later in Chapter 2 entitled ‘Inheritance’. Program 1.1 illustrates the use of classes and objects: Program 1.1 /*********************************************************************** Program 1(1).cpp Explaining the concept of classes and objects ***********************************************************************/ // Header files # include <iostream.h> # include <stdio.h> // Declration of class hospital class hospital { private: int patientno,wardno; //private data members char patient_name[20]; public: void register_patient(); //public member function prototype void showstatus() //public member function declaration { cout<<patientno<<":"<<patient_name<<":"<<wardno<<endl; } }; //semicolon is a must
- 291. 276 /*********************************************************************** Name: register_patient Class: hospital Purpose: Input patient information ***********************************************************************/ void hospital::register_patient() { cout<<"patient no ?"; cin>>patientno; cout<<"name ?"; gets(patient_name); cout<<"Ward no ?"; cin>>wardno; } // Main function void main() { hospital p1,p2; p1.register_patient(); p2.register_patient(); p1.showstatus(); p2.showstatus(); } // cin>>p1.wardno; ** not accessible as wardno is a private member // cin>>p1.patientno; **not accessible as patientno is a private member In the above example, one can see that a class Hospital is defined with i. Three private data members patientno, wardno and patient_name; and ii. Two public member functions register() and showstatus(). Member functions can be defined inside the class as well as outside the class as shown in the example, member function showstatus() is defined inside the class whereas the member function register() is defined outside the class by using the scope resolution operator (::) with the name of the class it belongs to. Figure 1.4 pictures the use the of the scope resolution operator. The objects of the class can call both these functions in the same way. Generally single lined functions are defined inside the class and multiple line functions or functions containing control structures (if then else or while loops) are defined outside the class to have clarity in understanding the behavior of the class.
- 292. 277Part II: Chapter 1 Object Oriented Programming Figure 1.4: A function declared outside the class Note: The size of the object (in bytes) depends upon the data members present in the class it belongs to. In the above example the object p1 occupies 24 bytes. COMMON PROGRAMMING ERRORS • Declaration of member functions outside the class definitions without the scope resolution operator (::) will lead to an error. • The class definitions must end with a semicolon. Ignoring this will lead to an error 1.3.1 Friend Function Only a member function of a class can access data declared in the private access specifier in a class declaration as a result of which there is no chance of accidental change of data. However there are cases where the user wants to access data of an object by a function which is not a member function of the class to which the objects belong. This is possible by declaring a function as a friend function of the class. The syntax for such a function is given below class class_name { private: ………………………… //private members ………………………... public: ……………………… // public members
- 293. 278 ……………………... friend data_type function_name(); // Declaration of friend function }; 1.3.2 Objects as arguments Objects like any other data may be passed as arguments to a function. The function call may be called by value or called by reference. In call by value the function cannot effect permanent changes to the data received as argument. However, in call by reference the changes made to the data are permanent. Program 1.2 explains this concept. Program 1.2 /*********************************************************************** Program 2(2).cpp The program displays objects being passed as arguments for class time and objects undergoing basic arithmetic operations. ***********************************************************************/ // Header Files # include <iostream.h> # include <conio.h> class time { int hrs,mins; public: // Function to get time from the user void gettime() { cout<<"Enter the hours !"; cin>>hrs; cout<<" nEnter the minutes !! "; cin>>mins; } void settime (int h ,int m) { hrs=h; mins=m; } // Function to show time on the screen
- 294. 279Part II: Chapter 1 Object Oriented Programming void showtime() { cout<<hrs<<":"<<mins<<endl; } // Function to sum two objects of class time void sum (time t1,time t2) { int minutes=t1.mins +t2.mins; hrs = t1.hrs+t2.hrs +minutes/60; mins=minutes%60; } }; // Main function void main() { clrscr(); time t1,t2,t3; t1.gettime(); // Call to function gettime t2.settime(3,45); // Call to function settime t3.sum(t1,t2); // Objects being passed as arguments cout<<"nt1="; t1.showtime(); // Call to function showtime cout<<"t2="; t2.showtime(); cout<<"t3="; t3.showtime(); getch(); } OUTPUT SCREEN Enter the hours !10 Enter the minutes !! 10 t1=10:10 t2=3:45 t3=13:55
- 295. 280 In this program class time has hrs, mins as its private members and hence cannot be accessed from the function main. Class time also has public member functions through which these private members can be accessed. When this program is executed t1, t2, t3 objects of class time are created. The function ‘gettime’ is invoked for object t1 and the user is asked to enter the hours and the minutes for this object. The values entered by the user ‘10’ and ‘10’ are stored as this objects values for hours and minutes. Similarly for object t2 function settime is invoked and values ‘3’ and ‘45’ are stored as objects value for hours – ‘hrs’ and minutes-‘mins’ respectively. For object t3 however the function sum is invoked in which objects t1 and t2 are being passed as arguments. The values for t3s private members ‘hrs’ and ‘minutes’ is the sum of the values for private members of t1 ‘10’ ,‘10’ and that of t2 ‘3’,’45’ respectively. 1.3.3 Encapsulation To understand the concept of encapsulation better take an example of a simple quadratic equation, All quadratic equations constitute an example of a class. As all quadratic equations have unique identity (degree ‘2’), state (coefficients: A, B and C) and similar behavior (2 roots). In OOPS we create objects by combining data members and data functions for input, output and data manipulation. As it might be obvious each and every object of the class ‘Quadratic Equation’ will have three coefficients (A, B and C) and two roots and in order for data manipulation each object will also have some member functions to read these coefficients, calculating the roots and displaying them on the screen. This method or practice of wrapping up of data and functions in a single unit is known as ‘Encapsulation’. Figure 1.5 given below further enlightens this concept. PROGRAMING TIPS Declare all member function as const if object remains unchanged. This can help to remove many errors. Languages like C++ are still evolving languages. Avoid using loaded words like ‘Object’. Even though “Object” is not a keyword in future might become one. Ax2 + Bx + C = 0
- 296. 281Part II: Chapter 1 Object Oriented Programming Figure 1.5: Data and member functions required by a class quadratic equations 1.3.4 Data Abstraction and Data Hiding Data abstraction is defined as bringing the data down to its essentials. For example, consider that an automobile company has come up with a new design for an automobile to make it fuel efficient. Now before they launch their product in the market they want to make the customers aware of the fuel efficiency that this new automobile will provide them. For instance they want to make the customers aware of the fact that this newly designed automobile will be 40% more fuel-efficient than any other automobile available in the market. However, they do not want to give away the intricate details of the design owing to the fear of the design being replicated by the competitors in the market. Therefore, the management decides that the company should only give away those basic details, which are essential and only includes features like how would this new design enhance the fuel efficiency of the automobile keeping the intricate details regarding the engine design hidden. Now this process in which only the essential features of the new automobile design are highlighted and thus, represented to the outside world without including the intricate features is termed as abstraction. In C++, the private class members initialized remain hidden from the outside world and hence all the intricate details can be stored in these members whereas the public members form the interface by providing the essential and relevant information to the outside world. This process is also termed as data hiding.
- 297. 282 1.3.5 Polymorphism & Function Overloading One of the key feature of OOP is polymorphism. The process of using an operator or a function in different ways for different set of inputs given is known as polymorphism. Function overloading is an example of polymorphism, where the functions with the same name work with different set of parameters to perform different operations. Consider the following example, Now when this piece of code is executed and the compiler reaches the 1st call to the function, function 1 is called as this function call matches the function header of function 1 as it has the same number of parameters. Similarly when the compiler reaches the call for the 2nd function, function 2 is called as the function call matches the function header for the 2nd function. This is an example of function overloading, where same named functions behave differently for different kind of inputs given to them by the user. When defining functions having same names it is becomes absolutely necessary that they should differ in number, type and order. Figure 1.6: Function overloading Function 1 int maximum (int a, int b); Function 2 int maximum (int a, int b, int c) Call to function 1 maximum (10, 12); Call to function 2 maximum (10, 12, 13); float volume (float radius); function definition I int volume (int length, int breadth, char name); function definition II int volume(char type, int length, int breadth); function definition III
- 298. 283Part II: Chapter 1 Object Oriented Programming For example consider figure 1.6. This figure shows three function definitions. These function definitions differ in type, number and order as shown in the following table 1.1. Table 1.1: Function Definitions Function Defintion Number of parameters Type of parameters Order of parameters I 1 radius float datatype float II 3 length, breadth, name int datatype, int datatype, char datatype int, int, char III 3 type, length, breadth char datatype, int datatype, int datatype char, int, int The various functions of these function definitions will behave differently for different types of input given to them. However, sometimes the following also happens. For example, During execution of this program, when the compiler reaches the call to the function maximum the compiler is not sure which function out of above two to invoke, function 1 or function 2 as both the functions 1 and 2 satisfy the function call maximum(10,12); When this type of a situation occurs in which the compiler is not sure about which of the 2 functions to invoke it is referred to as Function Ambiguity. If such a case occurs during a program execution the program will hang during its execution. Another case of function ambiguity may take place when one is using ‘float’ and ‘double’ or ‘int’ and ‘long’ as data types for the local variables in the function header. function 1 int maximum(a, int b); function 2 int maximum(int a , int b , int c=5); maximum(10,12);
- 299. 284 1.3.6 Solved Examples Example 1.3.6.1 Define a class in C++ with the following description Private members • A data member Flight Number of type integer • A data member Destination of type string • A data member Distance of type float • A data member Fuel of type float • A member function CALFUEL() to calculate the value of fuel as per the following criteria Distance Fuel <=1000 500 more than 1000 and 1100 <=2000 more than 2000 2200 Public members A function FEEDINFO() to allow the user to enter values for flight number, destination, distance & call function CALFUEL() to calculate the quantity of fuel A function SHOWINFO() to allow user to view the content of all the data members Solution of 1.3.6.1 class FLIGHT { int Fno; // Represents Flight Number char Destination[20]; // Represents Destination float Distance; // Represents Distance float Fuel; // Represents Fuel void CALFUEL(); public: void FEEDINFO(); void SHOWINFO(); }; void FLIGHT::CALFUEL() { if(Distance<=1000 ) Fuel=500;
- 300. 285Part II: Chapter 1 Object Oriented Programming if(Distance>1000 && Distance<=2000) Fuel=1100; else Fuel=2200; } void FLIGHT::FEEDINFO() { cout<<" Flight No:"; cin>>Fno; cout<<" Destination:"; gets(Destination); cout<<" Distance:"; cin>>Distance; CALFUEL(); } void FLIGHT::SHOWINFO() { cout<<" Flight no:"<<Fno <<endl; cout<<" Destination:"<<Destination<<endl; cout<<" Distance:"<<Distance<<endl; cout<<" Fuel:"<<Fuel<<endl; } Example 1.3.6.2 Define a class TEST in C++ with the following description: Private members • Test code of type Integer • Description of type string • Nocandidate of type integer • Centerreqd (no of centers reqd) of type integer • A member function CALCNTR() to calculate the no of centers as number of candidates/100+1 Public members A function SCHEDULE() to allow user to enter the values of testcode. Description , nocandidate & call function CALCNTR to calculate the number of center A function DISPTEST() to allow user to view the content of all the data members
- 301. 286 Solution of 1.3.6.2 class TEST { int TestCode; char Description[20]; int Nocandidate; int Centerreqd; void CALCNTR(); public: void SCHEDULE(); void DISPTEST(); }; void TEST::CALCNTR() { Centerreqd=Nocandidate/100+1; } void TEST::SCHEDULE() { cout<<" Test code:"; cin>>TestCode; cout<<" Description:"; gets(Description); cout<<"Number:"; cin>>Nocandidate; CALCNTR(); } Example 1.3.6.3 WAP (Write a program) to create an array of objects of class stud class stud { int rno; char nam[10]; float avg; public: void in();
- 302. 287Part II: Chapter 1 Object Oriented Programming void out(); }; Write a user defined function to sort the array of class stud in ascending order of roll no? Solution of 1.3.6.3 In this class we will have to create a function retroll() which will return the Rollno (‘rno’) of the particular objecT class stud { int rno; char nm[10]; float avg; public: void in(); void out(); int retroll () { return rno; } }; void stud :: in() { cout<<” Roll no :”; cin>>rno; cout<<” Enter the name”; gets(nm); cout<<” Enter the avg”; cin>>avg; } void stud ::out() { cout<<” Roll no:”<<rno; cout<<” Name:”; puts(nm); cout<<”Avg:”<<avg; }
- 303. 288 void accept ( stud s[],int n); void display( stud s[],int n); void sortroll( stud s[],int n); void main() { stud a[50]; int n; cout<<”enter the no of students”; cin>>n; accept(a,n); cout<<” The contents of the array”; display(a,n); sortroll(a,n); cout<<” The array in sorted form of roll nos is “; display(a,n); } Example 1.3.6.4 Give the output for the following questions (a) class parts { char pnm[20]; float prc; public: void assign(char p[],float pr) { strcpy(pnm,p);prc=pr;} void show() {cout<<pnm<<”@”<<prc<<endl;} void price( int l=0) { prc=(l==0)?prc+90:prc-90;} }; void main() { parts p,q; p.assign(“bolts”,900); q.assign(“clamps”,345);
- 304. 289Part II: Chapter 1 Object Oriented Programming p.price(); q.price(1);p.show();q.show(); p.price(1);p.show(); } b) class account { int balance,accno; public: account(){balance=0;} account(account &s) { balance = ++s.balance; accno=0; } void assign( int u) { balance = 110; accno = u; } void disp() { cout<<”Balance”<<balance<<endl; cout<<” Acc no.”<<accno<<endl; } }; void main() { account a1; a1.assign(9); a1.disp(); account a2(a1); a2.disp(); account a3(a2); int y=10; a3.assign(-y); a3.disp(); }
- 305. 290 Solution of 1.3.6.4 a) OUTPUT SCREEN b) OUTPUT SCREEN 1.4 CONSTRUCTORS AND DESTRUCTORS Constructors and destructors determine how the objects of a class are created, initialized, copied, and destroyed. They are member functions whose names are distinguished from all other member functions because they have the same name as the class they belong to. Constructors and destructors have many of the characteristics of normal member functions. You declare and define them within the class, or declare them within the class and define them outside--but they have some unique features. 1.4.1 CONSTRUCTORS Constructors are special member functions, which can be included in the class for automatic initialization at the time of object creation, these special member function(s) is/are known as Constructors or Constructer function(s). Constructor functions have the unique following features: Balance110 Acc no. 9 Balance111 Acc no. 0 Balance110 Acc no. -10 bolts@990 clamps@255 bolts@900
- 306. 291Part II: Chapter 1 Object Oriented Programming • Name of the constructor function is same as that of the class they are part of. • No return type is required for the constructor function. • Constructor functions are automatically called at the time of object creation/declaration. • Constructor functions are always defined in the public type access specifier. • Constructor functions can be overloaded. With the use of a constructor variables can be given initial values which otherwise would be having junk values. Program 1.3 /*********************************************************************** Program 1(3).cpp Shows the declaration of a constructor ***********************************************************************/ // Header Files # include <iostream.h> # include <conio.h> // Declaration of class trial class Trial { int a, b; public: Trial() { a=0;b=0; } void disp() { cout<<a<<b<<endl; } void raise() { a+=10; b+=10; } void down() Constructor for class trial
- 307. 292 { a-=5; b-=5; } }; // Main Function void main() { Trial T; T.disp(); T.raise(); T.disp(); T.down(); T.disp(); } OUTPUT SCREEN In program 1.3 we have a initialized a class trial which has two private data members a and b. This class also has a public member function Trial() but if we notice carefully this is a special member function as this is a constructor. Member function Trial() has the same name as that of the class of which it is a part of. Now when an object of class is declared the constructor of the class is automatically called. In this case when T an object of class Trial is declared the constructor trial() is automatically called for and private data members of the class trial a and b are given initial values of zero respectively. 1.4.1.1 DEFAULT CONSTRUCTOR If a constructor is not provided in a class, the compiler automatically creates a constructor, for the class and the constructor created is known as the default constructor. Program 1.4: Showing the declaration of a Default constructor by the compiler class try Automatic call to the Constructor 00 1010 55
- 308. 293Part II: Chapter 1 Object Oriented Programming { int a, b; public: try(){} void getit(); void show(); }; 1.4.1.2 PARAMETERIZED CONSTRUCTOR A parameterized constructor can take values as arguments into the constructor function at the time of object declaration. A parameterized constructor is defined in the same way as any other constructor, the only difference between an ordinary constructor and a parameterized constructor is that the function headers of these constructors will have parameters and hence different initial values can be given to objects of the same class with the use of this constructor. Program 1.5: /*********************************************************************** Program 1(5).cpp Object Declaration of a parameterized constructor ***********************************************************************/ # include<iostream.h> # include<conio.h> class employee { private: int employeeNo; public: employee(int n ) // parameterized constructor try() is the default constructor of the class try. PROGRAMING TIPS Constructors should be initialized in a class even though it is not necessary it is important to do so as it removes away the risk of variables having junk values and restricting many errors. Destructors should also be initialized to free the memory space occupied by constructors.
- 309. 294 { employeeNo = n; } }; void main() { clrscr(); employee E1(10); getch(); } 1.4.1.3 COPY CONSTRUCTOR A constructor, like any other function can take arguments. A copy constructor can take as argument an object of the same class to which it belongs. To understand the copy constructor lets look at the following program Program 1.6: /*********************************************************************** Program 1(6).cpp Shows the declaration of a Copy Constructor ***********************************************************************/ // Header Files # include<iostream.h> # include<conio.h> // Declaration of class play class play { int count, number; public: play() // constructor { count=0; number=0; } play ( play &p) // copy constructor { count=p.count +10; number= p.number+20; }
- 310. 295Part II: Chapter 1 Object Oriented Programming void disp() { cout<<count; cout<<number<<endl; } void change ( int c, int n) { count = c; number=n; } }; void main() { play p; // call for the constructor p.disp(); p.change(90,80); play q(p); // copy constructor call q.disp(); } A copy constructor is an overloaded constructor function in which an object of the same class is/are passed as reference parameter(s). It is used when an objects data value is related to or is initialized using another objects data value, of the same class. In the above example the values of data members of object q are dependent on the values of data members of object p. 1.4.1.4 CONSTRUCTOR OVERLOADING Constructors are just like other functions, they can also be overloaded, that is same named constructers can be used to give different set of outputs depending upon the kind of input given to them. The concept of constructer overloading can be easily understood by the program given below. Program 1.7: /*********************************************************************** Program 1(7).cpp Depicts the concept of constructor overloading. ***********************************************************************/ // Header Files # include<iostream.h> # include<conio.h>
- 311. 296 class work { int X,Y; public: work() //1 { X=10;Y=30; } work(int c) //2 { X=c; Y=2*c; } work (int x,int y) //3 { X=x; Y=y; } }; void main() { work w; // 10: Call for constructor 1 work r( 30,60); // 20: Call for constructor 3 work rr(30); // 30: Call for constructor 2 } The class work initialized in this example has 3 different overloaded constructors, which behave, differently on the types of inputs given to them. When this program is executed and the compiler reaches line 10 in the program then the 1st constructor is called as it matches the function header of this constructor. Similarly as the compiler reaches the lines 20 and 30 constructor 3 and constructor 2 are called respectively. COMMON PROGRAMMING ERRORS • Declaring a constructor with a return type (even void) will lead to a compilation error. • Constructors should always be declared in public access specifier.
- 312. 297Part II: Chapter 1 Object Oriented Programming • In case of copy constructor the parameter passed has to passed as reference parameter. 1.4.1.5 Solved Examples Example 1.4.1.5.1 Find error, if any in the following program class num { int n; int num() { n=4; } public; void display () { cout<<n; } }; Solution of 1.4.1.5.1 i) Constructer can only be present in Public access specifier ii) Constructor does not have a return type iii) After public there should be a colon (:) instead of a semicolon (;) Example 1.4.1.5.2 Write a program to generate a Fibonacci series using constructor to initialize the data members. Hint: Fibonacci series is a series in which the data members present in the series can be obtained by adding the previous 2 data members. Solution of 1.4.1.5.2 /*********************************************************************** Solved Example 1.7 Fibonacci Series ***********************************************************************/ // Header Files
- 313. 298 # include<iostream.h> # include<conio.h> class series { int a,b,c,i,n; public: series() { a = 1; b = 1; i = 2; } void read(); void calc(); }; void series::read() { cout<<" Enter the value of n"; cin>>n; } void series::calc() { cout<<endl<<a; cout<<endl<<b; while(i<n) { c=a+b; a=b; b=c; cout<<"n"<<c; i++; } }; void main() { series f; f.read(); f.calc(); getch(); }
- 314. 299Part II: Chapter 1 Object Oriented Programming Example 1.4.1.5.3 Write the output of the following program: a) class sum { int x,y,total; public: sum( int a,int b) { x=a; y=b*2; } void display() { total = x + y; cout<<total; } }; void main() { sum s(9,5); s.display(); } a) OUTPUT SCREEN b) // Header Files # include<iostream.h> # include<conio.h> class variable { int x; 19
- 315. 300 int y; public: variable() { x = 10; y = 20; } variable (int a, int b) { x = 2 * a; y = b; } void increase () { x+=10; y++; } void decrease() { x--; y--; } void output() { cout<<" Value of x "<<x<<"t"; cout<<" Value of y "<<y<<endl; } }; void main() { clrscr(); variable v1 ; variable v2(10,20); v1.increase(); v2.decrease(); v1.output(); v2.output(); getch(); }
- 316. 301Part II: Chapter 1 Object Oriented Programming b) OUTPUT SCREEN 1.4.2 DESTRUCTORS The destructor is the counterpart of a constructor. It is a member function that is called automatically when a class object goes out of scope. It has same name as that of the class , proceeded by a tilde (~). A destructor destroys an object. Destructors are declared in the header files of their class, or are inherited from the base class of that class. Typical syntax of a destructor is as given below: A destructor has the same features as that of the constructor that are • Name of the destructor function is same as that of the class they belong to • No return type is required for the destructor function • Destructor functions are automatically called when the object goes out of scope • Destructor functions are always defined in the public type access specifier When an object is created, sufficient amount of memory is allocated to it. When the scope of an object comes to an end, resource allocated to it must be freed so that it can be used elsewhere. This process can be achieved by the use of a destructor. This function is automatically invoked when the scope of an object comes to an end. Note: In case the programmer does not provide for a destructor manually, the compiler provides a default destructor just as the default constructor. Program 1.8: /*********************************************************************** Solved Example 1.8 Elucidates the concept of Destructors. ***********************************************************************/ Value of x 20 Value of y 21 Value of x 19 Value of y 19 classname::~classname();
- 317. 302 // Header Files # include<iostream.h> # include<conio.h> class trial { private: int t1; int t2; public: trial (int a,int b) // constructor function { t1=a; t2=b; cout<<" This is a trial program"; } void output() { cout<<endl<<" The sum of the variables entered is "<<t1+t2; } ~trial() // destructor function { cout<<" n The trial program is over"; } }; void main() { clrscr(); trial t(1000,2000); t.output(); getch(); }
- 318. 303Part II: Chapter 1 Object Oriented Programming OUTPUT SCREEN 1.5 Review Examples & Questions Example 1.5.1 Find the errors in the following programs. a) # include<iostream.h> class xyz { int a; int b; public: xyz(int x,int y) { a=x; b=y; } void output() { cout<<" Variable 1 "<<a; cout<<" Variable 2 ">>b; } } void main() { xyz x1; xyz x2(15,15); x2.output(); getch(); } a) Errors o <conio.h> header file is missing. o Illegal class declaration, semicolon (;) is missing. This is a trial program The sum of the variables entered is 3000 The trial program is over.
- 319. 304 o Arguments have to be passed with object x1. o Extraction operators are to be used with cout. The correct program is given below # include<iostream.h> # include<conio.h> class xyz { int a; int b; public: xyz(int x,int y) { a=x; b=y; } void output() { cout<<" Variable 1 "<<a; cout<<" Variable 2 "<<b; } }; void main() { xyz x1(10,10); xyz x2(15,15); x2.output(); getch(); } b) # include<iostream.h> # include<conio.h> class Integer { private:
- 320. 305Part II: Chapter 1 Object Oriented Programming int n; public: Integer (Integer x) { n=x.n; } void show() { cout<<n; } }; void main() { Integer k(8); k.show(); } b) Error • The argument passed in copy constructer has to be passed as a reference parameter: That is, Integer ( Integer &x ). • Also when declaring ‘k’ the parameter passed to the constructor has to be an object of class ‘Integer’. c) class circle { private float radius; public: void getdata(); { cout<<” Enter the radius “; cout<<radius; } float area(); } circle:: float area() { return (3.14 *radius*radius); }
- 321. 306 c) Errors o Line 3 Colon(:) missing after key word private o Line 7 Extra semicolon following function name header o Line 13 Missing semicolon at the end of the class definition o Line 15 The function should be declared with return type preceding the class name the correct code is: float circle::area(). Example 1.5.2 Declare a class max, which has 3 integer data type members, a function read to input the values of the data members and a function display to display, the greatest number between them? Solution of 1.5.2 // Header Files # include<iostream.h> # include<conio.h> class maximum { int x,y,z; public: void read(); void display(); }; void maximum::read() { cout<<" Enter the three numbers"; cin>>x>>y>>z; } void maximum::display() { if ( (x>y) && (x>z)) cout<<" Maximum number is "<<x; else if ((y>x)&&(y>z)) cout<<" Maximum number is"<<y; else cout<<" Maximum number is"<<z; }
- 322. 307Part II: Chapter 1 Object Oriented Programming Example 1.5.3 What will be the output of the following questions? a) // Header Files # include<iostream.h> # include<conio.h> int area(int s) { return (s*s); } float area(int b,int h) { return (0.5*b*h); } void main() { clrscr(); cout<<area(5)<<endl; cout<<area(4,3)<<endl; cout<<area(6,(area(3)))<<endl; getch(); } a) OUTPUT SCREEN b) # include<iostream.h> # include<conio.h> struct salary { float gross; float GPF; 25 6 27
- 323. 308 float tax; float net; }; void main() { clrscr(); salary x= {5800,10,500,0}; float deductions; deductions=x.GPF+x.tax; x.net=x.gross-deductions; cout<<" Net amount payable ="<<x.net; getch(); } b) OUTPUT SCREEN c) # include<iostream.h> struct SCORE { int point,bonus; }; void calculate(SCORE &S,int N=10) { S.point++; S.bonus+=N; } void main() { SCORE SC={20,5}; calculate(SC); cout<<SC.point<<”;”<<SC.bonus<<endl; Net amount payable =5290
- 324. 309Part II: Chapter 1 Object Oriented Programming calculate(SC); cout<<SC.point<<”;”<<SC.bonus<<endl; calculate(SC,15); cout<<SC.point<<”;”<<SC.bonus<<endl; } c) OUTPUT SCREEN d) // Header Files # include<iostream.h> # include<conio.h> # include<string.h> # include<stdio.h> // Declaration of class student class student { char name[15]; int rollno; public: student() { strcpy(name,"Ankit"); rollno=1; } student(int a) { strcpy(name,"Neha"); rollno=a; } void output(); void increase(); }; void student::output() 21;15 22;25 23;40
- 325. 310 { cout<<" The roll no of the student "<<rollno; cout<<" The name of the student "; puts(name); } void student::increase() { rollno++; } // main function void main() { student s; student s1(5); s.increase; s1.increase; cout<<"##### ---- Student program ---- #####"; s.output; s1.output; getch(); } d) OUTPUT SCREEN e) # include<conio.h> # include<iostream.h> class class1 { public: int i,j; ##### ---- Student program ---- #####
- 326. 311Part II: Chapter 1 Object Oriented Programming class1(int a, int b) {i=a; j=b;} void show() { i = 2; i--; if (i==1) { cout<<"class1:i:"<<i<<"j:"<<j<<endl; return; } show(); } }; int main() { class1 x(20,20); x.show(); getch(); } e) OUTPUT SCREEN f) # include<iostream.h> # include<conio.h> class example { static int x; int y,z; public: example(int i,int j) class1:i:1j:20
- 327. 312 { y=i; z=j; x++; } void display() { cout<<"y="<<y<<endl; cout<<"z="<<z<<endl; } static void dis() { cout<<"x="<<x<<endl; } }; int example::x; void main() { clrscr(); example o1(7,5),o2(8,6),o3(2,0); o1.display(); example::dis(); o2.display(); o3.display(); getch(); } f) OUTPUT SCREEN y=7 z=5 x=3 y=8 z=6 y=2 z=0
- 328. 313Part II: Chapter 1 Object Oriented Programming g) # include<iostream.h> # include<conio.h> int c =0; // Global Variable // Declaration of class block class block { public: block() { cout<<"Total object created"<<endl; } ~block() { cout<<"Total object destroyed"<<endl; } }; void main() { clrscr(); block M1,M2,M3,M4; cout<<"Block1"<<endl; block m5; cout<<"Block 2"<<endl; block m6; cout<<"Block 3"<<endl; getch(); } g) OUTPUT SCREEN Total object created Total object created Total object created Total object created Block1 Total object created Block 2 Total object created Block 3 Total object destroyed Total object destroyed Total object destroyed Total object destroyed Total object destroyed Total object destroyed
- 329. 314 Question 1.5.4 Define a structure? Declare a structure in c++ with name, roll number and age as its components. Answer of 1.5.4 A structure is a collection of variable having different data types. The declaration as struct student { char name[15]; int roll; int marks; }; Exercise: Now think how classes and structures are different and similar to each other. Example 1.5.5 Declare a structure client containing the following members: The character quantity of the client The integer quantity called clientid The integer quantity called no_of_items The long integer quantity called Client_total Solution of 1.5.5 struct client { char name[15]; int clientid; int no_of_items; long client_total; }; Example 1.5.6 Write a program to add and subtract complex numbers using the structure given below: struct complex { int real; int imag; };
- 330. 315Part II: Chapter 1 Object Oriented Programming Solution of 1.5.6 // Header Files # include<iostream.h> # include<conio.h> // Structure definition struct complex { int real; int imag; }; // Function definition void add(complex c1,complex c2) { complex c3; c3.real=c1.real+c2.real; c3.imag=c1.imag+c2.imag; cout<<" The Sum of the 2 complex numbers is "<<c3.real<<" + "<<c3.imag<<"i"; } // Function Definition void sub(complex c1,complex c2) { complex c3; c3.real=c1.real-c2.real; c3.imag=c1.imag-c2.imag; cout<<" The Sum of the 2 complex numbers is "<<c3.real<<" + "<<c3.imag<<"i"; } // Main Program void main() { int choice; char r; complex num1,num2; cout<<"Enter the real part"; cin>>num1.real; cout<<"n Enter the imaginary part"; cin>>num1.imag; cout<<"n Enter the real part"; cin>>num2.real; cout<<"n Enter the imaginary part";
- 331. 316 cin>>num2.imag; do { cout<<" #### MENU ####"<<endl; cout<<" 1) Add"<<endl; cout<<" 2) Subtract"<<endl; cin>>choice; switch(choice) { case 1 : add(num1,num2); break; case 2 : sub(num1,num2); break; } cout<<" Do you want to continue"; cin>>r; }while(r!='n'); getch(); } Question 1.5.7 What is a class and are the private members of class accessible to the object? Answer of 1.5.7 A class is a user defined data type used to implement an abstract object, giving you the capability to use object oriented programming with c++. No the private members of a class are not accessible to the object. Question 1.5.8 What are the differences between the following? a) Object oriented programming and procedural programming b) Public and private members of a class Answer of 1.5.8 a) In object oriented programming the stress is given on the object, whereas in procedural programming the stress is given on the subject. b) The public members are accessible to objects of a class whereas private member are not accessible outside the class or to objects of the class. Example 1.5.9 Write a program to solve a quadratic equation? Use guidelines in section 1.3.3 to solve this particular problem?
- 332. 317Part II: Chapter 1 Object Oriented Programming Solution of 1.5.9 // Header Files #include<iostream.h> #include<conio.h> #include<math.h> // Definition of class q class q { int a,b,c; float r1,r2,d; public: void read(); void calculate(); }; void q::read() { cout<<" Enter the values of a,b,c"; cin>>a>>b>>c; } void q::calculate() { d= (b*b)-(4*a*c); if (d>=0) { r1= (-b + sqrt(d))/(2*a); r2= (-b - sqrt(d))/(2*a); cout<<"n Root 1 is"<<r1; cout<<"n Root 2 is"<<r2; } else { cout<<"n Roots are imaginary "; } } // main function void main() { q q1; // object of class q q1.read(); q1.calculate(); }
- 333. 318 Question 1.5.10 Define the following terms: a) OOP f) Private section k) Constructor b) Procedural Programming g) Data Hiding l) Default Constructor c) Object h) Scope Resolution Operator (::) m) Copy Constructor d) Class i) Abstraction e) Class declaration j) Friend Function Answer of 1.5.10 a) OOP Object oriented programming enables the programmers to group together data and code that uses data into discrete units. In OOP the stress is given on the object in spite of procedure. b) Procedural programming In procedural programming, we think about functions and the execution flow these functions. c) Object An object is a self-contained abstraction of an item. An instance of a class. d) Class A class is a user defined data type used to implement an abstract object. It contains members of three types: private, public and protected. By default the data members of the class are private. Classes enable the programmer to model objects with attributes and behaviors. Class types can be defined in c++ using the key word ‘class’ and ‘struct’, but key word class is normally used for this purpose. e) Class declaration The class is declared with a keyword class class <class_name> { private: data_members; member_functions; public: data members; member_functions; };
- 334. 319Part II: Chapter 1 Object Oriented Programming f) Private section Private members are available to only members of class and not accessible to the objects or functions outside the class. The default access mode for classes is private so all members after the class header and before the first member access specifier are considered to be private. g) Data hiding The variables and functions declared in a class as private are not accessible to any outside function. This feature of a class is called as data hiding. h) Scope resolution operator (::) This operator is used in situations where a global variable exists with the same name as a local variable. Also, this is used in c++ class when the number functions are declared outside the class definition, the function name is preceded by the class name and the binary scope resolution operator (::) i) Abstraction The class members, which are private, remain hidden from the outside world, the public members form the interface by providing the essential and relevant information to outside world. Thus only the essential features are represented to outside world without including the background details, which is called abstraction. j) Friend Function A function outside of a class can be defined to be a friend function by the class. By doing so the class gives the friend function, free access to the private members of the class. k) Constructor A constructor is a special initialization function that is called automatically whenever instance of the class is declared. The name of the constructor is same as that of class and it returns no value. l) Default constructor A default constructor is one, which accepts no parameters: class student { private: int rollno; public: student() //Default constructor { rollno=1; } };
- 335. 320 m) Copy constructor A copy constructor is that constructor which is used to initialize one object with the values from another object of same class during declaration. Question 1.5.11 Is data hiding same as encapsulation? Answer of 1.5.11 No. Data hiding is the property of a program where vital data remains safely hidden from functions which do not use or manipulate them. On the other hand, encapsulation is the mechanism through which data hiding is achieved. Example 1.5.12 An automobile dealer has given identity number(idno) to all spare parts. The following class was defined to store information about identity number and listed price of a spare part: class product { private: int idno; public: float price; public: void getdata(int n, float p) { idno = n; price=p; } }; Write a program to edit the existing class definition for ‘class product’ to include tax in the listed price and then calculate and display the net price including tax for the following spare parts: (a) id no=1013 price=345.0 (b) id no=2314 price=125.0 Note: tax is 10% of the listed price
- 336. 321Part II: Chapter 1 Object Oriented Programming Solution of 1.5.12 class product { private: int idno; public: float price; float tax ; public: void getdata(int n, float p) { tax = .10 * p; idno = n; price=p+tax; } }; void main() { product p1,p2; p1.getdata(1013,345.00); p2.getdata(2314,125.00); cout<<" Net amount payable for spare part1:"<<p1.price<<"$"<<endl; cout<<" Net amount payable for spare part2:"<<p2.price<<"$"; } Example 1.5.13 Which function is called during each function call in the program given below: … long total(int x); //………. I char choose( char a, char b); //……….II char choose (char a); //………III float total (int x, float y); //………IV void main() { … total(n); // ……….1 choose(‘c’); // ……… 2 choose(‘c’,’d’); // ……….3 total(n,n1); // ……….4 }
- 337. 322 Solution of 1.5.13 Call 1 invokes function (I) Call 2 invokes function (III) Call 3 invokes function (II) Call 4 invokes function (IV) Example 1.5.14 Observe the following program … long total(int x); // ………. I int total (int x, int y=10); // ………II void main() { … total(n); //……….1 } Which function will the compiler invoke when the following program is executed? Solution of 1.5.14 When the following program is executed function ambiguity takes place as the compiler is not sure which function to invoke and as a result of this the compiler hangs. Note: The result of this program is compiler dependent. Example 1.5.15 Define a class student with the following specification: admno integer sname 20 character eng,math,science float total float ctotal() A function to calculate ‘total = eng + math + science’ Public member function of a class student takedata() To accept values for admno, sname, eng, math, science and invoke ctotal() to calculate total showdata() To display all the data members on the screen
- 338. 323Part II: Chapter 1 Object Oriented Programming Solution of 1.5.15 class student { int admno; char sname[20]; float eng,math,science; float total; void ctotal(); public: void takedata(); void showdata(); }; void student::takedata() { cout<<” n Enter the admission number of student “; cin>>admno; cout<<”n Enter the name of the student”; gets(sname); cout<<”n Enter the marks in English”; cin>>eng; cout<<”n Enter the marks in Maths”; cin>>math; cout<<”n Enter the marks in Science”; cin>>science; ctotal(); } void student ::ctotal() { total = eng + math + science; } void student::showdata() { cout<<”n Name ”<<sname; cout<<”n Admission no”<<admno; cout<<”n Total marks”<<total; }
- 339. 324 Example 1.5.16 Define a class batsman with the following specifications:- Private members bcode (4 digit code number) bname (20 characters) innings, notout, runs integer type batavg it is calculated according to the formula: batavg = runs/(innings – notout) calcavg() function to compute batavg public members readdata() function to accept values for bcode, name, innings, notout and invoke the function calcavg() displaydata() function to display the data members on the screen Solution of 1.5.16 class batsman { int bcode; char bname[25]; int innings, notout, runs; float batavg; float calcavg() { return (runs/(innings-notout)); } public: void readdata(); void displaydata(); }; void batsman::readdata() { clrscr(); cout<<”Enter the batsman code”; cin>>bcode; cout<<”Enter the batsman name”; gets(bname); cout<<”Enter total innings”; cin>>innings; cout<<”Enter total notout”; cin>>notout; cout<<”Enter total runs”; cin>>runs; batavg=calcavg(); } void batsman :: displaydata() {
- 340. 325Part II: Chapter 1 Object Oriented Programming clrscr(); cout<<” Batsman Code”<<bcode; cout<<endl<<” Batsman Name”<<bname; cout<<endl<<” Total innings ”<<innings; cout<<endl<<” Total notout ”<<notout; cout<<endl<<” Total runs”<<runs; cout<<endl<<” The batting average”<<batavg; } 1. 6 Review Exercise 1 Mark questions Q1) Which data members of a class is accessible to the outside world? Q2) What is data abstraction? Q3) What is the significance of the scope resolution operator (::)? Q4) What is wrong with the following declaration: Class{ int a,b,c,sum; private void read(); void display(); } Q5) How is the memory allocated to a class and to its objects? Q6) Is encapsulation a security technique? Q7) How does the compiler decide which function to invoke in case of function overloading? Q8) Is it possible to have an overloaded function with different return types but same number and type of arguments? Q9) Define an object? Q10) What is the relation between an object and a class?
- 341. 326 2 Mark questions Q1) How is object oriented programming different from procedural programming? Give proper examples for differentiating? Q2) Differentiate between a constructor and a destructor function? Q3) What is the purpose of the constructor function in a class? Give a suitable example of a constructor function in a class? Q4) Why is a destructor function required in a class illustrate with the help of a suitable example? Q5) Discuss the different ways by which we can access public members of an object? Q6) What do you understand by the term copy constructor give suitable examples to illustrate how it is useful? Q7) Differentiate between the following: a) a structure and a class b) an object and a structure variable Q8) Is polymorphism an example of function overloading, explain giving suitable examples? Q9) Can we use the same function name for a member function of a class and an outside function in the same program file? If yes, how are they distinguished? If no, give reasons. Q10) Find the syntax error(s), if any, in the following program: # include <“conio.h ”> # include<iostream.h> void Main() { int ch; cin<<ch; if ch>=9 cout<<ch; for (int i=0;i<2;i++) cout<<end; }
- 342. 327Part II: Chapter 1 Object Oriented Programming 4 Mark questions Q1) Declare a class employee having the following members: Data Members: Employee name Employee address Employee number Member Functions To read data members To display the data members Q2) Define a class teacher with the following specification: private members of class teacher name 20 characters subject 10 characters basic, hra, da float netpay float // Obtained by the sum of float variables ‘basic’ ‘hra’ and ‘da’ calculate A function computes and returns the net salary (‘netpay’) a public member function of class teacher readdata() Function to accept data values and invoke the calculate function displaydata() Function to prints the teacher information on the screen Q3) Declare a structure date having members as day - a string of size[2] represents day part of a date month - a string of size [20] represents month name of a date
- 343. 328 year - a string of size[4]; represents year part of a date Declare a class medicine with the following specifications private members of a class medicine_name - a string of size[25] priceperunit - float mfgdate - of type date expdate - of type date public members of the class medentry() - A function to accept values of the data members from user ShowWarning() - A function that displays the name of the medicine along with warning message as “Use within 2 Months”, if the expiry date is 2 months after manufacturing date ShowAll() - A function that displays above details of a medicine properly formatted in a single line Q4) Define a class to represent a bank account. Include the following members: Data members: Name of the depositor Type of the account Account number Balance amount in the account Member functions: To deposit an amount To withdraw an amount. To display the name, type of account and the balance of the account holder
- 344. 329Part II: Chapter 1 Object Oriented Programming Write the main() block to test the program. Use constructors to initialize the data members? Q5) What is function overloading? Use the concept of function overloading to compute the area of rectangle ( given length of two sides ) ,area of triangle (given length of the three sides using Heros formula) and area of a circle (given length of a radius) 1.7 Programming Project: Mechanical Engineering An automobile company, “FORD” is trying to launch a new car. The company is although not sure about which engine design to implement in their new car. The company currently has option of two engines (fig 1.7) A and B. Engine efficiency is determined by the following formula: E = ((x+y)*y)/z + ((x+z)/w)*i; The parameters for both engine designs are given below in table 1.2. Using the criteria’s given create a C++ object oriented program to determine the efficiency of the two engines. Which engine should the company use for the launch of its new Ford V9.12? (A) (B) Figure 1.7: Car Engines Engine Features Engine A Engine B Volume (x) 4.20 5.30 Horse Power (y) 190 260 Valves (z) Quad 4 Quad 4 Height (w) 0.98 0.89 HEAD C.C. (i) 45.5 46.6 Table 1.2: Engine Specifications
- 345. 330 1.8 Let us revise! A class is a user defined data type used to implement an abstract object. An object is as instance of a class. A class having no objects is known as an Abstract Class. A class definition has members which could be data members or member functions. A member function is a function which is a part of a class. A member function has access to private members of a class which other external functions do not have. A class has 3 kinds of access specifiers, private, public, protected. Private Members are those members which can only be accessed by the member functions. Public Members are those members which can be accessed by any function in the entire program. The practice of wrapping up of data and functions together in a single unit is known as Encapsulation. Data Abstraction is defined as bringing the data down to its essentials. The process of using an operator or a function in different ways for different set of inputs given is known as polymorphism. Constructors are special member functions which can be included in the class for automatic initialization at the time of object creation. Constructors have the same name as that of the class they are initialized in. They also have no return type and are automatically called at the time of object creation. A copy constructor is an overloaded constructor function in which an object of the same class is/are passed as reference parameter(s).
- 346. 331Part II: Chapter 1 Object Oriented Programming The destructor is the counterpart of the constructor. It is a member function that called automatically when a class object goes out of scope. OOPS has various advantages over procedural programming. OOPS has close relation between the real world objects and user created classes. Compilers for Object Oriented programs are able to detect conceptual errors hence increasing program reliability. Its various features like encapsulation, Inheritance, data abstraction, polymorphism make it much more advantageous than procedural programming hence object oriented languages like C++, Java are experiencing a rapid increase in popularity among other programming languages.
- 347. 332 Notes
- 348. Part II CHAPTER 2 INHERITANCE AIM OUTLINE • Introduce Inheritance, the most powerful concept of Object Oriented Programming • To learn about base classes and derived classes • To understand and use concept of inheritance and study its advantages 2.1 Introduction 2.2 Base and Derived Classes 2.3 Inheritance in C++ 2.4 Types of Inheritance 2.5 Solved Examples 2.6 Review Exercise 2.7 Programming Project 2.8 Let us revise
- 349. 334 2.1 Introduction We already know that modern Object Oriented languages provide three main capabilities, viz., ‘Data Abstraction’, ‘Polymorphism’ and ‘Inheritance’. We have already covered ‘Abstraction’ and ‘Polymorphism’ in previous chapters. In this chapter, we will deal with the most powerful concept in object oriented programming called ‘Inheritance’ Inheritance is the concept of creating a new class referred to as derived class from an already existing class (called Base Class). The derived class inherits the properties of the base class. In addition to this, the derived class can also have properties of its own. Let us understand this concept through an example involving development of software for ‘School Administration’. To do this, we would consider ‘Teachers’ and ‘Students’ classes - both these classes would be implemented differently as they have separate identity. However, there are some members which are common to both these classes, such as, ‘Name’, ‘Home Address’ and ‘Age’ including common functions for initiating these members. We can observe that both classes have common members and functions. We, therefore, need a concept which when implemented copies the code for these common members from one class to the other and then change the class name. This concept is implemented through ‘Inheritance’. Inheritance allows reusability of code of a class which is already debugged and tested. Inheritance also allows us to save on programming effort and enhances maintainability of the program. 2.2 Base and Derived Classes Inheritance is used to create hierarchical structures. Look at the category Animals in figure 2.1, there are three divisions of animals who eat only leaf and vegetables (Herbivores), animals who eat other animals as well as leaf and vegetables (Omnivores) and animals who eat only other animals for food (Carnivores). Each of these categories has further subcategories and under them even more subcategories exist. The important thing to note in this example is that once a category is defined all the elements in these categories and other subcategories satisfy the characteristic under which they were defined. The advantage of doing this is that once a member of a category is determined we do not need to learn again that whether the animal is an herbivore or an omnivore or a carnivore. In this example, ‘Animals’ is referred to as the base class and ‘Herbivores’, ‘Omnivores’ and ‘Carnivores’ are termed as derived classes. One may find that in other books, authors have also referred to ‘Base Classes’ as ‘Super Class’ and ‘Derived Classes’ as ‘Sub Class’. Derived classes also have the potential of becoming Base class, e.g., ‘Herbivores’ which is a derived class of class ‘Animals’ is also a base class for classes ‘Elephant’ and ‘Giraffe’. Generally derived classes are larger than their base classes as they contain data
- 350. 335Part II: Chapter 2 INHERITANCE members and member functions of their own as well as they may use all the functionality of the base class as well. Figure 2.1: Inheritance Hierarchy 2.3 Inheritance in C++ Let us look at the Program 2.1 to understand how inheritance is used in C++. Program 2.1: class cat_family { public: // The data and methods can be accessed directly char category[10]; char legs; int age; cat_family(); ~cat_family(); void food_habits(); void place_of_stay(); }; class panther : public cat_family { public: void sex_of_specie() { if (strcmp(category,”White Panther”)==0) { cout<<” White Panther is an endangered specie”;
- 351. 336 } } void color(); }; In this piece of code, we have created a derived class ‘panther’ from the base class ‘cat_family’. In order to create a derived class from an already existing class we use the following format: As you know, C++ class contains up to three sections: private, protected and public. Let us see how access specifiers work. • The data and member functions declared in the private section of a class are not accessible outside the class. Only the member functions declared within the class have access to the private data and functions of the class. • The data and member functions declared in the protected section of a class are not accessible outside the class except within sub-classes (or derived classes) given from the given parent (or base) class. • The data and member functions declared in the public section are of a class are available outside the class, i.e., when using the public type access specifier, all the public members of the base class are inherited by the derived class. These public members are also accessible by the objects of base and derived classes hence created. A hierarchy of abstractions can be established using derived classes. Objects of a derived class can be defined so as to inherit all the methods of the parent class or only a few selected methods of the base class. Further, new methods - not present in the base class - can also be specified that are peculiar to the derived class. Every object of a derived class contains field of data from the base class as well as its own private data. One can also observe in Program 2.1 that the derived class ‘panther’ accesses the public data members of its base class ‘cat_family’. Similarly, the protected members and functions under the ‘protected’ key word of the base class are also passed on as ‘protected’ members to the derived class. Though, these members cannot be accessed from instances or objects of this class.
- 352. 337Part II: Chapter 2 INHERITANCE The relationship between the access specifiers and the inheritance allowed by them is also summarized in the table (2.1) given below. Table (2.1): Inheritance with Access Specifiers PROGRAMING TIPS Create hierarchical models to understand the concept of inheritance for each of the sample problems provided. Hierarchical models allow us to understand better the flow of data members inherited by there derived classes. C++ also offers ways to save space and computer memory by using virtual base classes which allows us to save space and avoid data ambiguities in class hierarchies involving multiple inheritance. It eliminates the procedure in which each non virtual object contains a copy of all the data members defined in the base class.
- 353. 338 Now that we understand the concept of inheritance let us look at the following examples. 2.3.1 Solved Examples Example 2.3.1.1 Answer the questions (i) to (iv) on the basis of the following code: (i) Write the names of data members which are accessible from objects belonging to class ‘branch’. (ii) Write the names of all the member functions which are accessible from objects belonging to the class employee. (iii) Write the names of all the members which are accessible from member functions of class ‘employee’. (iv) How many bytes will be required by an object belonging to the class ‘employee’? class company { char website[12]; char name[20]; protected: int noemployees; public: company(); void register(); void status(); }; class branch:public company { long nocomputer; protected: float expense; public: branch(); void enter(); void show(); }; class employee:private branch {
- 354. 339Part II: Chapter 2 INHERITANCE int eno; char ename[20]; float salary; public: employee(); void joining(); void showdetail(); }; Solution of 2.3.1.1 (i) None of the data members are accessible from objects belonging to class branch. (ii) The names of member functions which are accessible from objects belonging to class ‘employee’ are joining(), showdetail(). (iii) The members accessible from objects belonging to class ‘employee’ are Data Members: eno, ename, salary, expense, noemployees Member Functions: register(), status(), enter(), show(), joining(), showdetails(). (iv) The amount of bytes required by an object belonging to class ‘employee’ is summation of all the bytes required by the objects of all the base classes and the class ‘employee’ which comes out to be ‘68’ bytes. Note: ‘register’ also happens to be a keyword in C. Hence some C/C++ compilers might give an error in executing the program. Here the author assumes that this is understood and lays stress only on the concept of inheritance. Example 2.3.1.2 Given the following class definitions answer the questions that follow: COMMON PROGRAMMING ERRORS Assuming that private members of a base class can be inherited directly by the derived classes and then using them as a part of the derived class is a logical and a compilation error. Declaring the constructors and destructors in the private access type specifier is an compilation error.
- 355. 340 class publisher { char pub[12]; double turnover; protected: void register(); public: publisher(); void enter(); void display(); }; class branch: protected publisher { char city[20]; protected: float employee; public: branh(); void haveit(); void giveit(); }; class author: private branch { int acode; char aname[20]; float amount; public: author(); void start(); void show(); }; (i) Write the names of data members which are accessible from objects belonging to class author. (ii) Write the names of all the member functions which are accessible from objects belonging to the class branch. (iii) Write the names of all the members which are accessible from member functions of class author. (iv) How many bytes will be required by an object belonging to class author?
- 356. 341Part II: Chapter 2 INHERITANCE Solution of 2.3.1.2 (i) None of the data members are accessible from objects belonging to class author. (ii) The names of the member functions which are accessible from objects belonging to class author are haveit(), giveit(). (iii) The members accessible from the member functions of class ‘author’ are: Data Members: employee, acode, aname, amount Member Function: register(), haveit(), giveit(), start(), show() (iv) The number of bytes occupied by an object belonging to class author are 70. At this point let me ask you a very interesting question. What happens to the constructors and destructors of the base class can they also be inherited by their derived classes? Well the constructors or destructors are not inherited by their derived classes but when even an object of a derived class is declared or initialized the default constructor for the base class is automatically called. Program 2.2 further enlightens the same concept. Program 2.2: // Header Files # include <iostream.h> # include <conio.h> // Class Definition for Class A class A { private: int a; int b; int c; public: A() { cout<<" This is the constructor for class A n"; } ~A() { cout<<" This is the destructor for class A n"; } Output Screen This is the constructor for class A This is the constructor for class B This is the destructor for class B This is the destructor for class A
- 357. 342 }; /* Class definition for Class B Base Class: A Access Specifier: public */ class B: public A { public: B() { cout<<" This is the constructor for class B n"; } ~B() { cout<<" This is the destructor for class B n"; } }; void main() { clrscr(); B object1; getch(); } In this program as soon as the ‘object1’ an object of class B is created, the default constructors for A and then B are called for. Similarly, as the scope of the class ends the message “This is the destructor for class B” is displayed followed by a similar message for class A. COMMON PROGRAMMING ERRORS Private members and Protected members of the base or derived classes are never accessible to the objects of these classes. The difference between the private and protected members lies only in the type of inheritance allowed by them. The members under the protected access specifier can be inherited by derived classes whereas the private members of a class can never be inherited properly.
- 358. 343Part II: Chapter 2 INHERITANCE 2.4 Types of Inheritance Inheritance exists in two forms depending upon the number of parent or base classes a derived class has. a) Single Inheritance (Figure 2.2): When a class inherits only from a single parent or in other words each derived class has a single parent. A good example of single inheritance is the tree structure [see Figure II (2.2(b))]. (a) (b) Figure 2.2: Single Inheritance (SI) b) Multiple Inheritance Multiple Inheritances are observed when the derived class inherit properties from more than one parent or base class. This is a very common practice used in object oriented programming as it allows each class to inherit and combine all useful features from a range of classes. Figure 2.3 shows few examples where multiple inheritances are observed.
- 359. 344 (a) (b) Figure 2.3: Multiple Inheritances Multiple inheritances are very similar to single inheritance. The piece of code given below displays the syntax to initiate multiple inheritances. Syntax: class A { … }; class B { … }; class C : public A, public B // Multiple Inheritance { … };
- 360. 345Part II: Chapter 2 INHERITANCE 2.5 Solved Examples Example 2.5.1 class mydata { int data1; protected: float val; void guess(); public: void get_mydata(); void manipulate(); }; class yourdata:protected mydata { int data2; protected: void access(); public: long value(); int catch(); void miss(); }; class ourdata:public yourdata { float data3; public: void have(); void give(); }; (i) Name the data members accessible from functions of class ‘ourdata’? (ii) How many bytes will be occupied by an object belonging to ‘classyourdata’? (iii) Define the function catch() which displays all the data members accessible to it. (iv) Name the member functions which are accessible from objects of class ‘ourdata’. Solution of 2.5.1 (i) The data members accessible from functions of class ‘ourdata’ are ‘data3’ of class ‘ourdata’ and value of class ‘yourdata’.
- 361. 346 (ii) An object belonging to class ‘yourdata’ will occupy 12 bytes. (iii) The function definition for ‘catch()’ which displays all the data members accessible to it is as follows. void yourdata::catch() { cout<<value<<data2<<val<<endl; } (iv) The member functions accessible class ourdata are catch(), miss() of ‘yourdata’ and have(), give() of class ‘ourdata’. Example 2.5.2 Given the following class definitions answer the questions that follow: class myclass { char data1[20]; int data2[5]; public: void read(); void show(); }; class myfirstclass : private myclass { int deriveddata1; float deriveddata2; protected: int standard; public: void read_firstderived(); void show_firstderived(); }; class mysecondclass : public myfirstclass { float deriveddata3; public: void read_secondderived(); void show_secondderived(); };
- 362. 347Part II: Chapter 2 INHERITANCE (i) Name the members which can be accessed inside class myfirstclass. (ii) Name the members which can be accessed by an object of mysecondclass. (iii) Name the members which can be accessed by an object of class myfirstclass. (iv) What will be the size of an object mysecondclass in bytes ? Solution of 2.5.2 (i) The members which can be accessed inside class myfirstclass: void read(); void show(); int deriveddata1; float deriveddata2; int standard; void read_firstderived(); void show_firstderived(); (ii) The members which can be accessed by an object of class mysecondclass. void read_firstderived(); void show_firstderived(); void read_secondderived(); void show_secondderived(); (iii) The members which can be accessed by an object of class myfirstclass are: void read_firstderived(); void show_firstderived(); (iv) The size of an object of class ‘mysecondclass’ is 42 bytes. Example 2.5.3 (a) Both private as well as protected members of a class are not accessible to the object of the class. So what is the significance of using members inside the protected specifier? (b) Will the following piece of code compile: # include <iostream.h> # include <conio.h> class X
- 363. 348 { protected: int var1; public: int var3; private: int var 2; }; void main() { X x1; getch(); } Solution of 2.5.3 (a) Although both, the private as well as protected members of a class are not accessible to the objects of the class, the significance of protected members lies in inheritance. The members declared in the protected specifier of the base class are accessible to the derived class, where as the derived class cannot access the members under the private specifier of the base class. (b) Yes, the piece of code given will compile as the order of placement of type specifiers does not matter and hence we can follow any format for type specifiers when defining a class. Example 2.5.4 Determine the output for the following piece of code: # include <iostream.h> # include <conio.h> # include <stdio.h> # include <string.h> class electronics { private: int equipment_no;
- 364. 349Part II: Chapter 2 INHERITANCE public: char equipment_name[15]; int equipment_price; electronics() { cout<<endl<<" An object of type Electronics has been declared"; } ~electronics() { cout<<endl<<" The scope of the created object is now over"; } void enter(); void display(); }; class computer: public electronics { public: computer() { cout<<endl<<" Computer is a derived class of electronics"; } char motherboard; int modelno; }; void electronics::enter() { cout<<" Enter Equipment Name"; gets(equipment_name); cout<<endl<<" Enter Equipment Price"; cin>>equipment_price; } void electronics:: display() { cout<<endl<<" Equipment Name ->"<<equipment_name; cout<<endl<<" Equipment Price ->"<<equipment_price; } void main() { clrscr(); computer c1; electronics e1;
- 365. 350 strcpy(c1.equipment_name,"Computer"); strcpy(e1.equipment_name,c1.equipment_name); c1.equipment_price = 2000; c1.display(); e1.equipment_price=c1.equipment_price; e1.equipment_price--; e1.display(); getch(); } Solution of 2.5.4 Output Screen An object of type Electronics has been declared Computer is a derived class of electronics An object of type Electronics has been declared Equipment Name ->Computer Equipment Price ->2000 Equipment Name ->Computer Equipment Price ->1999 The scope of the created object is now over The scope of the created object is now over 2.6 Review Exercises 1/2 Mark Questions Q1) What are the advantages of inheritance which make it the most powerful tool available in object oriented programming? Q2) What are the different types of inheritance, explain briefly by giving examples? Q3) How are members declared under private and protected access specifiers of a class different from outside the class declaration? Q4) Write the class declaration for a class ‘mouse’ derived from class ‘hardware’ using the public type access specifier, having a character array ‘type’ and an integer variable ‘cost’ as its data members. Q5) Write the class declaration for ‘Mirror’ from class ‘Glass’ and ‘Silica’. The class ‘Mirror’ should also include two character arrays ‘type’, ‘shape’ along with an integer variable ‘thickness’ as its data members.
- 366. 351Part II: Chapter 2 INHERITANCE Q6) Find the error(s) for the constructors in the following code fragments: i) class sample1 ii) class sample2 { { float a,b,c; float a,b,c; public: sample2(); float sample1(); }; }; Q7) Is it possible for a derived class to access the private members of the base class indirectly. If yes then create a base class and a derived class then use these classes to illustrate your point? 4/5 Mark Questions Q8) Given the following class definitions: class world { char continents[10]; protected: char oceans[10]; public: void input (char[]); void output(); }; class country : private world { char cname[20]; protected: char capital[20]; public: void indata(); void outdata(); }; class state: public country { char sname[20];
- 367. 352 public: void display(); }; (i) Name the base class and the derived class of the class ‘country’. (ii) What will be the size of an object of class ‘world’. (iii) What will be the size of an object of class ‘country’. (iv) What will be the size of an object of class ‘state’. (v) Name the data members that can be accessed from member functions of class ‘world’. (vi) Name the data members that can be accessed from member functions of class ‘country’. (vii) Name the data members that can be accessed from member functions of class ‘state’. (viii) Name the members that can be accessed by an object of class ‘world’. (ix) Name the members that can be accessed by an object of class ‘country’. (x) Name the members that can be accessed by an object of class ‘state’. 2.7 Programming Project The food chain shows how each living thing gets its food. Some animals eat plants and some eat other animals to survive. Your project is to create first inheritance hierarchy models of all the food chains that you can think of starting with ‘forest’ as the base class. Then create a menu driven program which displays the food chains of the various animals as per the ‘food chain’ hierarchy. An example of food chain of a ‘Wolf’ is given below. Use inheritance concept where ever possible. Food Chain of a ‘Lion’ WOLF SHEEP GRASS SUNLIGHT
- 368. 353Part II: Chapter 2 INHERITANCE 2.8 Let us revise! Inheritance is the concept of creating a new class, (derived class) from an already existing class (Base Class) The derived class can contain all the features of its base class and additional features of its own. A derived class is declared using the following format in C++: Syntax: class derived_className: Access_specifier Base_className. There are three type of access specifiers ‘Public’, ‘Protected’ and ‘Private’. The ‘Public’ type access specifier allows minimum amount of data hiding to take place. The ‘Protected’ type access specifier provides levels of data protection between that of ‘public’ or ‘private’ type access specifiers. The members under the ‘Protected’ type access specifier can be passed on to the derived classes but cannot be accessed by objects of classes. The ‘Private’ type access specifier provides highest level of data protection. Members under the ‘Private’ type access specifiers cannot be inherited by derived classes neither can they be accessed by objects of these classes. Constructors and Destructors of base classes cannot be inherited by derived classes. When ever an object of a derived class is declared or initialized the default constructor for the base class is automatically called for. A class can be derived from one or more base classes. When the derived class inherits its properties from a single base class it is defined as ‘Single Inheritance’ where as when a derived class derives its properties from more than one base class ‘Multiple Inheritance’ is said to be observed. The code for initiating multiple inheritances is given below: class A { … }; class B { … }; class C : public A, public B // Multiple Inheritance { … };
- 369. 354 Notes
- 370. 355 PPAARRTT IIIIII DDAATTAA SSTTRRUUCCTTUURREESS && FFIILLEESS
- 372. Part III CHAPTER 1 DATA STRUCTURES AIM OUTLINE • Introduce the concept of Data Structures • Learn to declare and use Structures, Arrays of Structures and Nested Structures. • Comparison structures and arrays. • Learn about Pointer to a structure • Pass and return structures to functions. • Appreciate the concept of ‘Structures’ 1.1 Introduction 1.2 Data Structures 1.3 Arrays 1.4 Operations on static data structures 1.5 Solved Examples 1.6 Review Exercise 1.7 Programming Project 1.8 Let us revise!
- 373. 358 1.1 Introduction A computer programmer creates programs which are series of instructions to manipulate and organize data in a desired manner. Organizing data and establishing a relationship among its members is called structuring data or in simpler words any mathematical and logical model of data is known as data structure. In programming, the term data structure refers to a scheme for organizing related pieces of information. The basic types of data structures include files, lists, arrays, records, trees and tables. Each of these basic structures allows operations to be performed on the data. Arrays are the simplest data structures. Some operations are common to all the data structures. These include accessing each member for some operation (traversal), search for locating a given record, insertion or deletion of records, sorting of records and merging of data structures. In the following sections, we shall discuss about the basic concepts of data structures, dynamic data structures and some more data structures known as linked lists, stacks and queues. 1.2 Data Structures Given below are some basic definitions related to data structures: Primitive data structure The data structures, whose definitions are available in the compiler, are known as the primitive data structures. Non Primitive data structure The data structures, whose definitions are not available in the compiler, are known as non primitive data structures. Linear data structure Linear data structures are those data structures where each element has access to a maximum of one predecessor and one successor element. For example: Stacks and Queues. Non-linear data structure They are the data structure in which each element can access any number of predecessor elements and any number of successor elements. For example: Tree, Graphs, etc.
- 374. 359 Part III: Chapter 1 Data Structures Static Data structure The data structure in which the numbers of elements are fixed is known as static data structure. Example: Arrays Dynamic data structures The data structures in which the numbers of elements are not fixed are known as dynamic data structure. Example: Linked List, Dynamic stack and Dynamic Queue. 1.3 Arrays Arrays are examples of static homogeneous data structures as the number of elements in array remain fixed. They are also commonly known as subscripted variables. We have already studied in previous chapters about how arrays are stored in consecutive memory locations. In this section, we will learn how we can determine the address of the memory locations occupied by an array. Let us begin with One Dimensional arrays. 1.3.1 One Dimensional Array Suppose we declare a one dimensional array ‘int A[i],’ where ‘i’ can take up any integer value. Now the number of elements in this array can also be found by subtracting the last index of the array also known as the ‘Upper Bound’ with the first index of the array also known as the ‘Lower Bound’. In this example, we have ‘i’ as the upper bound and ‘0’ as the lower bound. We can use the following general formula to determine the memory location of any index of a array. Memory location of A[i]: Location (A[i]) = Base_address (A) + W*(i-LB) Where, Number of elements = Upper Bound – Lower Bound = i - 0 = i Where ‘W’ is the size of the object created.
- 375. 360 1.3.2 Two Dimensional Arrays As discussed in Part I that the computer memory only supports a linear placement of data items akin to a one-dimensional array as a result of which the memory simulates the two/multi dimensional array as a one dimensional array only. Two dimensional arrays are also stored at contagious memory locations similar to that of the one dimensional array. The only difference between them being that two dimensional arrays can be stored in the following two ways: Row major Column Major Let us consider a two dimensional array A[i][j]. Number of elements = i * j = (UB(i) – LB(i) + 1) * (UB(j) – LB(j) +1) Where UB is the upper bound and LB is the lower bound. o Row Major Let us take a two dimensional array A[i][j] of size M*N that is having M rows and N columns (Figure 1.1). Figure 1.1: An array having M rows and N columns The computer always keeps track of the base address only. So, by knowing the base address (Base (A)) and size of each element (W) one can calculate the address of any [I,J] element. The algorithm for a row major two dimensional array is given below:
- 376. 361 Part III: Chapter 1 Data Structures Algorithm: For Row Major For Two dimensional array A[i][j] of size M rows and N columns. Base Address : Base (A), Address of the 1st element of the array No. of Column s: N = UB (j) – LB(j) UB(j) is the upper bound of column LB(j) is the lower bound of the column No. of Rows: M = UB(i) – LB(j) UB(i) is the upper bound of the row LB(j) is the lower bound of the column Size of object: W is the size of the object Memory Location: Loc(A[I][J]) = Base(A) + W*(M*I+J) Column Major ‘Column Major’ is also similar to the ‘Row Major’ order the only difference being that in the column major the elements are stored along the column. That is first the elements of the first column are filled and then the other ones are filled sequentially. To explain the functionality of column major order let us suppose an array A[i][j] with M rows and N columns.
- 377. 362 The algorithm given below explains the functionality of column major order. Algorithm: For Row Major For Two dimensional array A[i][j] of size M rows and N columns. Base Address : Base (A), Address of the 1st element of the array No. of Columns: N = UB (j) – LB(j) + 1 UB(j) is the upper bound of column LB(j) is the lower bound of the column No. of Rows: M = UB(i) – LB(j) + 1 UB(i) is the upper bound of the row LB(j) is the lower bound of the column Size of object: W is the size of the object Memory Location: Loc(A[I][J]) = Base(A) + W*(N*(J-LBJ) + (I-LBI)) If we assume LB(i), LB(j) as ‘0’ then we obtain a much simplified formula: Loc (A[I][J]) = Base(A) + W*(N*J+I) 1.3.3 Solved Examples Example 1.3.3.1 A one-dimensional array p[100] is stored in memory with a base address as 5000. Find out addresses of p[15] and p[40], if each element of the array requires 4 bytes. Solution of 1.3.3.1 Given Base(p) = 5000 W = 4 Loc (p[I]) = Base (p) + W* I Loc (p[15]) = 5000 + 4*15
- 378. 363 Part III: Chapter 1 Data Structures = 5000 + 60 = 5060 Loc (p[40]) = 5000 + 4*40 = 5000 + 160 = 5160 Solved Example 1.3.3.2 A one dimensional array A[-5,…,25] is stored in the memory with each element requiring 2 bytes. If the base address is 8000, find out the following: a) Address of A[5] and A[-3] b) Total no. of elements present in the array Solution of 1.3.3.2 Given, Base(A) = 8000 W = 2 LB = -5 Loc(A[I]) = Base(A) + W*(I-LB) Loc( A[5]) = 8000 + 2*(5-(-5)) = 8000 + 20 = 8020 Loc( A[-3]) = 8000 + 2*(-3-(-5)) = 8000 + 4 = 8004 Total No of elements = UB – LB + 1 = 25 – (-5) +1 = 31 Solved Example 1.3.3.3 A two dimensional array Q[5][15] is stored in memory along the row with each element requiring 2 bytes. If the base address is 6500, find out the following: a) Addresses of q[5][10] and q[3][5] b) Total no. of elements present in the array Solution of 1.3.3.3 Given, Base (q) = 6500 W = 2 M = 5
- 379. 364 Row major, Loc ( Q [I][J] ) = Base (Q) + W*(M*I+J) Loc ( Q[5][10]) = 6500 + 2 * (15 *5 +10) = 6500 +170 = 6670 Loc (Q[3][5]) = 6500 + 2 * (15*3 +5) = 6500 +100 = 6600 Total number of elements = N * M = 5 * 15 = 75 Solved Example 1.3.3.4 R[-4, …, 4, 7, …,17] is a two dimensional array, stored in the memory along the column with each element requiring 4 bytes. If the base address is 5000, find out the following a) Addresses of R[2][10] and R[3][15] b) Total no of elements in the array Solution of 1.3.3.4 Given, Base( R ) = 5000 W = 4 N = UBI – LBI + 1 = 4 – (-4) +1 = 9 M = UBJ – LBJ + 1 = 17 – 7 + 1 = 11 Column Major, Loc ( R[I][J]) = Base ( R ) + W * ( N * ( J – LBJ ) + ( I – LBI )) Loc ( R[2][10]) = 5000 + 4*( 9 * (10 -7 ) + ( 2 – (-4)) = 5000 + 132 = 5132 Loc ( R [3] [15] ) = 5000 + 4*(9*(15-7) + ( 3 – (-4)) = 5316 Total No of elements = N * M = 9 *11 1.4 Operations on static data structures 1.4.1 Insertion The insertion operation is performed by inserting an element at a desired location in an array. However, to add an element in between the array, all elements from
- 380. 365 Part III: Chapter 1 Data Structures the location of insertion need to be moved one location forward to accommodate the new element. This operation is shown in figure 1.2. Figure 1.2: Insertion of new element '19' in array INSERTION The program 1.1 gives the function to insert an element at a desired location into an array. Program 1.1: Function to insert Data at ‘Pos’ in the array ‘arr’. void insert (int arr[], int &l, int data) { if (l<max) { for (int c=l;c>pos;c--) arr[c]=arr[c-1]; arr[pos]=data; l++ ;// After insertion of an element increases L } else cout<<” Array is Full”<<endl; } Note: ‘Pos’ is presumably defined as global variable and ‘pos’ is the position from which insertion will take place. 1.4.2 Deletion As the name suggests the ‘deletion’ operation is used to delete an element from a specified position (‘pos’ in figure 1.3) in the static data structure. When the selected element is deleted from the data structure, then the elements stored at subsequent positions in the structure shift either one step forward or shift one step Location of Element for insertion New element inserted
- 381. 366 backward towards the position from where the element was deleted as also shown in the figure 1.3. Figure 1.3 Element 19 deleted from array INSERTION The program 1.2 given below enlists the piece of code used to delete an element from an array. Program 1.2: Function to delete data from POS in the array ‘arr’ void delete (int arr[], int &l, int pos) { for (int c=pos;c<l-1;c++) arr[c]=arr[c+1]; l--; } 1.5 Solved Examples Example 1.5.1 The base address of a two dimensional array A with 5 rows and 4 columns is 2000. If each element requires 4 bytes for its storage, find the address of element a[3][3] if arrangement is row major? Solution of 1.5.1 Now we have been given the Base address (A) = 2000 POS Array INSERTION before and after DELETION
- 382. 367 Part III: Chapter 1 Data Structures Size of each element (w) = 4 Number of columns (N)= 4 Now, the elements have been stored as row major Address of element A[i][j]= A + w(N*I + j) = 2000 + 4 * (4*3+3) = 2060 So, the required address is 2060. Example 1.5.2 For an array of real numbers Realarr[20][20] find the address of Realarr[10][12] if Realarr[1][1] is stored in location 1000. Assume each real number requires 4 bytes. Show the various steps of calculation? Solution of 1.5.2 We use following formula to compute the address: A[i][j] = B + w(n(I-LI) + (J-L2)) Now, Base address (B) = 1000 Size of each element = 4 L1=L2=1 A[10][12]= 1000 + 4(20(20-1) + (20-1)) = 1000 + 4(20(19) + 19) = 1000 + 4(380 +19) = 1000 + 4(399) = 1000 + 1596 = 2596 Example 1.5.3 Write an algorithm which finds the location and values of largest and second largest elements in two dimensional array DATA with N rows and M columns. Solution of 1.5.3 Let us assume an array A[][] having M rows and N columns. Step : 1. Max = A[0][0] 2. Min = A[0][0]
- 383. 368 3. i=0 4. while (i<m) repeat 5 to 10. 5. j=0 6. while (j<N) repeat step 7 to 9 7. if (Max<A[i][j]) Max= A[i][j] 8. if (Max>A[i][j]) Max=A[i][j] 9. j= j+1 10. i= i+1 11. cout<<” Maximum value is”<<Max 12. cout<<” Maximum value is”<<Min 13. End. Example 1.5.4 Each element in an array DATA[20][50] requires 4 bytes of storage. Base address of data is 2000, determine the location of DATA [10][10] when the array is stored as (a) Row major (b) Column major Solution of 1.5.4 Base Address = 2000 Size of the element = 4 (i) Row major Order N=50 DATA[10][10]= 2000 + 4(50(10-1) +(10-1)) = 2000 + 4(50 (9) + 9) = 2000 + 4(454) = 2000 + 1816 = 3816 (ii) Column major Order M = 20 DATA[10][10]= 2000 + 4(20(10-1) + (10-1)) = 2000 + 4(20(9) + 9) = 2000 + 4(180 +9) = 2000 + 4(189) = 2000 + 756 = 2756
- 384. 369 Part III: Chapter 1 Data Structures Example 1.5.5 Given an array A[6][16] whose base address is 100. Calculate the location A[2][5] if each element occupies 4 bytes and array is stored row wise? Solution of 1.5.5 Base address = 100 W=4 L1 = 1 L2 = 1 N = 16 A[2][5] = B + W(n(I- L1) + (J – L2)) = 100 + 4(16(2-1) + 5-1) = 100 + 4(16 + 4) = 100 + 80 = 180 Example 1.5.6 Explain how bubble sort will sort the following array: 11 6 26 9 8 24 18 Solution of 1.5.6 Bubble sort compares adjacent elements and swaps them if necessary and then compares the next pair of adjacent elements. The method when applied to the given array gives the following result:- Pass 1: Swap numbers 11 and 6 to give 6 11 26 9 8 24 18 Swap numbers 26 and 9 to give 6 11 9 26 8 24 18 Swap numbers 26 and 8 to give 6 11 9 8 26 24 18 Swap numbers 26 and 24 to give 6 11 9 8 24 26 18 Swap numbers 26 and 18 to give 6 11 9 8 24 18 26 Pass 2: Swap numbers 9 and 11 to give 6 9 11 8 24 18 26 Swap numbers 11 and 8 to give 6 9 8 11 24 18 26 Swap numbers 24 and 18 to give 6 9 8 11 18 24 26 Pass 3: Swap numbers 9 and 8 to give 6 8 9 11 18 24 26
- 385. 370 Example 1.5.7 It is required to sort the following array by method of insertion sort. Show the status of the array after each insertion: 40 23 15 67 55 84 32 Show the status of the array after each insertion? Solution of 1.5.7 Example 1.5.8 The following numbers (89, 20, 31, 56, 20, 64, 48) are required to be sorted using selection sort showing how the list would appear at the end of each pass. Solution of 1.5.8 The example has the following solution: 1.6 Review Exercise 1/2 Mark Questions Q1) Define data structure and give its physical significance? Q2) Describe the following operations on an array with suitable examples: 0 23 40 15 67 55 84 32 1 15 23 40 67 55 84 32 2 15 23 40 67 55 84 32 3 15 23 40 55 67 84 32 4 15 23 40 55 67 84 32 5 15 23 32 40 55 67 84 89 20 31 56 20 64 48 20 89 31 56 20 64 48 20 20 31 56 89 64 48 20 20 31 56 89 64 48 20 20 31 56 89 64 56 20 20 31 48 56 64 89 Iteration Array
- 386. 371 Part III: Chapter 1 Data Structures Traversal Linear search Bubble sort Merging Q3) Which of the two (‘bubble’ and ‘selection’) sort is the most efficient and why? Q4) The base address of a two dimensional array X with 15 rows and 4 columns is 2000. If each element requires 4 bytes for its storage, find the address of element X[10][4] if arrangement is row major? Q5) The following numbers are required to be sorted using bubble and selection sort showing how the list would appear at the end of each pass. a) 20 40 30 10 50 b) 10 2 3 4 -2 1 5 Q6) If 2 dimensional array C[5…10,-5…9] is stored using column major representation, then calculate the address if C[8,-2], if the base address is 10 and each array element requires 2 bytes of memory. 2/3 Mark Questions Q7) Each element of an array DATA[20][50] requires 2 bytes of storage. Base address of DATA is 2000. Determine the location of DATA[10][22] when the array is stored as (a) Row Major (b) Column Major Q8) Given two dimensional array A[10][20], base address of A is 100 and width of each element is 4 bytes. Find the location of A[12] Q9) If two dimensional array C[5, …, 9,-4,…, 10] is stored using column major representation, then calculate the address if C[8,-2], if the base address is 10 and each array element requires 2 bytes of memory. 4/5 Mark questions Q10) Write a menu driven program in C++ to sort an array using selection and bubble sort. Q11) Write a menu driven program to search the array using linear and binary search? Q12) Explain the following terms primitive and non primitive data structures and differentiate between them giving suitable examples.
- 387. 372 1.7 Programming Project The programming project for this chapter involves the creation of a telephone directory for a set of 50 subscribers. The directory should include subscriber names listed in alphabetical order, together with their street address and telephone numbers. The program created should also have an option to search for any specific data entry and enlist the details involved with that data entry. Hint: Use ‘Selection’ sort and ‘Binary’ search for better results
- 388. 373 Part III: Chapter 1 Data Structures 1.8 Let us revise! The data structure, which is available in the compiler, is known as a primitive data structure. The data structure, which is not available in the compiler, is known as non primitive data structure. The data structure in which each element has access to maximum of one predecessor element and maximum of one successor element is known as linear data structure. Example: stack, queue, etc. The data structure in which each element can access any number of predecessor elements and any number of successor elements known as non linear data structure. Example: Tree, Graphs, etc. The data structure in which the numbers of elements are fixed is known as static data structure. Example: Arrays The data structures in which the numbers of elements are not fixed are known as dynamic data structure. Example: Linked List One dimensional array Number of elements N= UB – LB + 1 Where LB- lower bound, UB – upper bound Memory location of A[i]: Loc (A[i]) = Base (A) + W*(i-LB) Loc (A[i])=Base(A) + W*I Where N is given as in c++, LB is assumed as 0 Two dimensional array A two dimensional arrays can be stored in the following two ways. Row major Column Major Number of elements = N * M = (UBI – LBI + 1) * (UBJ – LBJ +1)
- 389. 374 Binary search The prerequisite condition to search as element from an array using binary search technique is that the array should be sorted in ascending or descending order. The deletion operation is used to delete an element from any position (pos). The insertion operation is used when a new element is inserted into an existing array, but to insert a data entry in between the array all elements from the inserted place will move one place forward to accommodate the new data element.
- 390. Part III CHAPTER 2 LINKED LISTS AIM OUTLINE • Introduce the concept of ‘Linked Lists’. • Able to create and perform various operations on ‘Linked Lists’ • To create ‘Linked’ data structures using pointers, self referential structures. • Use applications of ‘Link Lists’ such as ‘stacks’ and ‘queues’. 2.1 Introduction 2.2 Linked lists 2.3 Linked lists as stacks 2.4 Linked lists as queues 2.5 Solved examples 2.6 Review exercise 2.7 Programming project 2.8 Let us revise!
- 391. 376 2.1 Introduction Till now, we have only studied about linear data structures like arrays and structures. This chapter introduces dynamic data structures which have distinct advantages over traditional linear data structures. The most obvious being that the size of elements in an array was limited to the number provided at the time of declaration. The next being that in order to avoid the danger of running out of allocated memory we would declare huge array sizes but hardly ever, would use the entire space allocated to these arrays. Hence, we were deprived of using a lot of potential memory which could have been allocated to and utilized by other applications. Further, operations like ‘insertion’ and ‘deletion’ of data were also very time consuming and inefficient, as order of all the elements had to be changed depending upon the operation applied. ‘Linked Lists’ introduce practical solutions to all these problems. This chapter introduces the concept of ‘Link Lists’ and describes algorithms and their implementation through programs. 2.2 Linked Lists Linked Lists are dynamic data structures which can shrink and expand during program execution. In theory, they are very similar to arrays. The difference lies in the construction of their structures. An array can be described as a huge unit of data where all its elements are grouped together as a single block of memory. Where as, linked lists allocate different blocks of memory to each element called ‘node’ where each node contains two fields the ‘data’ field to store data parameters and ‘next’ field which is a pointer which points to the next node in the list. The figure 2.1 given below shows the cases of sequential (array type) and linked memory allocations. Figure 2.1: Sequential and Linked memory allocation
- 392. 377 Part III: Chapter 2 Linked Lists Linked lists have various advantages over traditional arrays. Firstly, when declaring linked lists, we are not confined to a fixed space as in arrays. For example, if one observes carefully in the figure given above the pointer part in node ‘A4’ of the linked lists contains a NULL value. If we want to add another node to the list one can replace this null value with the address of the next node. Hence, we need to allocate memory only when required. In addition to this, one can also group more data in the data part of the node to save additional memory. Linked lists are also very useful as the process of ‘insertion’ or ‘deletion’ of data is fairly simple and can be achieved without shifting the entire data structure depending upon the operation applied. Figure 2.2 depicts how the deletion operation works. In this figure, we have deleted the node ‘A2’ by just changing the address stored in the pointer part of the node ‘A1’ from ‘A2’ to ‘A3’. Figure 2.2: Deletion of node As described earlier, linked lists consist of ‘data’ part for storing data and pointer variables which store the address of the next node in the list, so they can also be expressed with the help of pointer diagrams. For example, figure 2.3 shows pointer diagrams for the linked list declared in figure 2.1. Figure 2.3 Pointer Diagram
- 393. 378 In this figure, each box represents a ‘node’ and each arrow represents a ‘pointer’. Let us now learn how to create and use linked lists. 2.2.1 Creation of Linked Lists We already know that linked lists are a linear collection of self referential structures. So in order to generate a linked list, we first have to declare a self referential structure. Program 2.1 illustrates the creation of a linked list. Program 2.1: /******************************* Structure Name: list Structure Variables: data, link (pointer to structure) ******************************************************/ struct list { int data; list *link; }data1,data2,data3; data1.link=&data2; // link to data 2 data2.link=&data3; // link to data 3 data3.link=NULL; // Marks the end of the list In this program, we have declared a self referential structure ‘list’ having ‘data1’, ‘data2’ and ‘data3’ as its variables which have been used to generate a linked list. It is always easier to understand and keep track of linked lists with the use of figures so it is recommended to make pointer diagrams to understand the working of the program. Figure 2.4 shows two representations of the link list created in program 2.1. Figure 2.4 (a): Pointer Diagram
- 394. 379 Part III: Chapter 2 Linked Lists Figure 2.4 (b): Linked List Table PROGRAMMING TIPS Using dynamic data structures is far more efficient than using traditional data structures as dynamic data structures can grow or shrink in size as per the need of the program. Insertion and deletion of elements in an array is time consuming as the entire array has to be shifted accordingly. Hence, using link list is a more efficient method for data storage. 2.2.2 Traversing a linked list To access the data contained in the nodes of the list, we take the pointer to the first node of the list and then use the pointer in the link field of the first node to locate the second node. This procedure is repeated until we encounter the last node which has the entry NIL in its link field. The process of going through Node1, Node 2 and so on of the list is called as traversing a linked list. Now that we understand what one means by traversing a linked list let us develop an algorithm to perform this operation. Algorithm Prerequisite: This algorithm works for a linked list in which each node stores the address of its preceding node.
- 395. 380 Steps 1. Declare a pointer, pointing to the last node. 2. Assign the pointer variable declared in step 1, the address stored in the pointer part of the self referential structure. 3. Repeat the process described in step 2 until the pointer is assigned a NULL value which marks the end of the link list. It might be easier to understand this algorithm with the help of an example. Program 2.2 creates a linked list which fulfils the prerequisite for our algorithm and then traverses the link list using the algorithm given above. Program 2.2: Traversing a linked list /************ Program 2.2: Traversing a linked list *****************************************/ // Header Files # include<iostream.h> # include<conio.h> /**************** Self Referential structure Name: List Data Members: integer variable 'data', pointer variable 'ptr' *************************************************************/ struct list { int data; list *ptr; }; // Main Function void main() { clrscr(); int i=1; // Pointer variable pointing to memory block of type List list *l1 = new list; // Pointer variable l2 of type list
- 396. 381 Part III: Chapter 2 Linked Lists list *l2; l1->ptr=NULL; int counter=1; char ch = 'y'; while (ch != 'n') { cout<<" n Enter data for node"<<counter; cin>>l1->data; l2 = l1; l1=new list; l1->ptr=l2; cout<<"n Do you want to add another node (y/n)"; cin>>ch; } // While loop traversing the Linked List while ( i != 0) { if (l2->ptr == NULL) { i=0; } l2=l2->ptr; } getch(); } In this program we have created a linked list as a collection of self referential structures of type ‘list’ using two pointer variables ‘l1’ and ‘l2’. At first when these pointers are declared ‘l1’ points to a dynamically allocated block of memory or to the first node of type ‘list’ (figure 2.5(a)). The pointer part ‘ptr’ of this node is initialized to NULL and data value say ‘10’ is inserted in the data part ‘data’ of the node by the user. A copy of this nodes address is then assigned to the other pointer variable ‘l2’ which then also starts pointing to the this node (figure 2.5(b)).
- 397. 382 (a) (b) Figure 2.5: Creating link lists using pointer variables When this is done, the pointer ‘l1’ is then intialized to point to another memory block of type ‘list’ or create the second node of the ‘list’ (figure 2.6(a)) and the address stored in the pointer variable ‘l2’ is assigned to pointer part ‘ptr’ of the second node (figure 2.6(b)). Figure 2.6(a): Creating Linked Lists Figure 2.6 (b): Creating linked lists
- 398. 383 Part III: Chapter 2 Linked Lists This entire process can continue till the user wants and allows us to create link lists of any size but what do we achieve by applying this complicated process? Well, with the help of this piece of code we have created a linked list in which each node stores the address of its preceding node as per the prerequisite required in the algorithm we studied in section 2.2.2. Let us now try to understand how the program given above traverses the link list that was created. while ( i != 0) { if (l2->ptr == NULL) { i=0; } l2=l2->ptr; } Program 2.2 uses the above piece of code to traverse the link list. In this piece of code the pointer ‘l2’ is used to traverse the linked list and during each loop it is assigned the address of the node to which it points to, which happens to be the address of the preceding node. This process is carried on till the point the first node containing ‘NULL’ value stored in the pointer part is encountered where it comes out of the while loop achieving its purpose. COMMON PROGRAMMING ERRORS Not initializing the pointer in the last node of the link list can lead to a runtime error. Loosing the memory address of a node in a ‘Linked List’ can lead to a logical error. 2.2.2.1 Solved Examples Example 2.2.2.1.1 Create a linked list for the structure ‘Employee’ given below. The program should input the ‘Name’ and the ‘Age’ of the Employee and then display the linked list on the screen. The program should keep inserting nodes till the user wants.
- 399. 384 strcut Employee { char Name[15]; int Age; Employee *ptr; }; Solution of 2.2.2.1.1 /************ Solved Example: 2.2.2.1.1 *****************************************/ // Header Files # include<iostream.h> # include<conio.h> # include<stdio.h> /**************** Self Referencial structure Name: Employee Data Members: character array Name[] integer variable 'Age' pointer variable 'ptr' ******************************************************/ struct Employee { char Name[15]; int Age; Employee *ptr; }; // Main Function void main() { clrscr(); int i=1; // Pointer variable pointing to memory block of type Employee Employee *E1 = new Employee ; // Pointer variable l2 of type Employee
- 400. 385 Part III: Chapter 2 Linked Lists Employee *E2; E1->ptr=NULL; int counter=1; char ch = 'y'; while (ch != 'n') { cout<<" n Enter data for Employee "<<counter; cout<<" n Enter Name of Employee"; gets(E1->Name); cout<<" Enter the age of the Employee"; cin>>E1->Age; E2 = E1; E1=new Employee; E1->ptr=E2; cout<<"n Do you want to add another Employee (y/n)"; cin>>ch; counter++; } // While loop printing the Employee List cout<<" n n ######### EMPLOYEE LIST ########"; while ( i != 0) { if (E2->ptr == NULL) { i=0; } cout<<"n Employee Name :"<<E2->Name; cout<<" t t Age :"<<E2->Age; E2=E2->ptr; } getch(); }
- 401. 386 In this program, we have created a link list using the structure ‘Employee’. The important thing to notice in this example is that it uses two pointers ‘E1’ and ‘E2’ where ‘E1’ is used to point to the new dynamic memory blocks or the new nodes and the pointer ‘E2’ is used to store the address of the previous node and assign it to the pointer part ‘ptr’ of the node just as we did in program 2.2. 2.2.3 Inserting nodes in linked lists In this section, we will study about insertion of a node in the middle of a linked list given in figure 2.7. Figure 2.7: Link List Let us insert a node between ‘Node 3’ and ‘Node 4’. To do so, we have to keep two things in mind. Firstly, we need the address of the preceding node ‘Node 3’ and the node which is being inserted say ‘Node X’. Further, we need to use two pointers: one which will hold the address of ‘Node3’ and the other which will hold the address of the newly inserted node ‘Node X’ so that we can maintain the chain of links in the link list. Figure 2.8 shows how the link list would look like when ‘Node X’ is inserted in the link list. Figure 2.8: Insertion of Node ‘X’ in the already created link list 2.2.4 Deleting nodes from linked lists The process of deletion of a node from the linked list is quite simple. Let us again take the same link list which we took to understand the insertion operation (Figure 2.7). Say, we want to delete ‘Node 3’ of the link list then all one has to do is change the address
- 402. 387 Part III: Chapter 2 Linked Lists stored in ‘Node 4’ from 300 to 200 and ‘Node 3’ will be deleted. Figure 2.9, shows how the link list would look after ‘Node 3’ has been deleted from it. Figure 2.9: Deletion of nodes from a link list 2.2.5 Circular or ring linked lists Circular or ring linked lists are special linked lists. They have the special property that the last node links to the first node (see figure 2.10). Therefore, this link list does not have a node having a NULL pointer. The advantage of this structure is that the list can be accessed at any given node. There is no first or last node. The node structure is similar to linked list and can also be easily updated. Insertion of a new node at a desired place can be done by rearranging the links to include the new node in the ring. Deletion of a node is performed by adjusting the links as in the case of ordinary linked lists. Figure 2.10: Circular linked lists Since there is no first or last node, the problem of locating the head of the list must be solved. The following solutions are possible: While moving through the list from one node to the next, we may stop when we return to the node from where we started. We can also include a special recognizable node into each circular list as a convenient stopping place. This special node is often called the list head, as shown in figure 2.11.
- 403. 388 Figure 2.11: Circular linked lists HEAD COMMON PROGRAMMING ERRORS Declaring a NULL pointer in the last node of a ringed link list can lead to an logical error Loosing tracks of pointers while inserting or deleting link list well result in a logical or a runtime error. 2.2.6 Multi-Lists Singly linked lists pose problems as they are one-directional. One can only traverse these linked lists in the direction of the links but what happens if we want to traverse linked lists in both directions? For this purpose, we need to define a more general list structure - one in which data elements belong to several lists. This type of structure is called a multi-list. The nodes of such lists have more than one link. For example, we can include two links in each node, pointing to the items on both sides of the node. This structure is referred to as a doubly- linked (two-way) list. As shown in figure 2.12, each node has three fields. Two link fields, one each for forward & backward links and a data field. Figure 2.12: Doubly linked list As one might observe in the figure above, both the left and the right nodes of the list have NIL links. We can also have a ringed doubly linked ring structure as shown in the figure
- 404. 389 Part III: Chapter 2 Linked Lists 2.13. The only change we will observe in such a structure is that no node will contain NIL values stored in the link part of their structure. Figure 2.13: A doubly linked ring structure The cost of creating such structures is the need for extra memory to keep the second link. The process of deletion and insertion in doubly linked lists is very similar to the one we have already followed for singly linked lists. Figure 2.14 shows the modification in the linked list when a new node X is inserted into the link list and Figure 2.15 shows the effect on the link list when this element X is deleted from the linked list. Figure 2.14: Insertion operation in Double Linked Lists Figure 2.15: Deletion operation in doubly linked lists These sections describe linked lists and the various operations possible on them. Later in this chapter, we will learn about the practical applications of linked lists like creating stacks and queues which involve singly linked and doubly linked lists.
- 405. 390 2.3 Linked Lists as Stacks Stacks can be compared to a ‘pile of plates’ where we stack one plate on the top of another. Now, if we want to stack another plate in the pile we would stack it on the top of the pile as stacking a plate at the bottom of the pile will require far more effort. Similarly, when a plate is being removed, again it is removed from the top of the pile for simplicity. To further understand stacks, we can also take up the example of a ‘Pile of letters’ lying in a mail box where a new letter can only added on ‘Top’ of the pile and an old letter can be removed from the same point ‘Top’ as also shown in figure 2.16. Figure 2.16: An example of a stack Stacks are non primitive linear data structures in which insertion and deletion of data takes place from only one end known as the ‘TOP’. They are homogeneous collection of data of one type. Stacks are also an example of a LIFO structure that is ‘Last in First out’ data structure. Data is added in to the stack using the Push operation, and removed using the Pop operation. These topics will be covered in the following sections but first let us learn about static arrays. 2.3.1 Static Stacks Stacks can also be implemented as one dimensional arrays known as ‘Static’ stacks. ‘Static’ stacks have a fixed amount of data which can be entered into them and generally used when we know how much space the data would occupy. ‘Static’ stacks are created by declaring a linear one dimensional array in which insertion or ‘Pushing’ and deletion or ‘Popping’ of data is allowed only from one single location known as the ‘TOP’. Let us
- 406. 391 Part III: Chapter 2 Linked Lists take an example of a static stack by declaring an array ‘Static_Stack’ which can store 8 integer data values. What happens if we want to insert or ‘push’ data into our static stack? Say we push the data value ‘5’ into our stack then our linear array would look like figure 2.17. Figure 2.17: Pushing data into the stack Now after the insertion of the data value the position from where insertion and deletion takes place ‘TOP’ shifts to the proceeding memory location, just like in our ‘Pile of Plates’ example where once a plate is inserted or pushed in the pile the next plate can only be inserted on top of the previous plate. Now let us see what happens if insert another data value say ‘10’ into the array what will be the effect of this insertion in our stack. Figure 2.18: Insertion of data entry into the stack Our stack would then appear as figure 2.18 given above. Again if we notice carefully ‘TOP’ is shifted to the next memory location. Similarly, we can add more data values in our stack but what happens if we want to pop or delete a data entry? When using stacks, we can only delete the preceding entry that is the data entry at the location ‘TOP – 1’. So if we want to delete the data entry ‘5’, we will first have to delete the data entry ‘10’ as insertion and deletion in a stack is only allowed from a single location. It might be more apparent now that why stacks are also known as LIFO structures. Figure 2.19 shows the effect on the stack when the data entry ‘10’ is deleted or in other words ‘popped’ from the stack. After the data entry has been deleted the position ‘TOP’ is decremented and then signifies the preceding index of the linear array.
- 407. 392 Figure 2.19: Deleting a data value from a stack We can also develop a C++ code to carry this procedure. The program 2.3 given below represents stack as an array. Program 2.3 // Header Files # include<iostream.h> # include<conio.h> // Maximum No. of data entries allowed in the stack const int max=8 /******************************************* Function Name: Push Purpose: To insert a data entry in the array *******************************************/ void Push(int s[],int &t) { if (t<max) { cout<<" Data to be entered"; cin>>s[++t]; cout<<endl<<"Data has been entered"; cout<<endl<<"_____________________"; } else cout<<"n The stack is full"<<endl; } /******************************************* Function Name: Pop Purpose: To delete a data entry from the array
- 408. 393 Part III: Chapter 2 Linked Lists *******************************************/ void Pop(int s[],int &t) { if (t!=-1) { cout<<endl<<" The data entry "<<s[t]<<" has been deleted"; t--; } } /******************************************* Function Name: display Purpose: To display data entries in the stack *******************************************/ void display(int s[],int &t) { cout<<endl<<"Displaying data entries in the stack"; for (int i=t;i>=0;i--) cout<<endl<<s[i]; } //Main Function void main() { clrscr(); int static_stack[max], top=-1; int ch; do { cout<<endl<<"n 1) Push"; cout<<"n 2) Pop"; cout<<"n 3) Display"; cout<<"n 4) Quit"; cin>>ch; switch(ch) { case 1:Push(static_stack,top); break; case 2:Pop(static_stack,top); break; case 3:display(static_stack,top); break; }
- 409. 394 } while (ch!=4); getch(); } Let us closely examine this program. In this program, we have declared a one dimensional array ‘static_stack’ which can hold 8 integer data values represented by the constant variable ‘max’. The variable ‘Top’ is initialized to value ‘-1’ and is used to keep track of the position from where data values are inserted and deleted. The function ‘Push’ is given below: Function ‘Push’ void Push(int s[],int &t) { if (t<max) { cout<<" Data to be entered"; cin>>s[++t]; cout<<endl<<"Data has been entered"; cout<<endl<<"_____________________"; } else cout<<"n The stack is full"<<endl; } This function is used to insert data entries into the stack. The linear array and the variable ‘Top’ are passed to it as reference parameters. Which means that any changes made to the formal parameters ‘s’ and ‘t’ are also reflected back to the actual parameters. The function inserts the data entry at the index ‘t+1’ of the linear array. This is also the reason why the variable ‘top’ is initialized to -1 when it is being declared. This process carries on as long as the number of data entries inserted is less than the maximum number of data entries allowed in the stack. Function ‘Pop’ void Pop(int s[],int &t) { if (t!=-1) { cout<<endl<<" The data entry "<<s[t]<<" has been deleted"; t--;
- 410. 395 Part III: Chapter 2 Linked Lists } } This function is used to delete or pop data entries from the static stack. It also receives the static array and the variable ‘top’ as reference parameters which are used to pop a data entry from the stack. This process can carry on as long as the stack has data entries in it or in other words as long as the value of ‘t’ is not -1. The important thing to note in this function deletes only one data entry at a time and it always deletes the last data entry. Function ‘Display’ void display(int s[],int &t) { cout<<endl<<"Displaying data entries in the stack"; for (int i=t;i>=0;i--) cout<<endl<<s[i]; } All this function does is that it prints all the data entries present in the stack on the screen. 2.3.2 Dynamic Stacks Stacks which are implemented as arrays are very useful if we know the number of data entries from before, in other words they are useful when a fixed amount of data is to be accessed. To get an idea of this imagine a recursive loop. A menu driven stack program might be a good example. Now in this loop, we do not know how many times this loop might recur or how many data entries the user is willing to enter. The decision entirely rests on the user and might vary every time the program is executed. Using a stack represented as an array might be a poor solution as one will have to declare a very large block of memory so that there is no danger of the program running out of memory. This can waste a significant amount of computer’s memory if the loop only repeats itself a few times during the execution of the program. How do we solve this problem? In such a case, we have to use the concept of dynamic memory allocation. As stacks are represented as arrays, they can also be represented as pointers by creating a linked list with the use of which one can create ‘Dynamic stacks’, whose size can increase or decrease during the execution of the program. Program 2.4 demonstrates a ‘Dynamic Stack’ with its various operations.
- 411. 396 Program 2.4: Dynamic Stack // Header Files # include <iostream.h> # include <conio.h> /**************************** Self Referential Structure Name: Node Members: integer variable 'data' : pointer 'ptr' to type node ***********************************/ struct node { int data; node *ptr; }; /******************************************** Class: Stack Private Members: top: pointer to type node Public Members: void push() void pop() void display() Constructor: stack() Destructor: ~stack() ********************************************/ class stack { node *top; public: stack(){ top = NULL;} void push(); void pop(); void display(); ~stack(){}; }; /*************************** Name: push Parameters: None Purpose: Insert data entry in to dynamic stack ***********************************************/
- 412. 397 Part III: Chapter 2 Linked Lists void stack::push() { node *temp= new node; cout<<endl<<"Enter Value"; cin>>temp->data; temp->ptr=top; top=temp; } /*************************** Name: pop Parameters: None Purpose: Delete data entry from dynamic stack ***********************************************/ void stack::pop() { node *temp=top; if (temp != NULL) { cout<<endl<<top->data<<" Deleted"; top=top->ptr; delete temp; } else { cout<<"n The stack is empty"; } } /*************************** Name: display Parameters: None Purpose: Display data entries of the dynamic stack ****************************************************/ void stack::display() { node * temp= top; while (temp != NULL) { cout<<temp->data<<endl; temp=temp->ptr; } }
- 413. 398 /****************** Main Function ******************/ void main() { clrscr(); stack dynamic_stack; int choice; do { cout<<endl<<" 1. Push"; cout<<endl<<" 2. Pop"; cout<<endl<<" 3. Display"; cout<<endl<<" 4. Exit"<<endl; cout<<"n Enter Choice"; cin>>choice; switch(choice) { case 1: dynamic_stack.push(); break; case 2: dynamic_stack.pop(); break; case 3: dynamic_stack.display(); break; } } while (choice != 4); getch(); } Let us start from the very beginning of this program. At first, we have declared a self referential structure ‘node’ which contains an integer variable ‘data’ and a pointer variable ‘ptr’ of type ‘node’. What we have done with this declaration is that we have created a pointer which can hold an address of structure variable of type ‘node’ as also shown in figure 2.20.
- 414. 399 Part III: Chapter 2 Linked Lists Figure 2.20: Self referential structure Following the structure declaration, we have declared a class stack consisting of a pointer variable of type ‘node’ as its private member and contains ‘push’, ‘pop’ and ‘display’ as its public member functions. We will take up these functions in the next section of this chapter. The important thing to remember here is that whether one represents stacks as arrays or stacks as pointers it does not change the concept of ‘stacks’. 2.3.2.1 Functions of a dynamic stack The ‘push’ function void stack::push() { node *temp= new node; cout<<endl<<"Enter Value"; cin>>temp->data; temp->ptr=top; top=temp; } When this function is called then a pointer variable ‘temp’ is declared pointing to type ‘node’. Figure 2.21: Pointer ‘temp’ of type ‘node’ For understanding the concept, let us say we enter the data value ‘5’ then how does this data entry get pushed on to the array? To comprehend how insertion of a data entry takes place in a dynamic stack let us look
- 415. 400 at the following statements: cout<<endl<<"Enter Value"; cin>>temp->data; To begin with, we have stated ‘cin>>temp->data’, this statement is used to enter data entries into the ‘data’ part of the new node. We can also visualize this by looking at the figure 2.22 assuming that the user enters ‘5’ into the dynamic stack. Figure 2.22: Insertion of a data entry in a node In the next few statements, we have assigned the address ‘NULL’ stored in pointer variable ‘top’ initially to the pointer variable ‘ptr’ of the new node and then we have assigned the address stored in the pointer variable ‘temp’ to the pointer variable ‘top’ temp->ptr=top; top=temp; as shown in figure 2.23. Figure 2.23: Insertion of data entries in the stack
- 416. 401 Part III: Chapter 2 Linked Lists Now what happens if we insert another data entry ‘10’ into the stack? To know how this works, we will again follow the same steps as we did when we pushed our first data entry into the stack. The pointer ‘temp’ is again initialized pointing to a new memory block of type node in which the new data entry ‘10’ (see figure 2.24) is entered into the stack. Figure 2.24: Insertion of data entry into the stack Again, we execute the same set of statements given below which assign the address stored in the pointer ‘top’ to the pointer part of the new node and the address stored in pointer ‘temp’ to the pointer variable ‘top’. temp->ptr=top; top=temp; The only change one will observe with this data entry is that now the new node with the memory address ‘108’ will start pointing to the preceding node with the memory address ‘100’ ( see figure 2.25).
- 417. 402 Figure 2.25: Insertion of data entry into a Dynamic stack If one enters another data say ‘15’ in the stack, one would observe similar effect with only an additional level of reference as also shown below in figure 2.26. Figure 2.26: Insertion of data entry into a Dynamic stack The ‘pop’ function void stack::pop() { node *temp=top; if (temp != NULL) { cout<<endl<<top->data<<" Deleted"; top=top->ptr; delete temp; } else {
- 418. 403 Part III: Chapter 2 Linked Lists cout<<"n The stack is empty"; } } To understand the working of this function, we will carry on with the same link list which we created using our push function. So how does this function work? When this function is called at first, this program creates a pointer variable ‘temp’ of type ‘node’ and assigns it the address of the latest data entry in the stack stored in the pointer variable ‘top’ to the last node or the last data entry. Let us now come back to the stack we created in the previous section with integers ‘5’, ‘10’, ‘15’ as data entries. Say, now we want to pop a data entry then how does this work and how does the stack look like at each step? The entire process of how a data entry is popped and how does the stack look like in each step is explained in the figure 2.27 given below:
- 419. 404 Figure 2.27: Popping of data entry from dynamic stack The ‘display’ function void stack::display() { node * temp= top; while (temp != NULL) { cout<<temp->data<<endl; temp=temp->ptr; } } All this function does is that it prints all the data entries stored in the stack on the screen until it approaches the first node in the stack having ‘NULL’ value in its pointer part. The working of this function is left as an exercise to understand? Hint: Create pointer diagrams to understand working of any programs involving linked lists PROGRAMMING TIPS Using the delete operator, to delete the unused memory blocks is a good programming practice as it not only retrieves memory for other uses but also reduces potential errors. Constructors should always be accompanied with destructors to save memory.
- 420. 405 Part III: Chapter 2 Linked Lists 2.3.3 Uses of stacks One might wonder where stacks are used in programming languages. One common use of stacks is for converting an infix expression to postfix and finding the value of an postfix expression but what does one mean by an infix or postfix expression? An infix expression is an expression which we all have commonly observed in conventional algebra, for example ‘4+3’, which is an expression representing the sum of two integers ‘4’ and ‘3’. Infix expressions place their (binary) operators between the two values to which they apply. In the above example, the addition operator was placed between the ‘4’ and the ‘3’. On the other hand, a post fix expression places each operator after the two operands to which it applies as its name might suggest as ‘post’ means after. So if we want to convert our infix statement ‘4+3’ as a postfix statement it would become ‘4 3 +’. Figure 2.28 given below enlists few examples of ‘Infix’ and ‘Postfix’ operations. Infix: Postfix: 8 / 2 8 2 / (8+7) * 2 8 7 + 2 * (2+4)/(3+4) 2 4 +3 4 + / 8 * (5 + 4) 8 5 4 + * Figure 2.28: Infix and Postfix expressions The important thing to note in the table given above is that postfix expressions do not require () parenthesis for operations. Infix statements on the other hand require parenthesis to force certain order of evaluation. At this point, it might be interesting to develop an algorithm for evaluating a postfix expression. Let’s start with an empty stack. Whenever one observes a data entry, one uses the push function to push the data entries into the stack. On the other hand whenever an operator is observed, we pop the last two numbers say A and B. Then, we push the value obtained using the formula ‘B operator A’ on to the stack. This process should carry on as long as only one number is left which is popped out of the stack. That would be the result of the post fix operation provided one used a valid postfix expression. Let’s take up an example to understand the given algorithm better. Say, one wants to evaluate the post fix statement ‘1 7 + 2 *’. All one needs to do is to follow the algorithm we developed, i.e., we push the first two data entries ‘1’ and ‘7’ into the stack. Now, once this is done we move to the next point in our postfix expression. This time we observe a ‘+’ operator so we pop the last two data
- 421. 406 entries ‘7’ and ‘1’ where ‘7’ would correspond to A and ‘1’ would correspond to ‘B’ and we apply the formula ‘B operator A’, ‘1+7’ to obtain the result as ‘8’ which being a number is again pushed back into the stack as also shown in the figure 2.29. Figure 2.29: Postfix expression Now we again move to the next point and observe the number ‘2’ in the stack which is again pushed into the stack. Next, we observe the operator ‘*’. So we again pop the last two numbers ‘2’ and ‘8’ and obtain the result ‘16’ using the formula ‘B operator A’. Which is also the result of the postfix expression ‘1 7 + 2 *’ (Figure 2.30). Figure 2.30: Postfix expression
- 422. 407 Part III: Chapter 2 Linked Lists COMMON PROGRAMMING ERRORS Not initializing a constructor to assign NULL values to starting pointers when declaring class stack for dynamic stacks might result in runtime errors Not using the correct order of parenthesis while solving infix to postfix problems will result in an error. 2.3.4 Solved Examples Example 2.3.4.1 Write a function in C++ to perform the Push operation on a dynamically allocated stack containing real numbers. Solution of 2.3.4.1 struct node { float data node *link; } class stack() { node *top; public: stack(); void push(); void pop(); }; void stack::push() { node *temp; temp=new node; cin>>temp→data; temp→link=top; top=temp; }
- 423. 408 Example 2.3.4.2 Convert the expression given in infix form to postfix form: A+B*C^D-(E/F – G) Show the status of the stack after each step. Solution of 2.3.4.2 STEP INPUT STACK STATUS OUTPUT - 1. A EMPTY A 2. + + A 3. B + AB 4. * +* AB 5. C +* ABC 6. ^ +*^ ABC 7. D +*^ ABCD 8. - +* ABCD^ + ABCD^* EMPTY ABCD^*+ - 9. ( -( ABCD^*+ 10. E -( ABCD^*+E 11. / -(/ ABCD^*+E 12. F -(/ ABCD^*+EF 13. - -(- ABCD^*+EF/ 14. G -(- ABCD^*+EF/G 15. ) -(- ABCD^*+EF/G- 16. EMPTY ABCD^*+EF/G- Example 2.3.4.3 Evaluate the following postfix expression using stack and show the contents of the stack after execution of each step. 50, 40, +, 18, 14, -, 4, *,+
- 424. 409 Part III: Chapter 2 Linked Lists Solution of 2.3.4.3 STEP INPUT STACK STATUS 1. 50 50 2. 40 40 3. + 40 is popped 50 is popped Empty 90 4. 18 90 18 5. 14 90 18 14 6. - 14 is popped 18 is popped 90 90 4 7. 4 90 4 4 8. * 90 4 90 90 16 9. + 90 Empty 106 10. End of Input EMPTY Result 106 2.4 Link Lists as Queues A queue is an example of an abstract data type structure similar to a stack. The difference being that the first data element to be inserted is the first to be removed which can be abbreviated as FIFO (first in, first out) or LILO (last in, last out). Thus, items are removed from the queue in the very same order in which they were put into the queue in the first place. A queue is a homogeneous collection of elements in which placing of data elements is called as ‘insertion or enqueue’ which is done at the end of the list called the ‘rear’ and ‘deletion or dequeue’ is done at the other opposite end known as the ‘front’. The figure 2.31 given below shows various features present in a queue.
- 425. 410 Figure 2.31: An empty Queue One of the major applications of queue is to create job queues for ‘job scheduling’ by an equal-priority operating system. A job queue has the following features: Jobs present in a job queue are processed in order they enter the system; To delete later jobs, one has to delete previous jobs first; and As jobs enter and leave the system, the queue gradually shifts to the right. An example of ‘job scheduling’ can be observed when we command computer say, to print 4 articles stored in the computers memory then the computer prints them by first creating a job queue having the jobs assigned to it and then it uses the above defined features of ‘job scheduling’ to complete each job individually. 2.4.1 Static Circular Queues Queues can also be implemented as one dimensional arrays known as ‘Static circular’ Queues. ‘Static’ queues have a fixed amount of data which can be entered into them and generally used when we know how much space the data would occupy. ‘Static’ queues are created by declaring a linear one dimensional array in which insertion or ‘enqueue’ of data is allowed from a single location known as the ‘Rear’ and deletion or ‘dequeue’ of data elements is allowed only from one single location known as the ‘Front’. Let us take an example of a static queue by declaring an array ‘static_queue’ (shown in figure 2.32) which can store 5 integer data values. Figure 2.32: A static queue One of the important things to note from the figure above is that in an empty queue both the ‘Front’ and the ‘Rear’ ends of the queue share the same memory address or point to
- 426. 411 Part III: Chapter 2 Linked Lists the same index of the array. Let us now observe, what happens if we insert a data entry say ‘5’ into the queue? As we have learnt before that in a queue all the data insertions takes place from the ‘Rear’ end so the data entry ‘5’ is assigned to the 1st index of the array ‘static_queue[]’ (see Figure 2.33), which is the index the ‘Rear’ end points to. At this point, one might wonder, why is the data entry added to the 1st index of the array and not to the zeroth index of the array? This question will be answered later in this section. For now just try to understand how stacks work. Figure 2.33: Insertion of a data element in a queue We can further use this operation to enter more data entries like ‘10’ and ‘15’ in the queue. Figure 2.34 shows the effect of this operation on the queue. (a)
- 427. 412 (b) Figure 2.34: Enqueue operation Now that we have a rough idea of about how the ‘enqueue’ operations work, let us now learn how the delete or the ‘dequeue’ operation is performed. For explaining the working of this function, we will take the same queue we created in figure 2.34. If we recall queues are FIFO structures, so when this operation is executed the position ‘Front’ starts pointing to 1st index of the array and the data entry ‘5’ is deleted (see figure 2.35). Similarly, if we apply this operation again then the position ‘Front’ starts directing to the 2nd data entry and then the data entry ‘10’ is dequeued. How these operations exactly work and what role is played by the position variables ‘front’ and ‘rear’ during both operations ‘queue’ and ‘dequeue’ is explained in the program 2.5 given below, which declares and uses a dynamic circular queue with all the above operations applicable on it. Figure 2.35: Dequeue operation
- 428. 413 Part III: Chapter 2 Linked Lists Program 2.5: Static circular queue /*********************** Program 2.5 Static Circular Queue *********************** /* Header Files */ # include <iostream.h> # include <conio.h> const int max=10; /******************* Name: qinsert Return Type: void Purpose: To enter data into the queue **************************************/ void qinsert (int q[], int &r, int f) { if ((r+1)%max != f); { r = (r+1)%max; cout<<"n Data"; cin>>q[r]; cout<<"n Data has been entered"; } } /******************* Name: qinsert Return Type: void Purpose: To delete data from the queue **************************************/ void qdelete( int q[], int r, int &f) { if (r!=f) { f=(f+1)%max; cout<<endl<<q[f]<<"has been Deletedn"; } else {
- 429. 414 cout<<" Queue is empty n"<<endl; } } /******************* Name: qdisplay Return Type: void Purpose: To display the contents of the queue **********************************************/ void qdisplay(int q[], int r, int f) { int cn=f; cout<<"n Displaying the contents of the queue n"; while (cn!=r) { cn = (cn+1)%max; cout<< q[cn]<<endl; } } // Main Function void main() { clrscr(); // Initialization steps int queue[max], rear=0, front=0; int ch; do { cout<<"n === MENU ==="; cout<<"n 1: Insertn"; cout<<" 2: Deleten"; cout<<" 3: Shown"; cout<<" 4: Quitn"; cout<<" !: Enter choice"; cin>>ch; switch(ch) { case 1: qinsert(queue,rear,front); break; case 2: qdelete(queue,rear,front); break;
- 430. 415 Part III: Chapter 2 Linked Lists case 3: qdisplay(queue,rear,front); break; } } while (ch != 4); getch(); } Let us closely observe this program. In this program, we have declared a one dimensional array ‘queue’ which can hold 10 integer data values represented by the constant variable ‘max’ in the program. We have also declared two integer variables ‘rear’ and ‘front’ which are position variables. The variable ‘rear’ is the position variable which directs to the index of the array from where all the data entries are inserted and the other position variable ‘front’ directs to the index of the array from where all the dequeue operation is executed. This program also consists of the following functions to insert, delete and display the contents of the queue: Function ‘qinsert’ void qinsert (int q[], int &r, int f) { if ((r+1)%max != f); { r = (r+1)%max; cout<<"n Data"; cin>>q[r]; cout<<"n Data has been entered"; } } This function is used to insert data entries into the queue. The function receives the linear array ‘queue’, position variable ‘front’ as variable parameters and the position variable ‘rear’ as a reference parameter which are represented by ‘q’, ‘f’ and ‘r’ in the function. The function inserts the data entry at the index ‘r’ of the linear array provided the position variable ‘r’ and ‘f’ do not direct to the same index of the array or in other words it enters data in the queue until the queue is full. Let us take an example, say this function is called for the first time by the program to enter the first data entry then the expression ‘(r+1)%max’ would yield the value ‘1’ which is not equal to ‘f’ as ‘f’ was initialized to ‘0’ and hence the program would enter the ‘if’ loop. In the ‘if’ loop the function would insert the data entered by the user at ‘(r+1)%max’ index of the array, which happens to be the 1st index of the array.
- 431. 416 Function ‘qdelete’ void qdelete( int q[], int r, int &f) { if (r!=f) { f=(f+1)%max; cout<<endl<<q[f]<<"has been Deletedn"; } else { cout<<" Queue is empty n"<<endl; } } This function is used to delete or pop data entries from the static queue. It receives the static array ‘queue’, the position variable ‘rear’ as variable parameters and the other position variable ‘front’ as a reference parameter. The entries are deleted or dequeued until the queue is not empty which is performed by the if condition ‘r!=f’. To delete data entries the position variable ‘front’ represented by the formal parameter ‘f’ in the function is incremented by unity every time the program enters the if loop deleting the entry at this index of the array. Let us analyze figure 2.36. Figure 2.36: static array queue[] Assume this is the state of the static array ‘queue’ when this function is called for the first time by the program. First, before entering the ‘if’ loop it will check for equality between position variables ‘f’ and ‘r’. As ‘f=0’ is not equal to ‘r=8’ it will enter the ‘if’ conditional loop. After it enters the ‘if’ conditional loop, the position variable ‘f’ is assigned the value of ((f+1)%max, (0+1)%10=1) deleting the data entry ‘10’ at this index of the static array. Function ‘qdisplay’ void qdisplay(int q[], int r, int f) { int cn=f;
- 432. 417 Part III: Chapter 2 Linked Lists cout<<"n Displaying the contents of the queue n"; while (cn!=r) { cn = (cn+1)%max; cout<< q[cn]<<endl; } } All this function does is that it prints all the data entries present in the queue on the screen. 2.4.2 Dynamic Queue The concept of dynamic queues is similar to that of dynamic stacks. As queues can be represented as arrays they can also be represented with the use of pointers by creating a link list. Dynamic queues overcome the limitations present with static circular queues as they can grow and shrink during program execution, thus, save memory. Program 2.6 implements a ‘Dynamic Queue’. Program 2.6: Dynamic Queue /************************** Program Dynamic Queue **************************/ /* Header Files */ # include <iostream.h> # include <conio.h> /************************** Self referencial structure Name: node Members: 'data' of type integer : 'next' of type node ********************************/ struct node { int data; node *next; }; /****************************** Name: queue
- 433. 418 Members: Private members: Pointer 'rear' of type queue Pointer 'front' of type queue Public members: Constructor Function queue() Function qinsert() Function qdelete() Function qdisplay() Destructor Function queue **********************************************************/ class queue { node *rear, *front; public: queue() { rear = NULL; front = NULL; } void qinsert(); void qdelete(); void qdisplay(); ~queue(); }; /********************** Name: qinsert Parameters: None Return Type: void Purpose: To enter data entries into the queue **********************************************/ void queue::qinsert() { node *temp; temp = new node; cout<<"n Data:"; cin>>temp->data; cout<<" Data has been entered nn"; temp->next=NULL; if (rear == NULL) { rear = temp; front = temp; } else { rear->next = temp; rear = temp;
- 434. 419 Part III: Chapter 2 Linked Lists } } /********************** Name: qdelete Parameters: None Return Type: void Purpose: To delete data entries from the queue **********************************************/ void queue::qdelete() { if (front !=NULL) { node *temp=front; cout<<endl<<front->data<<" has been Deleted n"; front = front ->next; delete temp; if (front==NULL) { rear = NULL; } } else { cout<<" Queue is Empty n"; } } /********************** Name: qdisplay Parameters: None Return Type: void Purpose: To display the data entries of the queue **************************************************/ void queue::qdisplay() { cout<<"nDisplaying Queue............... n"; node *temp=front; while (temp != NULL) { cout<<temp->data<<endl; temp= temp ->next;
- 435. 420 } } /********************** Destructor Function Name:queue Parameters: None Return Type: void Purpose: To free the memory allocated to nodes of the queue ************************************************************/ queue::~queue() { while (front != NULL) { node *temp = front; front = front ->next; } } /* Main Function */ void main() { clrscr(); queue q1; int choice; do { cout<<"n $$$ MENU $$$n"; cout<<"1. Push n"; cout<<"2. Pop n"; cout<<"3. Display n"; cout<<"4. Quit n"; cout<<"! Enter Choice "; cin>>choice; switch(choice) { case 1: q1.qinsert(); break; case 2: q1.qdelete(); break; case 3: q1.qdisplay();
- 436. 421 Part III: Chapter 2 Linked Lists break; } } while (choice != 4); getch(); } Let us commence from the beginning of this program. At first, we have declared a self referential structure ‘node’ which contains an integer variable ‘data’ and a pointer variable ‘next’ of type ‘node’. Which means that this pointer can hold an address of a structure variable of type ‘node’? It might become more obvious if we glance at figure 2.37. Figure 2.37: Self referential structure After this we have declared a class ‘queue’. Which consists of two pointer variables ‘rear’ and ‘front’ of type ‘node’ as its private members and contains ‘qinsert’, ‘qdelete’ and ‘qdisplay’ as its public member functions. We will take up these functions in the next section. class queue { node *rear, *front; public: queue(){ rear = NULL; front = NULL; } void qinsert(); void qdelete(); void qdisplay(); ~queue(); }; One may note that we have used a constructor to assign the pointer variables ‘rear’ and ‘front’ default values of ‘NULL’ to prevent errors from garbage values and a destructor to free up the memory assigned to the memory blocks created by the queue. The
- 437. 422 important thing to remember here is that whether one represents queues as arrays or queues as pointers, the operation of ‘queues’ remains same. 2.4.2.1 Functions of a dynamic queue The ‘qinsert’ function void queue::qinsert() { node *temp; temp = new node; cout<<"n Data:"; cin>>temp->data; cout<<" Data has been entered nn"; temp->next=NULL; if (rear == NULL) { rear = temp; front = temp; } else { rear->next = temp; rear = temp; } } When this function is called then a pointer variable ‘temp’ is declared pointing to type ‘node’. Figure 2.38: Self referential structure For understanding the concept let us say we enter the data value ‘5’ then how does this data entry get pushed on to the queue?
- 438. 423 Part III: Chapter 2 Linked Lists To understand how insertion of a data entry takes place in a dynamic queue look at the following statements. cout<<"n Data:"; cin>>temp->data; ‘cin>>temp->data’, this statement is used to enter data entries into the ‘data’ part of the new node. We can also understand this by looking at the figure 2.39 assuming that the user enters ‘5’ into the dynamic queue. Figure 2.39: Insertion of data entry into the queue In the subsequent statements, we have assigned the address ‘NULL’ to the pointer part of the new node (see figure 2.40). cout<<" Data has been entered nn"; temp->next=NULL; Figure 2.40: Insertion of data entry in the queue Once this is done, we enter the if loop as ‘rear’ has been initialized to the NULL and in the if loop the pointer variables ‘rear’ and ‘front’ are also assigned the address stored in the pointer variable ‘temp’ as a result of which they too start pointing to the same memory location (see figure 2.41) as pointer variable ‘temp’.
- 439. 424 Figure 2.41: Insertion of data entry in the queue Now what happens if we insert another data entry say ‘10’ into the queue? We will follow the same steps as we followed to enter our first data ‘5’. The pointer ‘temp’ is again initialized pointing to a new memory block of type node in which the new data entry ‘10’ is entered into the data part of this node and the pointer part of the node is assigned a NULL value (see figure 2.42) Figure 2.42: Insertion of data entries in the queue The only difference is that the program would not enter the ‘if’ loop as the initial condition ‘rear==NULL’ will not be satisfied and instead the following statements will be executed resulting in assigning the pointer part of the node to the address of
- 440. 425 Part III: Chapter 2 Linked Lists the new node we just created and then finally assigning the pointer variable ‘rear’ the address stored in the variable ‘temp’ (see figure 2.43). Figure 2.43: Insertion of data entry in the queue Using the same concept function ‘push’ can be used again and again to enter many more data entries. Figure 2.4 shows the state of the queue once we have inserted the data entries ‘15’and ’20’ in the queue. Figure 2.44: Insertion of data entries in the queue
- 441. 426 The ‘qdelete’ function /********************** Name: qdelete Parameters: None Return Type: void Purpose: To delete data entries from the queue **********************************************/ void queue::qdelete() { if (front !=NULL) { node *temp=front; cout<<endl<<front->data<<" has been Deleted n"; front = front ->next; delete temp; if (front==NULL) { rear = NULL; } } else { cout<<" Queue is Empty n"; } } As the name suggests, this function is used to delete data entries from the queue. One can compare this function to the function ‘pop’ we learnt in dynamic stacks. To understand how this function works, we would be referring to the queue shown in figure 2.45. As we already know, queues are FIFO data structures so the data entries will be deleted one by one in the following manner 5, 10, 15, and 20. When this function is called it first checks if the queue has any data entries. If the queue is empty it enters the else structure and displays a suitable message on the screen. On the other hand, if the queue is not empty it enters the ‘if’ control structure. In this control structure we declare a pointer variable ‘temp’ and assign it the address stored in the pointer variable ‘front’.
- 442. 427 Part III: Chapter 2 Linked Lists Figure 2.45: Deleting data entries from a queue Using the next few statements, the function deletes the node to which both the pointer variables ‘front’ and ‘temp’ point to, but before deleting this node from the queue the pointer variable ‘front’ is assigned the value stored in the pointer part of the node to which it was pointing to as a result of which it starts pointing to the next adjacent node. node *temp=front; cout<<endl<<front->data<<" has been Deleted n"; front = front ->next; delete temp; The important thing to remember is that in this entire process, we do not delete the ‘new node’ we just delete the link to this memory location by deleting the pointer variable ‘temp’ which we created in the ‘if’ structure. The outcome of these statements is also shown in the figure 2.46. Figure 2.46: Deletion of data entries from the queue
- 443. 428 The ‘qdisplay’ function void queue::qdisplay() { cout<<"nDisplaying Queue............... n"; node *temp=front; while (temp != NULL) { cout<<temp->data<<endl; temp= temp ->next; } } This function is used to display the data entries present in the queue. It works on the logic that each node in the queue points to the next adjacent node. Using the address of the node to which the pointer variable ‘front’ points to, as the starting point, the while loop is repeated displaying the data entries in each node till the temporary pointer variable ‘temp’ reaches the node having a NULL address stored its pointer part, which happens to be the last node or the last entry in the queue. The exact working of this function though is left to the reader as an exercise. PROGRAMMING TIPS The major difference between stacks and queues is the fact that queues are LILO or FIFO structures and there practical applications lie in creating job queues for job scheduling problems. Insertions and deletions in it are made from two different ends ‘rear’ and ‘front’. Creating dynamic queues are more useful than static queues as they are applications of linked lists which allow two directional traversing and can shrink and grow during program execution. Where as stacks are FILO or LIFO structures, insertions and deletions in a stack can be made only through one specific end ‘top’. As dynamic queues dynamic stacks are also more efficient data structures than static stacks but they allow only one dimensional traversing.
- 444. 429 Part III: Chapter 2 Linked Lists 2.4.2.2 Solved examples Example 2.4.2.2.1 Give the necessary declarations for a linked implementation queue containing float values. Write a function insert() to add an element to this queue? Solution of 2.4.2.2.1 struct node { float data; node *next; }; class queue { node *rear, *front; public: queue() { rear = NULL; front = NULL; } void qinsert(); ~queue(); }; void queue::qinsert() { node *temp=new node; cout<<"Data:"; cin>>temp->data; temp->next = NULL; if (rear == NULL) { rear = temp; front =temp; }
- 445. 430 else { rear->next=temp; rear=temp; } } Example 2.4.2.2.2 Write a function in C++ to perform insert operation on dynamically allocated queue containing names of students. struct node { char name[20]; node *link; }; class queue { node *rear,*front; public: queue(); void insert(); void delete(); }; void queue::insert() { node *temp; temp = new node; gets(temp->name); temp->link=NULL; if (front==NULL) { rear = temp; front =temp; } else { rear->link=temp; rear=temp; } }
- 446. 431 Part III: Chapter 2 Linked Lists 2.5 Solved Examples Example 2.5.1 Evaluate the following postfix notation of expression: True, False, AND, True, True, NOT, OR, AND Solution of 2.5.1 Example 2.5.2 Evaluate the following postfix notation of expression: 20, 30, +, 50, 40, -, * Solution of 2.5.2
- 447. 432 Example 2.5.3 Write a function in C++ to perform push operation on a dynamically allocated stack containing complex numbers? Solution of 2.5.3 struct node { int real; int imaginary; node *link; }; class stack { node *top; public: stack(); void push(); void pop(); }; void stack ::push() { node *temp; temp=new node; cout<<" Enter the real part of the complex number"; cin>>temp->real; cout<<endl<<" Enter the imaginary part of the complex number"; cin>>temp->imaginary; temp->link=top; top=temp; } Example 2.5.4 Write a function in C++ to perform insert operation for an Employee chart on a dynamically allocated queue using the following structure:
- 448. 433 Part III: Chapter 2 Linked Lists struct node { char Emp_name[20]; int Emp_code; int Emp_basic; node *link; }; Solution of 2.5.4 # include<iostream.h> # include<conio.h> # include<stdio.h> struct node { char Emp_name[20]; int Emp_code; int Emp_basic; node *link; }; class queue { node *rear,*front; public: queue(); void insert(); void delete(); }; void queue::insert() { node *temp; temp = new node; cout<<" Enter Employee Name"; gets(temp->Emp_name); cout<<endl<<" Enter Employee Code"; cin>>temp->Emp_code; cout<<endl<<" Enter Basic Salary"; cin>>temp->Emp_basic; temp->link=NULL; if (front==NULL)
- 449. 434 { rear = temp; front =temp; } else { rear->link=temp; rear=temp; } } 2.6 Review Exercise 1/2 Mark Questions 1 Mark/ 2 Mark Q1) A self referential structure a) holds the address of a structure. b) references address of a structure. c) is an example of a pointer to a pointer. d) both a and b are correct. e) none of the above Q2) A link list is a) a collection of self referential structures each representing a node. b) a collection of a list of pointers. c) an example of a primitive data structure. d) both a and c are correct e) none of the above Q3) Stack is an example of a) FILO data structure b) LILO data structure c) LIFO data structure d) Both a and c are correct e) None of the above Q4) The difference between a stack and a queue is a) Stack is a LIFO structure where as queue is an FILO structure b) Stack is a FILO structure where as queue is an LILO data structure c) Stacks are one directional data structures where as queues are multi directional d) Both a, b and c are correct e) None of the above
- 450. 435 Part III: Chapter 2 Linked Lists Q5) A circular link list is an example of a link list a) in which no node contains a NULL value in its pointer part. b) which are always one directional. c) having no head or tail. d) both a and b are correct. e) both c and b are correct. Q6) What are the various advantages of using dynamic data structures over traditional static data structures? Q7) Represent the following infix statements in postfix form: (182+9) / 2 ((10*7) * 2)/4 (3+4)/(3*2) 8 % (5 + 4) Q8) What are the various differences between a linked list and a stack? Q9) Evaluate the following postfix notation of expressions: a) False, True, And, False, False, Not, And, And b) True, False, False, Or, And, True, Not, Or c) 40,50,+, 60,90,-,* Q10) Write a function in C++ to perform the display operation on a dynamically allocated stack containing real numbers. 3 Mark/ 4 Mark Q11) Create a link list which inputs a set of 10 strings and copies them to another link list with all the strings reversed. Q12) Write a function in C++ to perform the ‘push’ and the ‘pop’ function for creating a circular static stack? Q13) Write a program to insert and delete nodes in an already existing link list using the following node. struct node nt data; node &ptr; } ;
- 451. 436 Q14) Create a menu driven program to ‘create’ a link list using the structure given below. The program should also ‘display’ and ‘sort’ the link list in ascending order of the structure member ‘Emp_Basic’. struct node { char Emp_Name[20]; int Emp_No; int Emp_Basic; node *ptr; }; Q15) Create two link lists storing integers say ‘A’ and B’, once this is done write a function merge to which these link lists ‘A’ and ‘B’ are passed to it as parameters. This function should merge these two link lists ‘A’ and ‘B’ and produce a third linked ‘C’ sorted in ascending order of integers. Q16) Write a function in C++ to perform insert and delete operation on a dynamically allocated queue using the following structure. struct node { char Book_Name[20]; char Book_Author[20]; int Book_Price; node Book_ptr; }; Q17) Given the following structure of a node that represents a clinic in a dynamic stack. struct clinic { char bednumber[10]; char patientname[20]; char disease[30]; char doc_attending[20]; int patientno;
- 452. 437 Part III: Chapter 2 Linked Lists clinic *nextclinic; }; Define a class ‘clinicstack’ that shall have functions to push a clinic into a stack of clinics, pop out a clinic from the stack of clinics and display the names of all the patients (‘patientname’) with there patient numbers (‘patientno’) on the screen. 5 Mark Q18) Write a program to create a dynamic circular queue using the following structure. struct node { int data; node *ptr; }; The queue should allow the user to enter data entries and also delete data entries. The program should also include functions to sort the nodes of the queue in ascending and descending order of integers (‘data’) entered into them. Hint: The hint lies in figure 2.47 given below. Figure 2.47: Dynamic circular stack Q17) The following figure 2.48 depicts a link list, fill the missing ‘ ’ memory addresses in each of them?
- 453. 438 Figure 2.48: Link List 2.7 Programming Project ‘Infix’ and ‘Postfix’ expressions are two different but equivalent ways of writing an expression. ‘Infix’ notation is one in which operators are written in between their operands for example A+B. On the other hand postfix notation is one in which operators are written after their operands. For example, A B + is a postfix expression.Our programming project involves a creation of menu driven program using dynamic memory allocation, which prompts the user with the choice to enter a postfix expression or an infix expression. The program should then convert the postfix expression to infix and vice versa. Once the program is ready you can test using the following infix/postfix statements given below. Note: Use Dynamic Stack to complete programming project Infix Postfix A * B + C / D A B * C D / + A * (B + C) / D A B C + * D / A * (B + C / D) A B C D / + *
- 454. 439 Part III: Chapter 2 Linked Lists 2.8 Let us revise! Linked lists are dynamic data structures which can shrink and expand during program execution. Linked lists are a collection of a group of self referential structures grouped together logically. A link list can grow till the point memory is available to it. Linked lists are very similar to arrays. The difference in them though lies in the creation of their structures for storage in memory. Arrays are stored as a single block of memory storing huge amounts of data. Linked lists allocate different blocks of memory to each element called ‘node’ where each node consists of two fields. The ‘data’ field to store all data parameters and the ‘next’ field which could be a pointer to a node. Linked lists are generated by creating a series of self referential structures logically linked together. Linked lists can be traversed using the following algorithm: Algorithm Prerequisite: This algorithm works for a linked list in which each node stores the address of its preceding node. 1. Declare a pointer, pointing to the last node. 2. Assign the pointer variable declared in step 1, the address stored in the pointer part of the self referential structure. 3. Repeat the process described in step 2 until the pointer is assigned a NULL value which marks the end of the link list. Singly linked lists in which traversing was one directional gave birth to ‘multi- lists’. The nodes of such lists have more than one link and hence allow atleast two directional traversing. Stacks are non primitive linear data structures in which insertion and deletion of data takes place from only one end known as the ‘TOP’. They are homogeneous collection of data of one type. Stacks are also an example of a LIFO structure that is ‘Last in First out’ data structure. Stacks can also be implemented as one dimensional arrays known as ‘Static’ stacks. Static stacks have a fixed amount of data that can be fed into them. Stacks are implemented using three basic functions. The function ‘push’ is used to enter data entries in the stack, ‘pop’ is used to delete data entries and function ‘display’ is used to display the contents of the stacks. Static stacks can cause wastage of significant amount of computer’s memory hence using dynamic stacks is a more viable solution.
- 455. 440 Stacks represented as pointers by creating a linked list with the use of which one can create ‘Dynamic stacks’. This can waste a significant amount of the computer’s memory if the loop only recurs a few times during the execution of the program. Stacks are used for converting an infix expression to postfix and finding the value of an postfix expression. Any algebraic statement is termed as an infix expression. For example 4+5. A post fix expression places each operator after the two operands to which it applies as its name might suggest as ‘post’ means after. For example 4 5 + A queue is an example of an abstract data type structure. It is also termed as an FIFO (first in first out) or a LILO (last in last out structure). One of the major applications of queue is to create job queues for ‘job scheduling’ Queues can also be implemented as one dimensional arrays known as ‘Static circular’ Queues. Queues comprise of two basic operations: Enqueue: It is the process by which data entries are added in the queue. Dequeue: It is a process by which data entries are deleted from a queue. Link Lists are very efficient method of data storage and are far more advantageous over traditional static arrays and should be used where ever possible to improve program efficiency.
- 456. Part III CHAPTER 3 STREAMS & FILES AIM OUTLINE • Introduce the concept of ‘Streams and Files’. • Understand how streams provide input/output capabilities. • Able to create and use ‘text’ files for purposes of data storage. • Create binary files and be able to read and write data on and from them 3.1 Introduction 3.2 Streams 3.3 Text files 3.4 Binary files 3.5 Solved examples 3.6 Review exercise 3.7 Programming project 3.8 Let us revise!
- 457. 442 3.1 Introduction The chapter focuses on the streams and files present in C++. Streams provide an extensive set of input/output capabilities. They control the flow of data informs of sequence of bytes some of which, that we have already been using in our programs. Files on the other hand are collection of related data or program records stored as a unit with a single name. In simple terms, files are used for data storage. In computer terminology we divide files into two categories ‘text’ and ‘binary’ which will be dealt with later in this chapter. The chapter also explains us how to create our own files, read and write data on and from them. 3.2 Streams In C++, input and output (I/O) occurs with the usage of streams. Stream is a general name given to any kind of flow of data say, in form of a sequence of bytes. Streams are represented by an object of a particular class. All of us have already used streams in programs that we have written through use of ‘cin’ and the ‘cout’ stream objects - to enter and display data using standard input and output devices. There exist many streams in C++ which determine the various kinds of data flow. For example, the ‘ofstream’ classes are used to write operations on a file. Figure 3.1 shows all the basic stream classes available in C++. Figure 3.1: Stream classes in C++ Let us begin with the ‘iostream’ class. We have already used this class while declaring the ‘iostream.h’ header file in our previous programs. The ‘iostream’ class - also known as ‘I/O
- 458. 443Part III: Chapter 3 Streams & Files stream’ - is used to manage most of the input and the output operations through C++ statements, such as, ‘cin’ and ‘cout’. The ‘iostream’ class is actually a derived class of the ‘istream’ and the ‘ostream’ classes. All the input or extraction operations, e.g., get(), getline() and read() are accessed through ‘istream’. The ‘ostream’ class manages all the output or insertion operations for example, the functions put() and write(). These classes in turn are also derived from the base class ‘ios’ which also serves as a base class for many other streams. Figure 3.2 shows the complex hierarchy of these classes. Figure 3.2: Stream class hierarchy Let us now look at the other set of classes shown in figure 3.1. File processing in C++, is carried out using the ‘fstream’ class. The fstream class provides two distinct classes for file processing. The ‘ofstream’ class derived from the ‘ostream’ class and the ‘ifstream’ class derived from the ‘istream’ class. The ‘ofstream’ class is used to write on a file by using constructors available in the class where as the ‘ifstream’ class, is used to read data from a file. The ‘fstream’ class is derived from the ‘iostream’ class and it coalesces the functions defined in both the ‘ifstream’ and the ‘ofstream’ classes. 3.3 Text Files Text files are files where bytes or a sequence of bytes represent ordinary textual characters such as digits and numbers which are represented by their ASCII codes. Text files are mainly used by computer programs for data storage. The Windows system regards a file to be a text file if the suffix of the name of the file is "txt". However, source code for computer programs are also text, but usually have file name suffixes indicating which programming language the source is written in. 3.3.1 Creating/Opening Text Files in OUTPUT mode In this section we will learn how to create and open already existing text files. To create or open a text file we have to take help of the ‘ofstream’ class. For achieving this purpose we can either include the ‘fstream.h’ header file or the ‘ostream.h’ header file as both these header files include the definition of the ‘ofstream’ class. A text file can be
- 459. 444 created/opened using the following statement. What these statements do is create objects of the ‘ostream’ or the ‘fstream’ class depending upon which set of statements are used and then the object is used to create or open an existing text file. For example, ofstream fil(“ABC.txt”); ofstream filout (“Notes.txt”); These statements create two objects of class ofstream ‘fil’ and ‘filout’. These objects are then used to create the ‘ABC.txt’ and ‘Notes.txt’ text files. We can reach the same outcome using the ‘fstream’ class as shown below. Where the statement ‘fil.open(“ABC.txt”,ios::out);’ signifies that we are creating/opening a text file ‘ABC.txt’ in the ‘out’ mode which allows us to write on the file. fstream fil; fil.open(“ABC.txt”,ios::out); Now that we have learnt how to create and open existing text files let us now learn how to write on a file. Program 3.1 given below creates a text file “ABC.txt” and allows the user to write or insert data into this file. Program 3.1: Writing on a text file /* Header Files */ # include <fstream.h> # include <conio.h> # include <stdio.h> void main() { clrscr(); ofstream fil; fil.open("ABC.txt"); char lin[80]; char ch; do { cout<<"n Enter lines of text in the text file"; gets(lin); fil<<lin<<endl; cout<<" Do you want to enter more lines of text (y/n)";
- 460. 445Part III: Chapter 3 Streams & Files cin>>ch; } while (ch!='n'); fil.close(); getch(); } As we already know that this program opens/creates a text file “ABC.txt” and allows the user to write on the text file, but how does this process take place? What we have done in this program is that we have initialized a buffer string ‘lin’ which is used to input data from the user. This data is then fed into the file using the following statement. ‘fil<<lin<<endl’. This statement might look familiar to the statement ‘cout’ with (‘<<’) predefined operators, but the only thing we have to know is that all this statement does is insert the data stored in the variable ‘lin’ into the file. This process is repeated until the user wants, inserting data into the text file. The program ends with the statement ‘fil.close()’ which calls the member function ‘close()’ which closes the text file we opened/created. Once this function is called the stream object created becomes free and can be used to open another file. 3.3.2 Opening a text file in INPUT mode In the previous section we learnt about how to create/open text files and then write on these files. In this section we will learn about how to read existing files and display their contents on the screen. A text file can be opened for reading using the ‘ifstream’ or the ‘fstream’ class. The process of opening a file for reading purposes is very similar to opening the file for writing purposes involving only minor changes. For opening a text file for reading purposes we use the following syntax. Syntax: ifstream <FileObject> (<FileName>); or fstream <FileObject>; <FileObject>.open (<Filename>,ios::in); What these statements do is create objects of the ‘ifstream’ or the ‘fstream’ class depending upon which set of statements are used and then the object is used to create or open an existing text file for reading purposes. For example, ifstream fil(“ABC.txt”); fstream fil; fil.open(“ABC.txt”,ios::in);
- 461. 446 When these statements are executed the file ‘ABC.txt’ is opened for reading purposes. Now that we know how to open text files for reading purposes let us learn how to display the contents stored in a text file. Program 3.2 given below opens the text file ‘ABC.txt’ for reading purposes and displays its contents on the screen. Program 3.2: Reading a text file /* Header Files */ # include <fstream.h> # include <conio.h> # include <stdio.h> // Main Function void main() { clrscr(); fstream fil; // Creates an object of class fstream fil.open("ABC.txt",ios::in); // Opens the file "ABC.txt" in Read Mode char lin[80]; // String buffer while (fil.getline(lin,80)) { puts(lin); } getch(); } In this program we have used the ‘fstream’ class and read the file from it one line at a time using the getline() function which is a member function of ‘fstream’ or the ‘ifstream’ class. This function reads characters, including blank spaces until it encounters the ‘n’ operator and places the resulting string in the buffer supplied to it as an argument. The maximum size of the buffer is passed as the second argument. This process is repeated until we reach the end of the file displaying the contents stored in the buffer each time the condition is checked for. In our program this process is carried out by the following statements. while (fil.getline(lin,80)) { puts(lin); }
- 462. 447Part III: Chapter 3 Streams & Files Where the character array ‘lin’ acts as the buffer string having a maximum size of 80 characters. This is only one of the ways in which the contents of a file can be read and displayed program 3.4 illustrates another method of performing this function. Program 3.3: The EOF function /* Header Files */ # include <fstream.h> # include <conio.h> # include <stdio.h> // Main Function void main() { clrscr(); fstream fil; // Creates an object of class fstream fil.open("ABC.txt",ios::in); // Opens the file "ABC.txt" in Read Mode char lin[80]; // String buffer while (!fil.eof()) { fil.getline(lin,80); puts(lin); } getch(); } Always try to use the ‘fstream’ class to declare stream objects as one can then use the same class for opening the file in input/output modes and the syntax happens to be very similar to what we will use for binary files as well Whenever a file is opened for writing command using the ‘ostream’ class or the ‘fstream’ class using the ‘out’ specifier all the previous content of the file gets deleted. In order to keep our previous data intact we should use the ‘app’ specifier. Programming Tips
- 463. 448 If we notice carefully, this program is very similar to program 3.3. The only change in this program is that we have used the ‘eof()’ function as the condition specifier in the while loop. The ‘eof()’ function is a member function defined in the ‘fstream’ class which checks for the end of file. Hence when this function is used as a condition specifier in the while loop the loop continues until the end of file is reached allowing us to read through the entire file. The rest of the statements have already been dealt with in the previous programs. 3.3.3 Opening a file in append mode When we are opening the file in output mode what was happening was that all the previous data stored in the file was being deleted. In simple terms we were left with a black file. The advantage of opening a file in append mode is that the earlier content if existing, is not over writer and the file write pointer moves to the end of the file for adding new lines at the bottom of the file. This can be done using the following syntax. Syntax: fstream <Fileobject>; <Fileobject>.open(<Filename>,ios::app); For example for opening the text file ‘ABC.txt’ in append mode we would use the following statements. Example: fstream fil; fil.open(“ABC.txt”,ios::app); When creating text files always remember to use the ‘.txt’ extension. Remember to always close created file objects as this closes the input/output stream and the stream objects can be used to open other streams. Always try not to open a file for reading and writing at the same time using different stream objects as this might lead to unpredictable outcomes. Common Programming Errors
- 464. 449Part III: Chapter 3 Streams & Files 3.3.4 Solved Examples Example 3.3.4.1 Write a function to transfer the content of a text file to another text file Solution of 3.3.4.1 void transfer() { char fn1[20],fn2[20]; fstream fil1,fil2; fil1.open(fn1,ios::in); fil2.open(fn2,ios::out); char lin[80]; while(fil1.getline(lin,80)) { fil2<<lin<<endl; } fil1.close(); fil2.close(); } Example 3.3.4.2 Write a function to transfer the content of a text file to another text file after changing the cases to uppercase with each line reversed. Solution of 3.3.4.2 void newchange() { char fn1[20],fn2[20]; fstream fil1,fil2; fil1.open(fn1,ios::in); fil2.open(fn2,ios::out); char lin[80]; while (fil1.getline(lin,80)) { for (int i=strlen(lin)-1;i>=0;i--) { lin[i]=toupper(lin[i]) } fil2<<lin<<endl; }
- 465. 450 fil1.close(); fil2.close(); } Example 3.3.4.3 Write a menu driven program to create a text file ‘Experiment.txt’. The program should also have an option to display the contents of the text file and also count the number of words starting from the letter‘s’. Solution of 3.3.4.3 /* Header Files */ # include<fstream.h> # include<conio.h> # include<stdio.h> # include<string.h> int counter=0; /************* Function Name: create Purpose: creates a text file 'Experiment.txt' **********************************************/ void create() { fstream fil; fil.open("Experiment.txt",ios::out); char choice; char lin[80]; do { cout<<"nn Enter the lines of text"; gets(lin); fil<<lin<<endl; cout<<"n Do you want to enter more lines of text (y/n)"; cin>>choice; } while (choice!='n'); fil.close(); } /************* Function Name: display Purpose: displays the text file 'Experiment.txt'
- 466. 451Part III: Chapter 3 Streams & Files **********************************************/ void display() { fstream fil; fil.open("Experiment.txt",ios::in); char lin[80]; while (fil.getline(lin,80)) { puts(lin); } fil.close(); } /************* Function Name: count Purpose: counts the number of words beginning with the letters 's' **************************************************************** ***/ void count() { counter=0; fstream fil; fil.open("Experiment.txt",ios::in); char lin[80]; int i=0; while (fil.getline(lin,80)) { if (lin[0]=='s') { counter++; } for (i=0;i<strlen(lin);i++) { if (lin[i]==' ' & lin [i+1] =='s') { counter++; } } } cout<<"nn The number of words starting from letter 's' are "<<counter; fil.close(); } // Main function
- 467. 452 void main() { clrscr(); int choice; do { cout<<"nn------------- MENU -----------n"; cout<<"1. Createn"; cout<<"2. Displayn"; cout<<"3. Countn"; cout<<"4. Quitn"; cout<<"! Enter Choicen"; cin>>choice; switch(choice) { case 1: create(); break; case 2: display(); break; case 3: count(); break; } } while (choice !=4); getch(); } Example 3.3.4.4 Write a menu driven program to perform the following operations for each of the following Create a text file with a user defined name. Count the number of vowels in the text file created Display the text file created Solution of 3.3.4.4 /* Header Files */ # include<fstream.h> # include<conio.h> # include<stdio.h> # include<string.h>
- 468. 453Part III: Chapter 3 Streams & Files int counter=0; /************* Function Name: create Purpose: creates a text file with a user defined name ********************************************************/ void create(char filename[]) { fstream fil; fil.open(filename,ios::out); char choice; char lin[80]; do { cout<<"nn Enter the lines of text"; gets(lin); fil<<lin<<endl; cout<<"n Do you want to enter more lines of text (y/n)"; cin>>choice; } while (choice!='n'); fil.close(); } /************* Function Name: display Purpose: displays the text file created **********************************************/ void display(char filename[]) { fstream fil; fil.open(filename,ios::in); char lin[80]; while (fil.getline(lin,80)) { puts(lin); } fil.close(); } /************* Function Name: count_vowels Purpose: counts the number of vowels in the text file ********************************************************/
- 469. 454 void count_vowels(char filename[]) { int vowels=0; fstream fil; fil.open(filename,ios::in); char lin[80]; int i=0; while (fil.getline(lin,80)) { for (i=0;i<strlen(lin);i++) { if (lin[i]=='a' || lin [i] =='e'|| lin [i] =='i'|| lin [i] =='o'|| lin [i] =='u') { vowels++; } } } cout<<"nn The number of vowels are "<<vowels; fil.close(); } // Main function void main() { clrscr(); int choice; char filename[15]; cout<<"Enter the file name"; gets(filename); do { clrscr(); cout<<"nn------------- MENU -----------n"; cout<<"1. Create the text filen"; cout<<"2. Count the number of vowels n"; cout<<"3. Display the text filen"; cout<<"4. Quitn"; cout<<"! Enter Choicen"; cin>>choice; switch(choice) { case 1: create(filename); break;
- 470. 455Part III: Chapter 3 Streams & Files case 2: count_vowels(filename); getch(); break; case 3: display(filename); getch(); break; } } while (choice !=4); getch(); } Example 3.3.4.5 Write a menu driven program using separate function to perform the following operations: Create a text file with a user defined name. Copy the contents of this text file into another text file ‘Copy.txt’ with all the blank spaces in the text file converted to ‘$’. Display both the text files. Solution of 3.3.4.5 /*************** Header Files ***************/ # include <fstream.h> # include <conio.h> # include <stdio.h> # include <string.h> /***************** Function Definitions ********************/ /***************** Function: create Parameters: filename[] Return Type: void Purpose: Creates a text file with a user defined name *******************************************************/ void create(char filename[]) { fstream fil; fil.open(filename,ios::out);
- 471. 456 char lin[80]; char ch; do { cout<<"nEnter lines of text"; gets(lin); fil<<lin<<endl; cout<<"nDo you want to enter more lines of text (y/n)n"; cin>>ch; } while (ch!='n'); fil.close(); getch(); } /***************** Function: display Parameters: filename[] Return Type: void Purpose: Displays the text file entered by the user *******************************************************/ void display(char filename[]) { fstream fil; fil.open(filename, ios::in); char lin[80]; while (fil.getline(lin,80)) { puts(lin); } fil.close(); getch(); } /***************** Function: copy Parameters: filename[] Return Type: void Purpose: Creates the text file 'Copy.txt' with all the blank spaces converted to '$' sign. ******************************************/ void copy(char filename[]) { fstream fil1,fil2; fil1.open(filename,ios::in); fil2.open("Copy.txt",ios::out);
- 472. 457Part III: Chapter 3 Streams & Files char lin[80]; while(fil1.getline(lin,80)) { for(int i=0;i<strlen(lin);i++) { if (lin[i]==' ') { lin[i]='$'; } } fil2<<lin<<endl; } cout<<"n The text file 'Copy.txt' has been created"; fil1.close(); fil2.close(); } // Main Function void main() { clrscr(); char filename[15]; char file[15]; int choice; do { cout<<"nn ----------- MENU ----------- n"; cout<<" 1. Creating the text file n"; cout<<" 2. Displaying the text filen"; cout<<" 3. Create the text file 'Copy.txt' n"; cout<<" 4. Exitn"; cin>>choice; switch(choice) { case 1: cout<<"nn Enter Filename:"; gets(filename); create(filename); break; case 2: cout<<"Enter the name of the file to be displayednn"; gets(file); display(file); break;
- 473. 458 case 3: copy(filename); break; } } while (choice!=4); getch(); } 3.4 Binary Files A file stored in binary format is known as a binary file. A binary file is only computer readable and cannot be read by humans. Operating systems store the files in binary format as computers can only deal with binary numbers. Binary files in contrast to the text files can include any kind of data. The sound and image files we encounter in our daily use when using computers are also examples of binary files 3.4.1 Opening binary files in output mode (For writing) In this section we will learn how to open already existing binary files. For opening a binary file one has to use the ‘fstream’ class. The syntax for opening an existing an already existing binary file is given below. Syntax: fstream<FileObject>; <FileObject>.open (<Filename>,ios::binary |ios::out); | Be careful on the usage of brackets when using the following statement: while (fil1.getline(lin,80)) Programming Tips
- 474. 459Part III: Chapter 3 Streams & Files For example, Example: fstream fil; fil.open(“Binary_File.DAT”,ios::binary|ios::out); OR char fname[20]; cout<<” Enter Filename:”; gets(fname); fstream fil; fil.open(fname,ios::binary|ios::out); The set of statements given above would create a binary file ‘Binary_File’. The ‘ios::binary’ statement in the piece of code given above signifies that we are opening a binary file where as the proclamation ‘ios::out’ tells the compiler that we are opening the binary file for writing purposes. This syntax should seem similar to the one we learnt for opening text files as binary files are nothing but a more general form of text files. 3.4.2 Opening binary files in input mode (For reading from a file) In the previous section we learnt about how to open binary files for writing purposes. Let us now learn how to open binary files for reading purposes. For opening binary files in input mode we again use the ‘fstream’class using the following syntax. Syntax: fstream<FileObject>; <FileObject>.open (<Filename>,ios::binary |ios::in); The example given below opens the binary file ‘Binary_File.DAT’ in input mode. Example: fstream fil; fil.open(“Binary_File.DAT”,ios::binary|ios::in); OR char fname[20]; cout<<” Enter Filename:”; gets(fname); fstream fil; fil.open(fname,ios::binary|ios::in);
- 475. 460 The only change we observe from the previous section is in the statement ‘fil.open(“Binary_File.DAT”,ios::binary|ios::in);’. Where we have changed the proclamation from ‘ios::out’ to ‘ios::in’ which signifies we are opening the binary file for reading purposes. 3.4.3 Opening binary files in APPEND mode (For writing on a file) When we are opening the binary file in output mode what was happening was that all the previous data stored in the file was being deleted. In simple terms we were left with a black file. The advantage of opening a file in append mode is that the earlier content if existing, is not over writen and the file write pointer moves to the end of the file for adding new lines at the bottom of the file. This can be done using the following syntax. Syntax: fstream<Fileobject>; <Fileobject>.open(<Filename>,ios::binary|ios::app); For example for opening the text file ‘ABC.txt’ in append mode we would use the following statements. Example: fstream fil; fil.open(“Binary_File.DAT”,ios::binary|ios::app); OR char fname[20]; cout<<” Enter Filename:”; gets(fname); fstream fil; fil.open(fname,ios::binary|ios::in); 3.4.4 Creation of binary files When creating binary files inputting and outputting data using operators like << and >> and functions like ‘getline’ does not make too much sense although it is possible to use them. The file streams include two member functions which are specially designed for input and output of data sequentially. They are the ‘write’ and the ‘read’ functions. The ‘write()’ function is a member function of the ‘ostream’ class and is inherited by ‘ofstream’ where as the function ‘read()’ is a member function of the ‘istream’ class and is inherited by ifstream. Both these functions can be accessed by creating objects of ‘fstream’ class. These functions have the following prototypes.
- 476. 461Part III: Chapter 3 Streams & Files write ( char * buffer, streamsize size ); read ( char * buffer, streamsize size ); The function prototype takes two arguments. The first one being ‘buffer’ which is the address of the memory block where the data is inserted by the user or the data to be written on the binary file is stored temporarily where as the second argument defines the maximum allowed size that can be fed into the buffer. Binary files can be created in C++ with the help of structure variables or objects of a class. Program 3.4 given below declares a function which creates a user defined binary file with a single record. Program 3.4: Insertion of a record in a binary file using structures and classes Using Structures Using Classes struct stock { int Ino; char Item[20]; float qty; }; void create(char fname[]) { fstream fil; stock s; fil.open(fname,ios::binary|ios::out); cout<<”Item No. ”; cin>>s.Ino; cout<<”Item Name”; gets(s.Item); cout<<”Quantity”; cin>>s.qty; fil.write((char *)&s,sizeof(s)); fil.close(); } class stock { int Ino; char Item[20]; float qty; public: void input() { cout<<”Item No.”; cin>>Ino; cout<<”Item Name”; gets(Item); cout<<”Quantity”; cin>>qty; } void output() { cout<<Ino<<”:”<<Item<<”:”<<qty; } int Rino() { return Ino; } }; /*********** Function: create Parameters: fname
- 477. 462 ****************/ void create( char fname[]) { fstream fil; stock s; fil.open(fname,ios::binary|ios::out); s.input(); fil.write((char*)&s,sizeof(s)); fil.close(); } The entire structure of this program might seem very seimilar to the one we have already dealt in text files. The only difference one will observe is in the writing and the reading of binary files in this program we have used the following piece of code to write on the binary file. Let us see how this statement works. For this purpose let us look at it its general form. stream_object.write((char*)&(structure_variable/class_object),sizeof(structure _variable/class_object)); The first part of the statement ‘(char*)&(structure_variable/class_object)’ typcasts the structure or the class to character type. Where as the second part of the statement ‘sizeof(structure_variable/class_object)’ allocates the number of bytes to be fed into the binary file. For example, in our statement given above. The first part of the statement ‘(char*)&s’ typcasts the class ‘s’ so that data from it can be inserted into the binary file and the second part of the statement ‘sizeof(s)’ allocates the amount of data interms of bytes to be fed into the file at one time. This is the procedure one has to follow in order to insert a single record into a binary file but what happens if we want to insert multiple records and not just a single record ? Program 3.5 declares a function ‘create’ which proposes a solution to the given problem. Program 3.5: Creation of a binary file with multiple records struct stock { int Ino; char Item[20]; float qty; }; class stock { int Ino; char Item[20]; float qty; public: fil.write((char*)&sizeof(s)); STREAMS&FILES
- 478. 463Part III: Chapter 3 Streams & Files 3.4.5 Reading a binary file In this section of the chapter we will learn how to dislay already created binary files. We have already learnt in the previous section that binary files can be displayed using the function ‘read()’ which is a member function of the ‘istream’ class’ inherited by ifstream and the fstream classes. The process of reading data from binary files is very similar to the process of writing on them. Program 3.6 given below displays the piece of code required to display a single record. void create (char Fname[]) { fstream fil; stock S; fil.open (Fname,ios::binary|ios::out); char choice; do { cout<<” Item No?”; cin>>S.Ino; cout<<”Item Name”; gets(S.Item); cout<<”Quantity”; cin>>S.qty; fil.write((char *)&S,sizeof(S)); cout<<”More(Y/N) ?”; cin>>choice; } while (choice==’y’); fil.close(); } void input() { cout<<”Item No.”; cin>>Ino; cout<<”Item Name”; gets(Item); cout<<”Quantity”; cin>>qty; } void output() { cout<<Ino<<”:”<<Item<<”:”<<qty; } int Rino() { return Ino; } }; void create(char Fname[]) { fstream fil; stock s; fil.open(Fname,ios::binary|ios::out); char choice; do { s.input(); fil.write((char*)&s,sizeof(s)); cout<<” More(Y/N) ?”; cin>>choice; } while (choice == ‘y’); fil.close(); }
- 479. 464 Program 3.6: Using Structures Using Classes struct stock { int Ino; char Item[20]; float qty; }; void display (char Fname[]) { fstream fil; stock S; fil.open(Fname,ios::binary|ios::in); fil.read((char*)&S,sizeof(S)); cout<<”ItemNo:”<<S.Ino; cout<<”Item_Name:”; gets(S.Item); cout<<”Item quantity”; cin>>S.qty; fil.close(); } class stock { int Ino; char Item[20]; float qty; public: void input() { cout<<”Item No.”; cin>>Ino; cout<<”Item Name”; gets(Item); cout<<”Quantity”; cin>>qty; } void output() { cout<<Ino<<”:”<<Item<<”:”<<qty; } int Rino() { return Ino; } }; void display(char Fname[]) { fstream fil; stock S; fil.open(Fname,ios::binary|ios::in); fil.read((char*)&S,sizeof(S)); S.output; fil.close(); }
- 480. 465Part III: Chapter 3 Streams & Files As it is clear we have used the 'read’ function to display records from the file. The function ‘read’ works in the same way as the function ‘write()’ we learnt in the last section the only difference being that this time data is being fed out of the file. Rest of the program should be easily understood as it follows the same logic as present in program 3.5. It is also possible to read various records from a binary file. Program 3.8 given below shows how this is done. Program 3.7: Reading Records from a binary File struct stock { int Ino; char Item[20]; float qty; }; void display(char Fname[]) { fstream fil; stock S; fil.open(Fname,ios::binary|ios::in); while(fil.read((char*)&S,sizeof(S))) { cout<<”Item No:”<<S.Ino<<endl; cout<<”ItemName:”<<S.Item<<”Qty:”<<S.qty<<en dl; } fil.close(); } class stock { int Ino; char Item[20]; float qty; public: void input() { cout<<”Item No.”; cin>>Ino; cout<<”Item Name”; gets(Item); cout<<”Quantity”; cin>>qty; } void output() { cout<<Ino<<”:”<<Item<<”:”<<qty; } int Rino() { return Ino; } }; void display(char Fname[])
- 481. 466 3.4.6 Solved Examples Example 3.4.6.1 Using the following class write functions to do the following: (i) Store the objects of clinic in a binary file (ii) Read all the records from the binary file and display them on the screen { fstream fil; stock S; fil.open(Fname,ios::binary|ios::in); while(fil.read((char*)&S,sizeof(S))) { S.output(); } fil.close(); } When declaring binary files make sure that one is using the ‘.dat’ extension. Remember to use the ‘ios::binary’ specifier with the fstream class when opening/creating binary files for reading or writing purposes. The major difference between text and binary files is that text files are human readable where bytes or a sequence of bytes represent ordinary textual characters such as digits and numbers which are represented by their ASCII codes where as binary files are not human readable Binary files in contrast to the text files can include any kind of data not necessarily just numbers digits and characters Programming Tips
- 482. 467Part III: Chapter 3 Streams & Files (iii) Accept a disease name from the user and display the details of all such patients who are suffering of that particular disease // class definition of class 'clinic' class clinic { char bednumber[10]; char patientname[20]; char disease[30]; char doc_attending[20]; public: void getdetails() { gets(bednumber); gets(patientname); gets(disease); gets(doc_attending); } void showdetails() { cout<<"Bed Number"<<bednumber; cout<<"Patient Name"<<patientname; cout<<"Disease"<<disease; cout<<"Doctor Attending"<<doc_attending; } char*retdisease() { return disease; } }; Solution of 3.4.6.1 (i) void store() { fstream fout; fout.open("clinic.dat",ios::out|ios::binary); clinic ob;
- 483. 468 char ch; do { ob.getdetails(); fout.write((char*)&ob,sizeof(ob)); cout<<"More Records...(y/n)"; cin>>ch; } while (toupper(ch)=='y'); fout.close(); } (ii) void readrecords() { fstream fin; fin.open("clinic.dat",ios::in|ios::binary); clinic ob; while(fin.read((char*)&ob,sizeof(ob))) { ob.showdetails(); } fin.close(); } (iii) void diseasedetails() { char dis[30]; cout<<"Enter Disease name"; gets(dis); fstream fin; fin.open("clinic.dat",ios::in|ios::binary); clinic ob; while (fin.read((char*)&ob,sizeof(ob))) { if (!strcmp(ob.retdisease(),dis)) { ob.showdetails(); } }
- 484. 469Part III: Chapter 3 Streams & Files fin.close(); } Example 3.4.6.2 Write a function in C++ to add new objects at the bottom of a binary file "student.dat" assuming the binary file is containing the objects of the following class: class stud { int rno; char name[20]; public: void enter() { cin>>rno; gets(name); } void display() { cout<<rno<<name<<endl; } }; Solution of 3.4.6.2 void addnew() { fstream fil; fil.open("student.dat",ios::binary|ios: :app); stud s; char ch; do
- 485. 470 { s.enter(); fil.write((char*)&s,sizeof(s)); cout<<"More(Y/N)?"; cin>>ch; } while(ch!='y'); fil.close(); } 3.5 Solved Examples Example 3.5.1 Write a program in C++ to create a text file with a user defined name. The program should also display the created text file and count the number of times the letters ‘‘g’‘h’‘o’‘s’‘t’’ appear together ? Solution of 3.5.1 /********** Header Files **********/ # include <fstream.h> # include <conio.h> # include <stdio.h> # include <string.h> /******** Function Definitions ***********/ /***************** Function: create Parameters: filename[] Return Type: void fstream fil; fil.open(filename,ios::out);
- 486. 471Part III: Chapter 3 Streams & Files char lin[80]; char ch; do { cout<<"nEnter lines of text"; gets(lin); fil<<lin<<endl; cout<<"nDo you want to enter more lines of text (y/n)n"; cin>>ch; } while (ch!='n'); fil.close(); getch(); } /***************** Function: display Parameters: filename[] Return Type: void Purpose: Displays the text file ******************************/ void display(char filename[]) { fstream fil; fil.open(filename, ios::in); char lin[80]; while (fil.getline(lin,80)) { puts(lin); } fil.close(); getch(); } /***************** Function: count Parameters: filename[] Return Type: void Purpose: counts the number of times the letters 'g''h''o''s''t' encountered
- 487. 472 void count(char filename[]) { fstream fil; fil.open(filename, ios::in); char lin[80]; int count=0; while (fil.getline(lin,80)) { for (int i=0;i<strlen(lin);i++) { if (lin[i]=='g'&lin[i+1]=='h'&lin[i+2]=='o'&lin[i+3]=='s'&lin[i+4]=='t') count++; } } cout<<" No. of times "<<count; fil.close(); getch(); } // Main Function void main() { clrscr(); char filename[15]; int choice; do { cout<<"nn ---------------------- MENU ----------------------- n"; cout<<" 1. Creating the text file n"; cout<<" 2. Displaying the text filen"; cout<<" 3. Number of times the letters 'g''h''o''s''t' are encountered together"; cout<<"n 4. Exitn"; cin>>choice; switch(choice) { case 1: cout<<"nn Enter Filename:"; gets(filename); create(filename); break;
- 488. 473Part III: Chapter 3 Streams & Files Example 3.5.2 Write a program to create a user defined text file and display alternate lines of the file? Solution of 3.5.2 case 2: display(filename); break; case 3: count(filename); break; } } while (choice!=4); getch(); } /********** Header Files **********/ # include <fstream.h> # include <conio.h> # include <stdio.h> # include <string.h> /******** Function Definitions ***********/ /***************** Function: create Parameters: filename[] Return Type: void Purpose: Creates a text file with a user defined name *******************************************************/ void create(char filename[]) {
- 489. 474 fstream fil; fil.open(filename,ios::out); char lin[80]; char ch; do { cout<<"nEnter lines of text"; gets(lin); fil<<lin<<endl; cout<<"nDo you want to enter more lines of text (y/n)n"; cin>>ch; } while (ch!='n'); fil.close(); getch(); } /***************** Function: display Parameters: filename[] Return Type: void Purpose: Displays alternate lines of the text file entered by the user **********************************************************************/ void display(char filename[]) { fstream fil; fil.open(filename, ios::in); char lin[80]; int even=0; while (fil.getline(lin,80)) { if (even%2==0) { puts(lin); } even++; } fil.close(); getch(); } // Main Function void main()
- 502. 487 PPAARRTT IIVV DDAATTAA BBAASSEESS && CC++++
- 504. Part IV Chapter 1 DATABASE CONCEPTS AIM OUTLINE To introduce database and its usage To learn about database Models and Relational Algebra To know what is normalization To understand use of databases through examples 1.1 Introduction 1.2 Databases 1.3 Database Management System 1.4 Purpose of Databases 1.5 DBMS Models 1.6 Relational Database Terminology 1.7 Relational Algebra 1.8 Normalization 1.9 Review Examples 1.10 Let us revise
- 505. 490 1.1 Introduction Computers are used in business and industry for diverse purposes. For example, an industry might be using computers for storing sales data, inventory control, and production scheduling and similar activities. An educational institute might be using computers for storing the information about the performance of students, school fees, scholarships granted or other forms of financial assistance. Software for such a system uses customized data files and reports specifically designed for the application. If these applications have been designed for the same organization then the same data may be used in each of the application for a different purpose. For example, the name, address and roll no of a student may be stored in both the result file as well as the payment file (which stores information about the payments made by a student as school fees). This leads to some duplication and redundancy of data. It would have been better had all the relevant information about an organization been centralized and accessible without the need for creating separate files for different purposes. The concept of database attempts to achieve this objective. This part of the appendix briefly discusses the concept of databases and its applications. 1.2 Databases A database is a collection of related files. Database is a collection of interrelated files containing collection of records, where each record of the file is a collection of logically related data items. It is required for keeping the information of different applications of an organization to enable manipulation of the data and generate reports for its operation as well as the management. Database should be made in the following way that • It can accept the required changes and grow in the future. • It can be used by more than one independent application. Organizations seeking computer assistance prefer to store their data in the form of an integrated database. A person of the organization is assigned the responsibility to maintain the database. He is usually designated as the database administrator (DBA). Any group of the organization can be granted access to the centralized database for a specific purpose. 1.3 Database Management System A database management system (DBMS), or simply a database system (DBS), consists of the following: A collection of interrelated and persistent data (usually referred to as the database (DB)). A set of application programs used to access, update and manage that data (which form the data management system (MS)).
- 506. 491Part IV: Chapter 1 Database Concepts The goal of a DBMS is to provide an environment that is both convenient and efficient to use in. 1.4 Purpose of Databases To see why database management systems are necessary, let's look at a typical ``file- processing system'' supported by a conventional operating system. 1. The application is a savings bank: o Savings account and customer records are kept in permanent system files. o Application programs are written to manipulate files to perform the following tasks: Debit or credit an account. Add a new account. Find an account balance. Generate monthly statements. 2. Development of the system proceeds as follows: o New application programs must be written as the need arises. o New permanent files are created as required. o Over a long period of time files may be in different formats, and application programs may be in different languages. 3. So we can see there are problems with the straight file-processing approach: o Data redundancy and inconsistency Same information may be duplicated in several places. All copies may not be updated properly. .. o Data isolation Data in different files. Data in different formats. Difficult to write new application programs. o Multiple users Want concurrency for faster response time. Need protection for concurrent updates. For example, two customers withdrawing funds from a common account at the same time - account has $500 in it, and they withdraw $100 and $50. The result could be $350, $400 or $450 if no protection is provided. o Security problems Every user of the system should be able to access only the data they are permitted to see. For example, payroll people only handle employee records, and
- 507. 492 cannot see customer accounts; tellers only access account data and cannot see payroll data. Difficult to enforce this with application programs. o Integrity problems Data may be required to satisfy constraints. For example, minimum account balance allowed $25.00. These problems and others led to the development of database management systems. 1.5 DBMS Models A DBMS is divided into 4 models given below: 1. Hierarchical data model 2. Network data model 3. Relational data model 4. Object based logical models A brief description of relationalobject based data models is given in the following section. 1.5.1 Relational data model Relational database is a collection of related tables (relations) where each table is a collection of rows (tuples) and columns (attribute). A relation among set of values (in each column) is represented by a row of a table, hence the term relational database. A relation in a database has the following characteristics: 1. Every value in a relation is atomic which means that it can not be further divided 2. The rows in the relation are not ordered. 3. Names of columns are distinct and order of columns is immaterial “Since relation is a set, and sets are not ordered hence no ordering is defined on tuples of relation” 1.5.2 Object based logical model Object based logical models describe data at the conceptual and view levels. It provides fairly flexible structuring capabilities. It allows one to specify data constraints explicitly. Over 30 such models, including the few given below:
- 508. 493Part IV: Chapter 1 Database Concepts Entity-relationship model. Object-oriented model. Binary model. Semantic data model. Info logical model. Functional data model. 1.6 Relational Database Terminology Relational database terminology consists of the following parameters: Entity An entity is an object which can be uniquely identified. For example, Chris a student with ‘R.NO’ 101(table 1.1) is an entity Attribute An entity can be presented with a set of attributes. Thus, a row in a table represents an entity and a particular column of this entity is an attribute. So, we can say that a set of attributes define characteristics of an entity. In the ‘student table’ (table 1.1) NAME, R.NO, CITY are all attributes. Domain Domain is a collection of possible values that an attribute can have. An atomic domain is that containing values representing individual units. Thus, values of NAME of the student entity are drawn from all valid names. Similarly the domain of attribute R.NO can be considered to be only a valid integer number. The percentage can never be more then 100. The domain of this attribute is a valid number in the range of 0 to 100. Relation A relation is most conveniently represented by a table. A row of the table is known as a tuple. Number of rows or records is defined as ‘cardinality’ of the relation. Consider the relation shown in table 1.1. Table 1.1: TABLE STUDENT NAME R.NO CITY Ankit 1000 Delhi Deepti 1100 Delhi Karan 1107 Patna Lucky 1111 Chennai Raghav 1007 Mathura
- 509. 494 The relation is named ‘STUDENT’. The cardinality of this relation is 5. A tuple consisting of ‘n’ number of attributes is called a n-tuple. In term of tuple definition, we can say that the number of tuples in a relation is known as ‘cardinality’ and the number of attributes in a tuple is known as ‘degree’ of a relation. Key Within a given relation there is one attribute or collection of attributes with values that are unique within the relation and thus can be used to identify a tuple from a relation, such attribute or collection of attributes of the relation is known as key. In a relation, we may have more than one key identifying a tuple uniquely. All such keys are known as candidate keys. From the candidate keys, one can be used as primary key and others will become alternate keys. In the relation, CONSUMABLE (table 1.2), Ccode and Item are candidate keys. If Ccode is taken to be the primary key then Item will become the alternate key ‘else if’ Item is taken as primary key than Ccode will become an alternate key. Table 1.2 CONSUMABLE Ccode Item Qty Stock C01 CD-ROM 50 C02 DISK 1.44 120 C03 DISK 1.20 50 C04 ST.80 COL. 30 C05 ST.132 COL. 120 1.7 Relational Algebra In relational algebra we study about the various operations that can be carried out on a relation. Some of the major operations are given below:- • Selection (Unary Operator) By using selection operation one selects the horizontal subset of a relation. • Projection (Unary Operator) By using projection operation one selects the vertical subset of a relation. • Cartesian product(Binary Operator) It operates on two relations and is denoted by X. For example Cartesian product of two relations R1 and R2 is represented by R=R1*R2. The degree of R is equal
- 510. 495Part IV: Chapter 1 Database Concepts to the sum of the degrees of R1 and R2. Similarly the cardinality of R is obtained by the product of cardinality of R1 and cardinality of R2. For Example, Relation: R1 Relation: R2 Resultant Relation: R= R1 * R2 • Union (Binary Operator) The union operation operates on two relations. The operation is indicated by the alphabet capital ‘U’. For example, R=R1 U R2 represents sum of cardinality of R1 and of R2 but before the union operation can be used the following have to be considered for the operation R1 U R2. Degree of R1 = Degree of R2 jth attribute of R1 and jth attribute of R2 must have a common domain. For Example, Relation: R1 Relation: R2 Roll no Student Name Class 1 Varun XII 4 Ankit X 10 Deepti XI Teacher Code Teacher Name 107 Mr Malik 109 Mr Chopra Col1 Col2 Col3 Col4 Col5 1 Varun XII 107 Mr Malik 1 Varun XII 109 Mr Chopra 4 Ankit X 107 Mr Malik 4 Ankit X 109 Mr Chopra 10 Deepti XI 107 Mr Malik 10 Deepti XI 109 Mr Chopra
- 511. 496 Resultant Relation: R = R1 U R2 Column 1 Column 2 S101 Raghav S102 Anisha R101 Anshuman R103 Gaurav R107 kartik 1.8 Normalization As we have observed earlier the main goal of relational database design is to generate a set of relational structures (schemes) that allow us to store information without necessary redundancy, yet allow easy retrieval of information, effective updating and manipulation of data at the same time. If the database design is not proper then it may have some undesirable characteristics like:- 1. Redundancy/ Duplicity of information 2. Inability to map certain information 3. Loss pf information A design is said to be good if one fact is stored at one place. Theory of normalization is a design aid which tries to approach the concept of one fact at one place. Thus, we can say that ‘normalization is the process of analyzing data for grouping entries and data elements into set of harmonious relations’. There are many forms of normalization the ones to learn are given as follows:- • First Normal form A database is said to be in the first normal form if each of its attributes contains atomic values and has no repeated values or no two rows are identical. Table 1.3 Raw Data ICod e INam e I_type S_Code 1 Suplier 1 Scity1 S_Code2 Supplier2 Scity2 S_code 3 Supplier3 Scity3 8102 Saree L_wear S101 Krishna Delhi S201 Yuvraj Mumbai 7945 Belt Accs S102 Kriti Mumbai S201 Yuvraj Mumbai S113 Parekh Mumbai 6789 Bag Accs S103 Gupta Kolkata 1060 Tie M.wear S104 Sinha Delhi S102 Kriti Student_Code Student_Name S101 Raghav S102 Anisha Student_ID Name R101 Anshuman R103 Gaurav R107 Kartik
- 512. 497Part IV: Chapter 1 Database Concepts Table 1.3 in First Normal Form ICode IName Item_type Supplier_Code Supplier_Name Supplier_city 8102 Saree L_wear S101 Mr. Krishna Mumbai 8102 Saree L_wear S201 Mr. Yuvraj Delhi 7945 Belt Accs S102 Mr. Kriti Madras 7945 Belt Accs S201 Mr. Yuvraj Delhi 7945 Belt Accs S113 Mr. Parekh Mumbai 6789 Bag Accs S103 Mr. Gupta Kolkata 1060 Tie M.wear S104 Mr. Sinha Delhi 1060 Tie M.wear S201 Mr. Kriti Madras • Second Normal Form A database is in the second normal form (2NF) if it is already in first normal form(1NF) and each table in the database contains data about one and only one entity (this can be optionally checked by looking at the primary key) and that all non key attributes of the table are dependent on the key value(s).If we convert the above relation into 2NF, table 1.4 ‘Item’ consists of complete details of ‘Item’ whereas table 1.5 ‘Supplier’ contains details of the supplier and the items he is supplying: Second Normal Form Table 1.4 Item Table ICode IName Item_type 8102 Saree L_wear 7945 Belt Accs 6789 Bag Accs 1060 Tie M.wear Table 1.5 Supplier Table ICode Supplier_Code Supplier_Name Supplier_city 8102 S101 Mr. Krishna Mumbai 8102 S201 Mr. Yuvraj Delhi 7945 S102 Mr. Kriti Madras 7945 S201 Mr. Yuvraj Delhi 7945 S113 Mr. Parekh Mumbai 6789 S103 Mr. Gupta Kolkata 1060 S104 Mr. Sinha Delhi 1060 S201 Mr. Kriti Madras
- 513. 498 • Third Normal Form A database is said to be in the third normal form if it is already in the second normal form and none of the non key attributes are dependent on any other non key attributes. In other words, all the non key attributes directly depend only on the primary key. So for bringing a table in the third normal form all attributes not dependent on the primary key attribute should be separated out to another table or removed. For converting the database that we converted into third normal form the table ‘Item’ will undergo no changes, where as the ‘Supplier’ table will be divided into two tables ‘Item_supply’ which consists of who the supplier is (Supplier_code) and what he is supplying (ICode) and the ‘Supplier’ table which has the details of the ‘Supplier’. Third Normal Form Table Item Table Item_supply ICode Supplier_code 8102 S101 8102 S201 7945 S102 7945 S201 7945 S113 6789 S103 1060 S104 1060 S201 ICode IName Item_type 8102 Saree L_wear 7945 Belt Accs 6789 Bag Accs 1060 Tie M.wear
- 514. 499Part IV: Chapter 1 Database Concepts Table Supplier Supplier_Code Supplier_Name Supplier_city S101 Mr. Krishna Mumbai S201 Mr. Yuvraj Delhi S102 Mr. Kriti Madras S201 Mr. Yuvraj Delhi S113 Mr. Parekh Mumbai S103 Mr. Gupta Kolkata S104 Mr. Sinha Delhi S201 Mr. Kriti Madras 1.9 Review Examples Example 1.9.1 What is data redundancy and how does redundancy of data lead to data inconsistency? Explain with an example? Solution of 1.9.1 Data redundancy is the storage of same data in more than one file. This leads to data inconsistency. This is because the same data might have been entered by different users at different times or it might have been updated by different users at different times. For example, the address of students in an educational institution may be stored in the students section of the institution. Some of these students might be residing in the hostel of the institution. Their addresses may be available in the hostel as well. Suppose the address of a student is changed. It is possible that the change in address is entered in the file maintained in the hostel. But the same may not be updated in the file of the student section. Thus duplication may lead to data inconsistency. Example 1.9.2 What is a relation? What is the difference between a tuple and an attribute? Solution of 1.9.2 A relation is like a table in which data are arranged in the form of rows and columns. The rows of the table are known as the tuples and the columns of the table are known as the attributes. Example 1.9.3 What is data security? Solution of 1.9.3 Data security is the protection given to the data against modification or destruction, intentional or inadvertent, by authorized users.
- 515. 500 Example 1.9.4 What is the degree and cardinality of the table products? Also check if the 2 tables given below are union compatible or not? Table 'PRODUCTS' Table ‘CUSTOMER’ Solution of 1.9.4 The degree of table product is 4 and the cardinality is 5 and the two tables specified are not compatible union. Example 1.9.5 Define the following terms:- • Tuple • Attribute • Domain • Degree • Cardinality Solution of 1.9.5 • Tuple: The rows of the table are known as tuples. • Attribute: The columns of the table are knows as the attributes. • Domain: It is a set to which each entry in the column of a relation belongs. • Degree: The number of columns (attributes) is a relation known as the degree of the relation. • Cardinality: The number of rows (tuples) in a relation are known as the cardinality of the relation. Example 1.9.6 Find the cartesian product of the two relations ‘R1’ and ‘R2’ given below:- Relation R1 Relation R2 VEDOR PRODUCT QTY. PRICE V1 P1 5 500 V1 P2 7 700 V2 P3 7 750 V3 P4 10 720 V4 P5 8 780 ORDER NO. CUSTOMER 1 C1 2 C2 3 C3 T1 T2 T3 R 1 25 J 2 24 P 3 23 T4 T5 A X B Y C Z
- 518. Part IV Chapter 2 STRUCTURED QUERY LANGUAGE (SQL) AIM OUTLINE To introduce SQL Commands To study the built-in functions To learn use of the SQL commands and functions through lot of solved examples 2.1 Introduction 2.2 Capabilities of SQL 2.3 SQL Data types 2.4 SQL Commands and Functions 2.5 Review Examples 2.6 Review Exercise 2.7 Let us revise
- 519. 504 2.1 Introduction SQL is the base language of all ‘Relational database management systems (RDBMS)’. SQL allows users to access data in relational database management systems, such as Oracle, Sybase, Informix, Microsoft SQL Server, Access, and others, by allowing users to perform operations, such as, create database tables, insert, delete, sort, and manipulate data. In this appendix, we will discuss elementary SQL commands. It may be noted that limited SQL commands are also available in recent versions of Dbase, Foxbase, Foxpro etc. 2.2 Capabilities of SQL DDL(Data Definition Language) A database schema (The overall design of the database) is specified by a set of definition that is expressed by a data definition language. The SQL DDL provides commands for defining relation schemes, creating indices, and modifying relation schemes. DML(Data Manipulation language) The SQL-DML includes those commands, which are based on both the relational algebra and the tuple relational calculus. DML is a language that enables users to access or manipulate data. By Data manipulation, we mean: o The retrieval of information stored in the database o The insertion of new information into the database o The deletion of information from the database o The modification of information stored in the database 2.3 SQL Datatypes SQL supports following data types for specification of attributes of a relation: 1. ‘char(n)’ datatype: An attribute can be declared of type char(n) which specifies that the attribute will store a string of n characters to a maximum of 240 characters. 2. Small ‘int’: An integer having a value between -99,999 to 99,999. 3. Date: A date value that is displayed is in accordance with the SET DATE BRITISH/AMERICAN and SET CENTURY ON/ OFF commands in the format {dd/mm/yy}, {mm/dd/yy, {dd/mm/yyyy} or {mm/dd/yyyy}
- 520. 505Part IV: Chapter 2 Structured Query Language 2.4 SQL Commands and Functions The SELECT Statement In a relational database, data is stored in tables. For example look at table 2.1 ‘EmployeeAddressTable’ which relates Social Insurance number (SIN), Name (FirstName, LastName) and Address (Address, city, state) of an employee. Now let's say, we want to see the address of each employee. For doing so, one will have to use the SELECT statement. The command for using the SELECT clause to view the addresses of all the employees is given below. SELECT FirstName, LastName, Address, City, State FROM EmployeeAddressTable; The following is the results of the query given above on the database: FirstName LastName Address City State Ankit Asthana 29 Sirifort Road Kanpur U.P. Deepti Asthana H-43 Anand Lok New Delhi Delhi Varun Beri E-39 Susant Lok Bhopal M.P. All we have done by executing this query is that we have asked for the entire data in the EmployeeAddressTable, Z which includes columns called ‘FirstName’, ‘LastName’, ‘Address’, ‘City’, and ‘State’. Note that column names and table names do not have spaces...they must be typed as one word; and that the statement ends with a semicolon (;). The general form for a SELECT statement, retrieving all of the rows in the table is given below: SELECT ColumnName, ColumnName, ... FROM TableName; 2.1 EmployeeAddressTable SIN FirstName LastName Address City State 512687458 Ankit Asthana 29 Sirifort Road Kanpur U.P. 758420012 Deepti Asthana H-43 Anand Lok New Delhi Delhi 102254896 Varun Beri E-39 Susant Lok Bhopal M.P.
- 521. 506 To view all the columns of a table without typing all column names, we can also use the following statement: SELECT * FROM TableName; o Conditional SELECT statement To further discuss the SELECT statement, let's look at a new example table (‘EmployeeStatisticsTable): EmployeeStatisticsTable EmployeeIDNo Salary Benefits Position 010 75000 15000 Manager 105 65000 15000 Manager 152 60000 15000 Manager 215 60000 12500 Manager 244 50000 12000 Staff 300 45000 10000 Staff 335 40000 10000 Staff 400 32000 7500 Entry-Level 441 28000 7500 Entry-Level Relational Operators There are six Relational Operators in SQL as given in table below: = Equal <> or != (see manual) Not Equal < Less Than > Greater Than <= Less Than or Equal To >= Greater Than or Equal To We can also use the ‘WHERE’ clause with the select statement. The ‘WHERE’ is used to specify that only certain rows of the table are to be displayed based on the
- 522. 507Part IV: Chapter 2 Structured Query Language criteria or conditions described in the WHERE clause. Let us take an example to learn how do we use the ‘WHERE’ clause. Suppose that from ‘EmployeeStatisticsTable’ we want to view the EmployeeIdNo’s of all the employees making over $0000, we would use the following query. SELECT EMPLOYEEIDNO FROM EMPLOYEESTATISTICSTABLE WHERE SALARY >= 50000; Notice that the >= (greater than or equal to) sign is used, as we wanted to view Ids of those employees who earn greater than or equal to $50,000. The result of this query is given below: The WHERE clause SALARY >= 50000, is known as a condition (an operation which evaluates to True or False). Let us take another example, suppose we want to display the Id numbers of all the emplyoee(s) having position ‘Manager’ assigned to them. This can be done using the following query: SELECT EMPLOYEEIDNO FROM EMPLOYEESTATISTICSTABLE WHERE POSITION = 'Manager'; This query displays the ID Numbers of all Managers. Generally, with text columns, stick to equal to or not equal to, and make sure that any text that appears in the statement is surrounded by single quotes ('). More Complex Conditions: Compound Conditions / Logical Operators o The ‘And’ Operator The AND operator joins two or more conditions, and displays a row only if that row's data satisfies ALL conditions listed in the query it is used in. For example, to display all the staff employees making over $ 40,000 the following query shall be used: EMPLOYEEIDNO ------------ 010 105 152 215 244
- 523. 508 SELECT EMPLOYEEIDNO FROM EMPLOYEESTATISTICSTABLE WHERE SALARY > 40000 AND POSITION = 'Staff'; o The ‘OR’ Operator The OR operator joins two or more conditions, but returns a row if ANY of the conditions listed hold true. To see all those who make less than $.40, 000 or have less than $10,000 in benefits, listed together, one would use the following query: SELECT EMPLOYEEIDNO FROM EMPLOYEESTATISTICSTABLE WHERE SALARY < 40000 OR BENEFITS < 10000; One can also combine the ‘AND’ & the ‘OR’ and use them to create queries. For example, SELECT EMPLOYEEIDNO FROM EMPLOYEESTATISTICSTABLE WHERE POSITION = 'Manager' AND SALARY > 60000 OR BENEFITS > 12000; When this query is accessed what SQL does is that it displays the ID numbers of those employees who are assigned the position ‘Manager’ and earning salary greater than $6000 or earning benefits greater than $12000. The important thing to note though is that the AND operation is done first. To generalize this process, SQL performs the AND operation(s) to determine the rows where the AND operation(s) hold true (remember: all of the conditions are true), then these results are used to compare with the OR conditions, and only display those remaining rows where any of the conditions joined by the OR operator hold true (where a condition or result from an AND is paired with another condition or AND result to use to evaluate the OR, which evaluates to true if either value is true). Mathematically, SQL evaluates all of the conditions, then evaluates the AND "pairs", and then evaluates the OR's (where both operators evaluate left to right). True AND False OR True AND True OR False AND False Step 1: Simplify the AND pairs: Result : False OR True OR False Step 2: Simplify the OR's from left to right in the statement Result: True OR False True
- 524. 509Part IV: Chapter 2 Structured Query Language The result of this query hence comes out to be ‘True’, and the row passes the query conditions. The logical operations in SQL follow the precedence shown in table 2.2. Table 2.2: SQL Order of Logical Operations (each operates from left to right) 1. NOT 2. AND 3. OR DISTINCT clause with SELECT statement Before understanding how does the DISTINCT clause work with the SELECT statement, let us first look at the following table: Table 2.3: ANTIQUES OwnerID OwnerLastName OwnerFirstName Ownercity 01 Jacob Sam New Delhi 02 Pathan Ram Kanpur 15 Dravid Dinesh Mumbai 21 Leander Ramesh Jhansi 50 Marley Lucas Kolkata SELECT DISTINCT Ownercity from ANTIQUES In this example, since everyone lives in a different city hence all of them will be displayed. Aggregate Functions in SQL There are five important aggregate functions: SUM, AVG, MAX, MIN, and COUNT which are used in SQL. They are known as aggregate functions because they summarize the results of a query, rather than listing all of the rows.
- 525. 510 • SUM () gives the total of all the rows, satisfying any conditions, of the given column, where the given column is numeric. • AVG () gives the average of the given column. • MAX () gives the largest figure in the given column. • MIN () gives the smallest figure in the given column. • COUNT(*) gives the number of rows satisfying the conditions. Let us take up a few examples using the ‘EmployeeStatisticsTable’ to understand how aggregate functions are used in SQL. EmployeeStatisticsTable EmployeeIDNo Salary Benefits Position 010 75000 15000 Manager 105 65000 15000 Manager 152 60000 15000 Manager 215 60000 12500 Manager 244 50000 12000 Staff 300 45000 10000 Staff 335 40000 10000 Staff 400 32000 7500 Entry-Level 441 28000 7500 Entry-Level SELECT SUM(SALARY), AVG(SALARY) FROM EMPLOYEESTATISTICSTABLE; This query shows the total of all salaries in the table, and the average salary of all of the Employees in the table. SELECT MIN(BENEFITS) FROM EMPLOYEESTATISTICSTABLE WHERE POSITION = 'Manager'; This query gives the smallest figure of the Benefits column, of the employees who are Managers, the result is ‘12500’. SELECT COUNT(*) FROM EMPLOYEESTATISTICSTABLE WHERE POSITION = 'Staff'; This query tells us how many employees have Staff status (3).
- 526. 511Part IV: Chapter 2 Structured Query Language Views In SQL, you might (check your DBA) have access to create views for yourself. A view allows you to assign the results of a query to a new, personal table, that you can use in other queries, where this new table is given the view name in the FROM clause. When you access a view, the query that is defined in your view creation statement is performed (generally), and the results of that query look just like another table in the query that you wrote invoking the view. To create a view we use the following syntax: Syntax: CREATE VIEW view_name as SELECT col1, col2 from table_name where condition; Example, CREATE VIEW temp as SELECT salary, benefits, position from EmployeeStatisticsTable Modification of database through views is possible (updates, deletion, insertion) since they are physically stored in the database, though this may have potentially disadvantageous consequences. Note: To delete a view from the database one can use the drop view command. Syntax: DROP VIEW View_Name; Creating New Tables All tables within a database must be created at some point in time...let's see how we would create the ORDERS table: Syntax: Create table table_name(col1 datatype(size), col2 datatype(Size,……… .); For Example, CREATE TABLE ORDERS (OWNERID INTEGER NOT NULL, ITEMDESIRED CHAR(40) NOT NULL); This statement gives the table name and tells the DBMS about each column in the table. Please note that this statement uses generic data types, and that the data types might be
- 527. 512 different, depending on what DBMS you are using. Some common generic data types are: • Char(x) - A column of characters, where x is a number designating the maximum number of characters allowed (maximum length) in the column. • Integer - A column of whole numbers, positive or negative. • Decimal(x, y) - A column of decimal numbers, where x is the maximum length in digits of the decimal numbers in this column, and y is the maximum number of digits allowed after the decimal point. The maximum (4,2) number would be 99.99. • Date - A date column in a DBMS-specific format. • Logical - A column that can hold only two values: TRUE or FALSE. NOT NULL means that the column must have a value in each row. If NULL was used, that column may be left empty in a given row. Altering Tables Let's add a column to the Antiques table to allow the entry of the price of a given Item: Syntax: ALTER TABLE Table_Name ADD(col1datatype(size), col2 datatype(size) …… .); ALTER TABLE ANTIQUES ADD (PRICE DECIMAL(8,2) NULL); The data for this new column can be updated or inserted using the ‘INSERT INTO’ clause explained in the following section. Adding Data To insert rows into a table, do the following: Syntax, INSERT INTO Table_Name (col1,col2,….) VALUES ( value 1, value 2…); INSERT INTO ANTIQUES VALUES (21, 01, 'Ottoman', 200.00); This query inserts the data into the table, as a new row, column-by-column, in the pre- defined order. Instead, let's change the order and leave Price blank:
- 528. 513Part IV: Chapter 2 Structured Query Language INSERT INTO ANTIQUES (BUYERID, SELLERID, ITEM) VALUES (01, 21, 'Ottoman'); Deleting Data Syntax: DELETE FROM Table_Name where condition; For Example, DELETE FROM ANTIQUES WHERE ITEM = 'Ottoman'; For deleting entries from the data base the ‘delete’ clause is used. The statement given above deletes all rows where ITEM is ‘Ottoman’. In order to delete the row we added in the previous section, we have to be more specific in our query. The following statement deletes the row; we created in the previous section. DELETE FROM ANTIQUES WHERE ITEM = 'Ottoman' AND BUYERID = 01 AND SELLERID = 21; Updating Data Let's update a Price into a row that doesn't have a price listed yet: Syntax: UPDATE Table_Name SET COL1 = expression 1, COL2= expression 2, where condition; Example, UPDATE ANTIQUES SET PRICE = 500.00 WHERE ITEM = 'Chair'; This sets all the Chair's Prices to 500.00. As shown above, more WHERE conditionals, using AND, must be used to limit the updating to more specific rows. Also, additional columns may be set by separating equal statements with commas.
- 529. 514 GROUP BY & HAVING Antiques SellerID BuyerID Item 01 50 Bed 02 15 Table 15 02 Chair 21 50 Mirror 50 01 Desk 01 21 Cabinet 02 21 Coffee Table 15 50 Chair 01 15 Jewelry Box 02 21 Pottery 21 02 Bookcase 50 01 Plant Stand One special use of GROUP BY is to associate an aggregate function (especially COUNT; counting the number of rows in each group) with groups of rows. First, assume that the Antiques table has the Price column, and each row has a value for that column. We want to see the price of the most expensive item bought by each owner. We have to tell SQL to group each owner's purchases, and tell us the maximum purchase price: SELECT BUYERID, MAX(PRICE) FROM ANTIQUES GROUP BY BUYERID; Now, say we only want to see the maximum purchase price if the purchase is over $ 1000, so we use the HAVING clause: SELECT BUYERID, MAX(PRICE) FROM ANTIQUES GROUP BY BUYERID HAVING PRICE > 1000; 2.5 Review Examples Example 2.5.1 What is difference between terms DDL and DML?
- 531. 516 6 Ketaki 16 ENT 24/02/98 250 F 7 Ankita 29 Cardiology 20/02/98 800 F 8 Zareen 45 Gynecology 22/02/98 300 F 9 Kush 19 Cardiology 13/01/98 800 M 10 Shilpa 23 Nuclear Medicine 21/02/98 400 F a) To select all the information of patients of the cardiology department. b) To list the names of female patients who are in the ENT department. c) To list names of all patients with their date of admission in ascending order. d) To display patients name, charges, Age for only the female patients. e) To count the number of patients with age<30. f) To insert a new row in the HOSPITAL table with the following data: 11,”Aftab”, 24,”Surgery”,{25/02/98},300,”M” g) Give the output of following SQL statements: i. Select COUNT(DISTINCT charges) from HOSPITAL; ii. Select MIN(Age) from HOSPITAL where sex = “f”; iii. Select sum(charges) from HOSPITAL where department = “ENT”; iv. Select AVG(charges) from HOSPITAL where Dateofadm<{12/02/98}. Solution of 2.5.3 a) SELECT * FROM hospital where department = “Cardiology”; b) SELECT Name FROM hospital where Department = “ENT” AND Sex=”F”; c) SELECT Name, Dateofadm FROM hospital ORDER BY Dateofadm; d) SELECT Name, Dateofadm FROM hospital WHERE Sex=”f”; e) SELECT COUNT(*) FROM hospital where Age<30; f) INSERT INTO hospital values (11,”Aftab”,24,”Surgery”,{24/02/98},300,”M”}; g) i) 5 ii) 16 iii) 500 iv) 380.00 Example 2.5.4 Write SQL statement to create a table EMP with the under mentioned structure:- EmpNo NUMBER(4) DeptNo NUMBER(2) EmpName CHAR(20) Job CHAR(10) Manager NUMBER(4) Hiredate DATE Salary NUMBER(7,2) Commission NUMBER(7,2)
- 532. 517Part IV: Chapter 2 Structured Query Language Solution of 2.5.4 Create table EMP (EmpNO NUMBER(4) NOT NULL PRIMARY KEY, DeptNo NUMBER(2), EmpName CHAR(20), Job CHAR(10), Manager NUMBER(4), Hiredate DATE, Salary NUMBER(7,2), Commission NUMBER(7,2)); Example 2.5.5 Discuss about the various types of datatypes in SQL? Solution of 2.5.5 • Char(x) - A column of characters, where x is a number designating the maximum number of characters allowed (maximum length) in the column. • Integer - A column of whole numbers, positive or negative. • Decimal(x, y) - A column of decimal numbers, where x is the maximum length in digits of the decimal numbers in this column, and y is the maximum number of digits allowed after the decimal point. The maximum (4,2) number would be 99.99. • Date - A date column in a DBMS-specific format. Example 2.5.6 What is a view? What are the advantages of this clause? Solution of 2.5.6 A view allows you to assign the results of a query to a new, personal table, that you can use in other queries, where this new table is given the view name in your FROM clause. When you access a view, the query that is defined in your view creation statement is performed (generally), and the results of that query look just like another table in the query that you wrote invoking the view. For example, to create a view: Syntax: create view view_name as select col1, col2 from table_name where condition;
- 534. 519Part IV: Chapter 2 Structured Query Language (iv) To display the coach’s name and ACode in ascending order of ACode from the table COACH. (v) SELECT COUNT(DISTINCT Number) from ACTIVITY; (vi) SELECT MAX(ScheduleDate),MIN(ScheduleDate)FROM ACTIVITY; (vii)SELECT SUM (PrizeMoney) FROM ACTIVITY; (viii) SELECT DISTINCT Number FROM ACTIVITY;
- 535. 520 2.7 Let us revise! Here are the general forms of the statements discussed in this chapter, plus some extra important ones (explanations given). REMEMBER that all of these statements may or may not be available on your system, so check documentation regarding availability: ALTER TABLE <TABLE NAME> ADD|DROP|MODIFY (COLUMN SPECIFICATION[S]...see Create Table); --allows you to add or delete a column or columns from a table, or change the specification (data type, etc.) on an existing column; this statement is also used to change the physical specifications of a table (how a table is stored, etc.), but these definitions are DBMS-specific, so read the documentation. Also, these physical specifications are used with the Create Table statement, when a table is first created. In addition, only one option can be performed per Alter Table statement-- either add, drop, OR modify in a single statement. CREATE [UNIQUE] INDEX <INDEX NAME> ON <TABLE NAME> (<COLUMN LIST>); --UNIQUE is optional; within brackets. CREATE TABLE <TABLE NAME> (<COLUMN NAME> <DATA TYPE> [(<SIZE>)] <COLUMN CONSTRAINT>, ...other columns); (also valid with ALTER TABLE) --where SIZE is only used on certain data types (see above), and constraints include the following possibilities (automatically enforced by the DBMS; failure causes an error to be generated): 1. NULL or NOT NULL (see above) 2. UNIQUE enforces that no two rows will have the same value for this column 3. PRIMARY KEY tells the database that this column is the primary key column (only used if the key is a one column key, otherwise a PRIMARY KEY (column, column, ...) statement appears after the last column definition. 4. CHECK allows a condition to be checked for when data in that column is updated or inserted; for example, CHECK (PRICE > 0) causes the system to check that the Price column is greater than zero before accepting the value...sometimes implemented as the CONSTRAINT statement. 5. DEFAULT inserts the default value into the database if a row is inserted without that column's data being inserted; for example, BENEFITS INTEGER DEFAULT = 10000 6. FOREIGN KEY works the same as Primary Key, but is followed by: REFERENCES <TABLE NAME> (<COLUMN NAME>), which refers to the referential primary key.
- 536. 521Part IV: Chapter 2 Structured Query Language CREATE VIEW <TABLE NAME> AS <QUERY>; DELETE FROM <TABLE NAME> WHERE <CONDITION>; INSERT INTO <TABLE NAME> [(<COLUMN LIST>)] VALUES (<VALUE LIST>); SELECT [DISTINCT|ALL] <LIST OF COLUMNS, FUNCTIONS, CONSTANTS, ETC.> FROM <LIST OF TABLES OR VIEWS> [WHERE <CONDITION(S)>] [GROUP BY <GROUPING COLUMN(S)>] [HAVING <CONDITION>] [ORDER BY <ORDERING COLUMN(S)> [ASC|DESC]]; --where ASC|DESC allows the ordering to be done in ASCending or DESCending order UPDATE <TABLE NAME> SET <COLUMN NAME> = <VALUE> [WHERE <CONDITION>]; --if the Where clause is left out, all rows will be updated according to the Set statement
- 537. 522 Notes
- 538. Part IV Chapter 3 CONNECTING DATABASES WITH C++ AIM OUTLINE Introduce the technique of connecting C++ programs to Databases To learn steps of connecting Borland C++ with databases To know how to connect Microsoft’s ‘Visual C++’ to databases 3.1 Open Database Connectivity (ODBC) 3.2 Setting up ODBC data source 3.3 Connecting from Delphi or C++ Builder using the ODBC data source 3.4 Connecting MS Visual C++ 5.0 to Databases 3.5 Code to Connect C/C++ to the ODBC data source
- 539. 524 3.1 Open Database Connectivity (ODBC) ODBC is a standard or open application programming interface (API) for accessing to database management systems (DBMS). This API is independent of any one programming language, database system or operating system. ODBC is based on the Call Level Interface (CLI) specifications from SQL, X/Open (now part of The Open Group), and the ISO/IEC. ODBC was created by the SQL Access Group and first released in September, 1992. By using ODBC statements in a C++ program, you can access files in a number of different databases, such as, MS Access, MS SQL, Oracle, DB2 and My SQL. In addition to the ODBC software, a separate module or driver is needed for each database to be accessed. 3.2 Setting up ODBC data source In order to make a connection from 3rd party products using ODBC, an ODBC data source is needed. A data source stores connection information such as user name, password, and location of database. Steps for setting up a data source: Step 1: ODBC Administrator can be started by going to Control Panel and double clicking on 32 bit ODBC or Starting the utility called "32bit ODBC Administrator” if you have the ODBC SDK installed. Step 2: Stay at the User Data Source tab and click on Add. This will bring up another window titled "Create New Data Source". Step 3: Pick the ODBC driver that to be used. Pick the InterBase driver which is called "InterBase 5.x driver by Visigenic (*.gdb)" and then click on Finish. This will bring up a new window with the title "InterBase ODBC Configuration". Step 4: Fill in the blank fields in this window Data Source Name: Make up a name for your data source. Description: This is the description of the data source. It's not required. Network Protocol: Choose the protocol from the drop down list.
- 540. 525Part IV: Chapter 3 Connecting Databases with C++ Database: Fill in the physical full path to the database including the database name to server. Server: Fill in the server name. If you choose the protocol "local", this will default to the local server. Username: Fill in the database user name. Password: Fill in the password corresponding to the above user name. Go to the Advanced tab and fill in the CharacterSet and Roles. Step 5: Clicking the OK button will bring back to the main form. You should see the newly added user data source there. Note: A user data source is a data source visible to the user whereas a system data source is visible to the system. 3.3 Connecting from Delphi or C++ Builder using the ODBC data source ODBC connection from Delphi is very similar to connecting using Borland Development Environment from Delphi. Here is an example of connecting using the Tquery component. This example will also display the results of a sql statement. Step 1: Drop a Tquery, a Tdatasource, and a Tdbgrid component on a Delphi form. Step 2: Set the following properties for the Tquery component: DatabaseName: Pick from the list the data source name you just created in ODBC Administrator. SQL: Input the SQL statement to be executed. For example: "select * from table1". Active: Set to True to connect. And supply user name and password on connection. Step 3: Set the following properties for the Tdatasource component: Data Set: Set to the name of the Tquery component, or "query1" in this case. Step 4: Set the following properties for the TDBGrid component: Data Source: Set to the name of the Tdatasource component, or "data source1" in this case.
- 541. 526 Step 5: Now you can see the returned results from select statement in the dbgrid area. 3.4 Connecting MS Visual C++ 5.0 to Databases Now we have to create an ODBC data-source, which points to your database. Step 1: Start up ODBC Administrator. Double click the "32bit ODBC" icon in Control Panel to start the ODBC administrator. Step 2: Click the "Add" button to create a new data-source, select the driver you wish to use (if your database is say, MS SQL select the "Microsoft SQL Driver" and so on) and click the "finish" button. Step 3: Enter a name for the data source in the "Data Source Name" field. "Description" field is optional. Step 4: Click on the "Select" button from the "Database" field. A browser box appears where you select the location of your MS SQL database file. When done click the "OK" button to return back to the previous screen and "OK" again to complete the creation of the data source. Now click the "OK" button to quit the ODBC administrator. Step 5: Now, we can start writing a C++ program that will open a connection to the database and perform an SQL query. The SQL query can be any SQL statement to search, insert, update or delete data in the database. This program will not do any validation of whether or not a record with the same value already exists in the database table and the SQL string is a legal SQL statement. A set of C/C++ functions are supplied with ODBC and are found in the two header files sql.h and sqlext.h. We would need to include both the header files in the program using the "#include" operation. 3.5 Code to Connect C/C++ to the ODBC data source The code for connecting C/C++ to OBDC data source (“ABC”) and performing SQL query is given in program 3.1. The code does not include error handling considered desirable when modifying the table or establishing and closing the connection to the database. You will need to change the name of the data source if you wish to use this code. The database may also require a user ID and password. Program 3.1 // ODBC data source is called "ABC". It then executes a SQL statement “SELECT Model // FROM Makes WHERE Make = ' Suzuki';" should pull out all models made by Suzuki stored in the database
- 542. 527Part IV: Chapter 3 Connecting Databases with C++ #include <windows.h> #include <sqlext.h> #include <stdio.h> #include <iostream.h> void main() { HENV hEnv = NULL; // Handle from SQLAllocEnv() HDBC hDBC = NULL; // Handle for Connection HSTMT hStmt = NULL; // Statement handle UCHAR szDSN[SQL_MAX_DSN_LENGTH] = "ABC"; // Data Source Name UCHAR* szUID = NULL; // User ID buffer UCHAR* szPasswd = NULL; // Password buffer UCHAR szModel[128]; // Model buffer SDWORD cbModel; // Model buffer bytes received UCHAR szSqlStr[] = "Select Model From Makes Where Make=''Suzuki"; // SQL string RETCODE retcode; // Return code // allocate memory for ODBC Environment handle SQLAllocEnv (&hEnv); // Allocate memory for the connection handle SQLAllocConnect (hEnv, &hDBC); // connect to the data source "ABC" using userid and password. retcode = SQLConnect (hDBC, szDSN, SQL_NTS, szUID, SQL_NTS, szPasswd, SQL_NTS); if (retcode == SQL_SUCCESS || retcode == SQL_SUCCESS_WITH_INFO) { // allocate memory for the statement handle retcode = SQLAllocStmt (hDBC, &hStmt); // prepare the SQL statement by assigning it to the statement handle retcode = SQLPrepare (hStmt, szSqlStr, sizeof (szSqlStr)); // execute SQL statement handle retcode = SQLExecute (hStmt); // Project only column 1 which is the models SQLBindCol (hStmt, 1, SQL_C_CHAR, szModel, sizeof(szModel), &cbModel); // Get row of data from the result set defined above in the statement retcode = SQLFetch (hStmt); while (retcode == SQL_SUCCESS || retcode == SQL_SUCCESS_WITH_INFO) { cout ("t%sn", szModel); // Print row (model) retcode = SQLFetch (hStmt); // Fetch next row from result set } // Free the allocated statement handle SQLFreeStmt (hStmt, SQL_DROP); // Disconnect from datasource SQLDisconnect (hDBC);
- 543. 528 } // Free the allocated connection handle SQLFreeConnect (hDBC); // Free the allocated ODBC environment handle SQLFreeEnv (hEnv); }
- 546. APPENDIX A ASCII TABLE The American Standard Code for Information Interchange is a standard seven-bit code that was finalized by ANSI in 1968. ASCII is a standard code used for information interchange and communication between data processing systems (including Internet). ASCII which is pronounced as ‘ask key’ represents ‘English’ characters like alphabets, decimal numbers, control and graphic characters as unique numbers from 0 to 127. A.1 CONTROL CHARACTERS The 7 bit ASCII table is given in table A.1 and A.2. The first part of the table includes descriptions of the first 32 non-printing also referred to as control characters. ASCII was designed for use with teletypes as a result control characters are not as frequently used as intended. Table A.1 ASCII TABLE Name/Meaning Ctrl char Dec Hex ctrl-@ 0 00 ctrl-A 1 01 ctrl-B 2 02 ctrl-C 3 03 ctrl-D 4 04 ctrl-E 5 05 ctrl-F 6 06 ctrl-G 7 07 ctrl-H 8 08 ctrl-I 9 09 ctrl-J 10 0A ctrl-K 11 0B ctrl-L 12 0C ctrl-M 13 0D ctrl-N 14 0E Char NUL SOH ETX EOT ENQ ACK BEL BS HT LF VT FF CR SO STX Null start of heading start of text end of text end of xmit Enquiry acknowledge Bell Backspace horizontal tab Line feed vertical tab Form feed carriage feed shift out
- 547. 532 shift in ctrl-O 15 0F SI data line escape ctrl-P 16 10 DLE device control 1 ctrl-Q 17 11 DC1 device control 2 ctrl-R 18 12 DC2 device control 3 ctrl-S 19 13 DC3 device control 4 ctrl-T 20 14 DC4 Neg acknowledge ctrl-U 21 15 NAK synchronous idel ctrl-V 22 16 SYN End of xmit block ctrl-W 23 17 ETB Cancel ctrl-X 24 18 CAN end of medium ctrl-Y 25 19 EM Substitute ctrl-Z 26 1A SUB Escape ctrl-[ 27 1B ESC file separator ctrl- 28 1C FS group separator ctrl-] 29 1D GS record separator ctrl-^ 30 1E RS unit separator ctrl-_ 31 1F US A.2 PRINTABLE CHARACTERS The ASCII table also includes of 96 printable characters which are given in table A.2. For example, the capital letter ‘A’ is always represented by the order number ‘65’, Table A.2: ASCII TABLE Dec Hex Char Dec Hex Char Dec Hex Char 32 20 Space 64 40 @ 96 60 ` 33 21 ! 65 41 A 97 61 a 34 22 " 66 42 B 98 62 b 35 23 # 67 43 C 99 63 c 36 24 $ 68 44 D 100 64 d 37 25 % 69 45 E 101 65 e 38 26 & 70 46 F 102 66 f 39 27 ' 71 47 G 103 67 g 40 28 ( 72 48 H 104 68 h APPENDIXA
- 548. 533Appendix A APPENDIXA 41 29 ) 73 49 I 105 69 i 42 2A * 74 4A J 106 6A j 43 2B + 75 4B K 107 6B k 44 2C , 76 4C L 108 6C l 45 2D - 77 4D M 109 6D m 46 2E . 78 4E N 110 6E n 47 2F / 79 4F O 111 6F o 48 30 0 80 50 P 112 70 p 49 31 1 81 51 Q 113 71 q 50 32 2 82 52 R 114 72 r 51 33 3 83 53 S 115 73 s 52 34 4 84 54 T 116 74 t 53 35 5 85 55 U 117 75 u 54 36 6 86 56 V 118 76 v 55 37 7 87 57 W 119 77 w 56 38 8 88 58 X 120 78 x 57 39 9 89 59 Y 121 79 y 58 3A : 90 5A Z 122 7A z 59 3B ; 91 5B [ 123 7B { 60 3C < 92 5C 124 7C | 61 3D = 93 5D ] 125 7D } 62 3E > 94 5E ^ 126 7E ~ 63 3F ? 95 5F _ 127 7F DEL A.3 EXTENDED CHARACTER SET The Extended ASCII Character Set (table A.3) also consists of 128 decimal numbers and ranges from 128 through 255 (using the full 8-bits of the byte) representing additional special, mathematical, graphic, and foreign characters.
- 549. 534 Table A.3: ASCII table DEC HEX CHAR- ACTER DEC HEX CHAR- ACTER DEC HEX CHAR- ACTER 128 0x80 € 171 0xAB « 214 0xD6 Ö 129 0x81 172 0xAC ¬ 215 0xD7 × 130 0x82 ‚ 173 0xAD 216 0xD8 Ø 131 0x83 ƒ 174 0xAE ® 217 0xD9 Ù 132 0x84 „ 175 0xAF ¯ 218 0xDA Ú 133 0x85 … 176 0xB0 ° 219 0xDB Û 134 0x86 † 177 0xB1 ± 220 0xDC Ü 135 0x87 ‡ 178 0xB2 ² 221 0xDD Ý 136 0x88 ˆ 179 0xB3 ³ 222 0xDE Þ 137 0x89 ‰ 180 0xB4 ´ 223 0xDF ß 138 0x8A Š 181 0xB5 µ 224 0xE0 à 139 0x8B ‹ 182 0xB6 ¶ 225 0xE1 á 140 0x8C Œ 183 0xB7 · 226 0xE2 â 141 0x8D 184 0xB8 ¸ 227 0xE3 ã 142 0x8E Ž 185 0xB9 ¹ 228 0xE4 ä 143 0x8F 186 0xBA º 229 0xE5 å 144 0x90 187 0xBB » 230 0xE6 æ 145 0x91 ‘ 188 0xBC ¼ 231 0xE7 ç 146 0x92 ’ 189 0xBD ½ 232 0xE8 è 147 0x93 “ 190 0xBE ¾ 233 0xE9 é 148 0x94 ” 191 0xBF ¿ 234 0xEA ê 149 0x95 • 192 0xC0 À 235 0xEB ë 150 0x96 – 193 0xC1 Á 236 0xEC ì 151 0x97 — 194 0xC2  237 0xED í 152 0x98 ˜ 195 0xC3 à 238 0xEE î 153 0x99 ™ 196 0xC4 Ä 239 0xEF ï 154 0x9A š 197 0xC5 Å 240 0xF0 ð 155 0x9B › 198 0xC6 Æ 241 0xF1 ñ 156 0x9C œ 199 0xC7 Ç 242 0xF2 ò 157 0x9D 200 0xC8 È 243 0xF3 ó 158 0x9E ž 201 0xC9 É 244 0xF4 ô 159 0x9F Ÿ 202 0xCA Ê 245 0xF5 õ 160 0xA0 203 0xCB Ë 246 0xF6 ö 161 0xA1 ¡ 204 0xCC Ì 247 0xF7 ÷ 162 0xA2 ¢ 205 0xCD Í 248 0xF8 ø 163 0xA3 £ 206 0xCE Î 249 0xF9 ù 164 0xA4 ¤ 207 0xCF Ï 250 0xFA ú 165 0xA5 ¥ 208 0xD0 Ð 251 0xFB û 166 0xA6 ¦ 209 0xD1 Ñ 252 0xFC ü 167 0xA7 § 210 0xD2 Ò 253 0xFD ý 168 0xA8 ¨ 211 0xD3 Ó 254 0xFE þ 169 0xA9 © 212 0xD4 Ô 255 0xFF ÿ 170 0xAA ª 213 0xD5 Õ
- 550. APPENDIX B Standard Template Library (STL) B.1 Introduction The Standard Template Library (STL) is a general-purpose C++ library of algorithms and data structures, container classes, and iterators. The STL has derived its name “template library as it is implemented by deploying template mechanism of C++. The STL is a generic library, meaning that its components are heavily parameterized. Almost every component in the STL is a template. You must know how templates work in C++ as some aspects of the library are complex; however, it is straightforward to use most of its features. STL use facilitates reuse of the sophisticated data structures and algorithms it contains. Although not part of the STL, the string class provides a commonly needed facility (character string handling) and is also part of the ANSI C++ standard library. The STL documentation classifies components in two ways. 1. Categories are a classification by functionality. The categories are: o Container o Iterator o Algorithm o Function Object o Utility o Adaptor o Allocator 2. Component types are a structural classification: one based on what kind of C++ entity (if any) a component is. The component types are: o Type (i.e. a struct or class) o Function o Concept B.2 Containers and Algorithms The main purpose of container classes in STL is to contain other objects. The STL includes classes, viz., vector, list, deque, set, multiset, map, multimap, hash_set,
- 551. 536 536 hash_multiset, hash map, and hash_multimap. Each class in STL is a template, and can be instantiated to contain any type of object. For example, you can use a vector<int> in the same way as you would use an ordinary C array, except that if you use vector you don’t need to manage dynamic memory allocation by hand. vector <int> v(3); // Declare a vector of 3 elements. v[0] = 1; v[1] = v[0] + 2; v[2] = v[0] + v[1]; // v[0] == 1, v[1] == 3, v[2] == 4 The data stored in containers is manipulated through a large collection of algorithms included in STL, e.g. to reverse the order of elements in a vector, reverse algorithm described below can be used: reverse(v.begin(), v.end()); //result will now be that v[0] == 4, v[1] == 3, v[2] == 1 You need to observe two important points in reverse algorithm: 1) it is a global function, not a member function and 2) it takes two arguments which define a range of elements. Range can be a few selected elements or the entire container v. STL algorithms are decoupled from the STL container classes. This means that reverse can be used not only to reverse elements in vectors, but also to reverse elements in lists, and even elements in arrays defined in the C++. The following program shows how reverse algorithm works equally well on a array: double A[4] = { 1.0, 2.4, 3.6, 1.5}; reverse(A, A + 6); //range is denoted [A, A + 6); for (int i = 0; i < 4; ++i) cout << "A[" << i << "] = " << A[i]; B.3 Iterators Iterators are the mechanism that makes it possible to decouple algorithms from containers. Algorithms are templates, and are parameterized by the type of iterator, so they are not restricted to a single type of container. Let us revisit the reverse algorithm in section D 1.2. You may note that arguments to reverse are iterators, which are a generalization of pointers. Pointers themselves are iterators, which is why it is possible to reverse the elements of an array using the same reverse algorithm. In the example above, the type returned by v.begin() and v.end() is vector<int>::iterator. There are also some iterators, such as istream_iterator and ostream_iterator, that aren't associated with containers. Consider, for example, how to write an algorithm that performs linear search through a range (see Table B.1).
- 552. 537Appendix B Table B.1: Algorithm for linear search STL Find Algorithm Equivalent Function in C++ Comments template <class InputIterator, class T> type int, InputIterator find(InputIterator first, InputIterator last, const T& value) { int* find(int* first, int* last, const int& value) { Find has three arguments: two iterators that define a range, and a value to search for in that range. while (first != last && *first != value) ++first; while (first != last && *first != value) ++first; return first; } return first; } Variables ‘First’ and ‘last’ are of type InputIterator which is a template parameter. As you know, there isn't actually any type called InputIterator: when you call find, the compiler substitutes the actual type of the arguments for the formal type parameters InputIterator and T. See C++ function in Table B. 1.1. Algorithms in the STL, in many cases, come in two versions, a plain and one that uses a predicate. Let’s take an example of the algorithm find. In its plain version, it looks for a given element in a container: it = find(first, last, elem); Here, first and last are iterators defining a sequence, possibly stored in a container, and elem is an object of the same type as the elements of the sequence. The algorithm returns an iterator to the first position in the sequence between first and last — including first but not last — where the element of the sequence equals elem. If no such element exists, the function returns last. If the elements of the sequence are complex structures, such as a customer-information structure as given below: typedef struct _customer_info_tag { std::string str_name; long l_id; double order_value; } CUSTOMER_INFO, *LPCUSTOMER_INFO ;
- 553. 538 538 It is obvious that you will ever be searching for a CUSTOMER_INFO element as you know it beforehand. Most probably, you will be searching for an element whose str_name member or whose l_id member you know. This operation with the plain algorithm find is not possible and you will need to use the related algorithm find_if. To this end, you write a suitable predicate in the form of a "functional" — a simple class that may be applied to a CUSTOMER_INFO object (via its defined operator()) and that returns true if the str_name member equals a specified name. Figure B.1 shows how to write such a predicate. Now you can search the container for an element whose str_name member equals a given customer name: it = find_if(first, last, pred("Ankit")); Figure B.1: Tests a CUSTOMER_INFO object for a match to a given name #include<functional> class name_pred : public std::unary_function<CUSTOMER_INFO, bool> { pubic: name_pred(string str_name) : m_str_name(str_name){} bool operator() (CUSTOMER_INFO arg) const { return m_str_name == arg.m_str_name; } private: string m_str_name; }; /* End of File */ B.4 Concepts and Modeling Concepts are, in fact, not a part of the C++ language. Nevertheless, concepts are an extremely important part of the STL. Using concepts, it is possible to write programs that separate interface from implementation: for example, if you want to implement find function then you only have to consider the interface specified by the concept InputIterator. Similarly, if you want to use find, you need only to ensure that the arguments you pass to it are compatible to the models of InputIterator. This is the reason why find and reverse can be used with lists, vectors, C arrays, and many other types: programming in terms of concepts, rather than in terms of specific types, makes it possible to reuse software components and to combine components together. It is interesting to note that find implicitly defines a set of requirements on types, and that it may be instantiated with any type that satisfies those requirements. Whatever type is substituted for InputIterator must provide operations, such as, 1) compare two objects of that type for equality; 2) increment an object of that type and 3) dereference an object of that type to obtain the object that it points to, and so on.
- 554. 539Appendix B Each STL algorithm has a set of requirements like find algorithm, e.g., the arguments to for_each and count, and other algorithms, must satisfy the same requirements. These requirements are sufficiently important that we give them a name: we call such a set of type requirements a concept, and we call this particular concept Input Iterator. We say that a type conforms to a concept, or that it is a model of a concept, if it satisfies all of those requirements. We say that int* is a model of Input Iterator because int* provides all of the operations that are specified by the InputIterator requirements. B.5 Refinement An InputIterator needs to perform not all but only a subset of pointer arithmetic. For example, to implement find algorithm it will suffice if an Input Iterator supports increment operation using prefix and postfix operator ++). In case of Reverse algorithm, you need both decrement and increment operation for its arguments as it uses the prefix operator in the expression ‘- -last’. In terms of concepts, reverse algorithm's arguments must be models of BidirectionalIterator rather than InputIterator. Therefore, the types that are models of BidirectionalIterator are a subset of the types that are models of InputIterator. Int*, e.g., is both a model of BidirectionalIterator and a model of InputIterator, but istream_iterator, is only a model of InputIterator as it does not conform to the more stringent BidirectionalIterator requirements. One may, therefore, interpret that BidirectionalIterator is a refinement of InputIterator. Refinement of concepts is in some way identical to inheritance of C++ classes; but a different word is used instead of calling it "inheritance", is to emphasize that refinement applies to concepts rather than to actual types. There are five iterator concepts, viz., Output Iterator, Input Iterator, Forward Iterator, Bidirectional Iterator, and Random Access Iterator. Forward Iterator is a refinement of Input Iterator, Bidirectional Iterator is a refinement of Forward Iterator, and Random Access Iterator is a refinement of Bidirectional Iterator. Container classes, like iterators, are organized into a hierarchy of concepts. All containers are models of the concept Container; more refined concepts, such as Sequence and Associative Container, describe specific types of containers. B.6 Other parts of the STL The STL does, however, include several other types of components: 1. utilities: very basic concepts and functions that are used in many different parts of the library; 2. some low-level mechanisms for allocating and deallocating memory, and
- 555. 540 540 3. Finally, the STL includes a large collection of function objects, also known as functors. Just as iterators are a generalization of pointers, function objects are a generalization of functions. There are several different concepts relating to function objects, including Unary Function say, F(x) and Binary Function requiring two arguments, i.e. f(x, y). Function objects are an important part of generic programming because they allow abstraction not only over the types of objects, but also over the operations that are being performed. TableB.2: Describes STL Categories and Classes Categories Sub-categories Classes Containers Concepts Container Forward, Reversible and Random Access Container Sequences Front Insertion and Back Insertion Associative Containers Simple, Pair, Sorted, Hashed, Unique, Multiple, Unique Sorted, Multiple Sorted, Unique Hashed, Multiple Hashed Associative Container and Hash Function Container classes Sequences vector, deque, list, slist and bit_vector Associative Containers set, map, multiset, multimap, hash_set, hash_map, hash_multiset, hash_multimap and hash String package Character Traits, char_traits and basic_string Rope Container adaptors stack, queue and priority_queue Bitset Iterators Concepts Trivial, Input, Output, Forward, Bidirectional and Random
- 556. 541Appendix B Access Iterator Iterator Tags iterator_traits, iterator_category, distance_type, value_type Iterator tag classes Input, Output, Forward, Bidirectional and random access Iterator base classes Input, output, forward, bidirectional and random access iterator Iterator functions Distance and advance Iterator classes istream_iterator ostream_iterator front_insert_iterator back_insert_iterator insert_iterator reverse_iterator reverse_bidirectional_iterator raw_storage_iterator sequence_buffer Algorithm Non-mutating algorithms for_each, find, find_if , adjacent_find, find_first_of, count, count_if, mismatch, equal, search, search_n and find_end Mutating algorithms Copy Swap transform Replace fill, fill_n, generate, generate_n, Remove unique unique_copy reverse reverse_copy rotate rotate_copy random_shuffle random_sample Copy, copy_n, copy_backward Swap, iter_swap, swap_ranges, replace, replace_if, eplace_copy, and replace_copy_if, remove, remove_if, remove_copy, and remove_copy_if
- 557. 542 542 random_sample_n partition stable_partition Sorting sort, stable_sort, partial_sort, partial_sort_copy, is_sorted and nth_element Binary search lower_bound, upper_bound, equal_range and binary_search merge inplace_merge Set operations on sorted ranges includes, set_union, set_intersection, set_difference and symmetric_difference Heap operations push_heap, pop_heap, make_heap, sort_heap and is_heap Minimum and maximum min, max, min_element and max_element lexicographical_compare lexicographical_compare_3way, next_permutation, and prev_permutation Generalized numeric algorithms iota, accumulate, inner_product, partial_sum, adjacent_difference and power Function Objects Concepts Concepts Generator Unary Function Binary Function Adaptable Generator Adaptable Unary Function Adaptable Binary Function Concepts - Predicates Predicate, Binary, Adaptable, Adaptable Binary, and StrictWeakOrdering
- 558. 543Appendix B Concepts - Monoid Operation Concepts - Random Number Generator Predefined function objects Arithmetic operations plus minus multiplies (formerly called "times") divides modulus negate Comparisons equal_to, not_equal_to, less greater less_equal greater_equal Logical operations logical_and logical_or logical_not Generalized identity operations identity project1st project2nd select1st select2nd subtractive_rng Function object adaptors binder1st, binder2nd, ptr_fun, pointer_to_unary_function, pointer_to_binary_function, unary_negate, binary_negate, unary_compose, and binary_compose Member function adaptors mem_fun mem_fun_ref mem_fun1 mem_fun1_ref Utilities Concepts Assignable, Default Constructible, Equality Comparable and LessThan
- 559. 544 544 Comparable and LessThan Comparable Functions Relational Operators Classes pair Memory Allocation Classes Allocators and raw_storage_iterator Functions Construct, destroy, uninitialized_copy, uninitialized_copy_n, uninitialized_fill, uninitialized_fill_n, temporary_buffer, get_temporary_buffer, and return_temporary_buffer B.7 Links to other resources For more information on the STL, try the following links: • Reference and tutorial information o Title: C++ Programmer's Guide to the Standard Template Library Author: Mark Nelson Publisher: IDG Books Worldwide, 1995. Pages: 875, ISBN: 1-56884-314-3 o Title: STL Tutorial and Reference Guide; C++ Programming With the Standard Template Library Authors: David R. Musser and Atul Saini Publisher: Addison-Wesley Publishing Company, Inc., 1996. Pages: 400, ISBN: 0-201-63398-1
- 560. APPENDIX C Setting Up the C++ Development Environment C.1 Introduction To get started with setting up the C++ programming environment one needs to download a C/C++ compiler. A compiler is a computer program that translates a computer program written in a computer language (called the source language) into an equivalent program written in another computer language (called the output or the target language). Source files written in C++ have ‘.cpp’ extension. When these files are compiled they turn into object files with ‘.obj’ extension which when linked with necessary library routines and header files become executable files (‘.exe’ extension). It is these executable files, we access when we are running our programs. There are various compilers present today in the programming world. In this appendix, we will take up the most commonly used compilers which are efficient for both beginners and sophomores. C.2 Borland’s C++ Builder Borland C++ builder is built on Borland’s proven C++ compiler technology and shares Delphi’s award winning Visual component Library giving you great interoperability with both Borland C++ and Delphi. Code generated on C++ builder is a visual front end to Borland C++ and code written with Borland C++ can be ported to other platforms. Borland has recently released their C++ Compiler (version 5.5) as a freeware. The freeware includes the command line compiler, incremental linker, resource compiler and binder, C++ Win32 preprocessor, import library utility, librarian, obj/lib/exe dump utility, etc. The C/C++ runtime library and ANSI/ISO Standard Template Library (STL) are also included with this compiler. The CD provided with the book includes this compiler consists of all the instructions and help - you might need to get started. The compiler can also be downloaded by accessing the link given below. http://www.borland.com/downloads/download_cbuilder.html On accessing the link the following page should appear. Access the compiler link and traverse the corresponding links to download the compiler.
- 561. 546 546 Figure C.2: Borland C++ compiler 5.5 With the Borland C++ builder one also requires a programming notepad. This can be easily found on the net. The CD also provides with a few downloading sties for these notepads on the internet. C.3 Microsoft Visual C++ Microsoft Visual C++ has for long been an industry standard for C++ development on Windows and its applications. Earlier versions of Microsoft Visual C++ technology preview enabled us to start on developing for internet explorer by accessing the power and flexibility of Dynamic HTML and the new Common Controls. The Standard Editions (Visual C++ 6.0) present today are a complete integrated development environment including debugging support with a look and feel that should be familiar to users of other Microsoft products. Microsoft Visual C++ Toolkit 2003 launched earlier included the command line versions of the optimizing C++ compiler, linker, and static libraries (including the Standard C++ Library and STL) that ships with the commercial Visual Studio .NET Professional 2003 for about a hundred dollars. Recently Microsoft Visual C/C++ command line compiler, along with C#, VB.NET and JScript.NET, appears to be available from Microsoft for download for free. Accompanying the compiler one also needs to download the Microsoft Platform SDK which contains the Windows headers and libraries for the compilers. The latest version of Visual C++ compiler is the ‘Visual C++ Express Edition Beta 2’ launched in 2005. The edition is accompanied with professional-grade compiler, code editor, debugger and a project system. The system requirements necessary for the express edition are enlisted in
- 562. 547 547 Appendix C ‘http://lab.msdn.micros oft.com/express/readme/’. This edition can be downloaded using the following link. http://lab.msdn.microsoft.com/express/visualc/default.aspx The following page should appear on accessing the above mentioned link. When the link is accessed it leads to downloading a file ‘vcsetup.exe’ (2.66 Mb) which acts as a setup for the entire installation process.To conclude Visual C++ Standard Edition is an excellent choice for beginners looking to learn C++ on Windows systems as it contains all one needs to begin to compile and test programs.
- 563. 548 548 Notes
- 565. 550 Graphics can produce graphics on display as well as printer or any other Windows supported device. Therefore, it becomes device dependent. Memory management Memory blocks are dynamic, and can pre- load or load on call. They can be fixed, moveable, discardable, or non-discardable. Memory blocks are static. The integrated Windows environment provided by C and Pascal compilers from Microsoft, Borland & many other vendors enables a programmer to carry out individual steps in compilation and linking (see figure D.1) automatically through the use of development tools. Windows programs have separate resource file that contain static data such as menu and bitmaps. Resourcedata is supplied by a separate compiler called resource compiler. Figure D.1: Steps in compilation and linking Section D.3 in this chapter describes graphics programming under DOS. This section will be useful to those users who still prefer to operate under DOS.
- 566. 551Appendix D D.2 Windows Graphics Programming Functions from the Windows Graphics Device Interface (GDI) are used for displaying graphics in an application's window. These can be accessed through the TGdiBase class hierarchy in OWL in Borland C. We briefly review some of the important concepts applicable to programming in Windows. Many uppercase identifiers are used. Several of these identifiers contain a two-letter or three-letter prefix followed by an underscore, e.g., WM_PAINT, DT_SINGLELINE, CS_VREDRAW, IDC_ARROW etc. These are simply numeric constants and indicate a general category to which the constant belongs, as indicated in table D.2. Handles are used quite frequently in Windows. A handle is a 16-bit number that refers to an object. The handles for icon, cursor, and brush are called HICON, HCURSOR, and HBRUSH, respectively. The actual value of the handle is unimportant to a program, since the Windows module that gives the program the handle knows how to use it to reference the object. One also encounters uppercase identifiers for various types of handles, such as HANDLE, HWND, and HDC for general handle, handle to a window and handle to a device context, respectively. Table D.2: Prefix and category We will use the ObjectWindows application framework for graphics programming. ObjectWindows supports the following classes of Window: Windows TWindow is the base class for all windows. It represents the functionality of all windows e.g. dialog boxes, controls, Multiple Document Interface (MDI) windows and so on. Prefix Category CS Class style IDI ID for an icon IDC ID for a cursor WS Window style CW Create Window WM Window message DT Draw text
- 567. 552 Frame windows TFrameWindow is derived from Window class. Frame windows manage application- wide tasks,like menus and tool bar,and support a client window which can be specialized to do a single task. Changes (i.e. adding tool or status bars) introduced in a frame window do not affect client windows. MDI windows MDI manages multiple documents or windows in a single application. Decorated windows Several classes, such as TLayoutWindow and TLayoutMetrics provide support for decoration controls like tool bars, status bars, and message bars. Dialog Box Classes TDialog is a derived class of TWindow. It is used to create dialog boxes that handle a variety of user interactions, such as choosing files, fonts, colors, and printing options as well as searching and replacing text, inputting text, and aborting printing jobs. Control Classes TControl is a class derived from TWindow for supporting behaviour common to all controls. ObjectWindows supports standard controls from Windows, such as widgets, gadgets band decorations. Graphics Classes Windows supports a graphics library called Graphics Device Interface (GDI). ObjectWindows encapsulates GDI to make it easier to use device context (DC) classes (TDC) and GDI objects (TGDIObject). Printing classes TPrinter and TPrintout encapsulate the communications with printer devices and the task of printing a document respectively. Module and Application Classes TApplication is responsible for initializing the windows and directing messages received from Windows to the proper window. Application objects are used for deriving an appli- cation object from the TApplication class, creating an application object, overriding base class function to customize the application behavior, using Borland custom control and Microsoft control 3D libraries. To create and use TApplication object one must include the correct header files create an application object, and call the application’s call function. The following example shows a small program representing an application of ObjectWindows.
- 568. 553Appendix D Program D.1 // p3_1.cpp - An application creating a main window with the caption “trial”. It can //resize, move, minmize, maximize, and close the generated window. #include <owlapplicat.h> // This header file has definition of Tapplication which is //derived from TModule and includes owlmodule.h. int OwlMain(int /*argc*/, char* /*argv*/ []) { return TApplication("trial").Run(); } It is necessary to derive a new class from TApplication to introduce more functionality in the application. Initializing an ObjectWindows application requires the following steps: i. Constructing the application object Member functions are InitApplication, InitInstance, and InitMainWindow. One can override any of these members to customize initialization of one's application. The constructor for the TrialApp class derived from TApplication shown in the following example takes application name as only argument. The default value is zero indicating no name. The application name is referenced by a char far* member of the TModule base class called Name. Program D.2 // p3_2.cpp - Creates and shows a window entitled, “ Trial”. #include <owlapplicat.h> class TrialApp: public Tapplication { public: TrialApp(const char far* name = 0): TApplication(name) {} }; int OwlMain(int /*argc*/, char* /*argv*/ []) { return TrialApp("trial").Run(); }
- 569. 554 ii. Initializing the application There are three initialization functions - InitApplication initializes the first instance of the application; InitInstance initializes the each instance of the application; and InitMainWindow initializes the application’s main window. iii. Initializing each new instance One can run multiple instances (copies) of an application simultaneously. iv. Initializing the main window Tapplication::InitMainWindow creates a frame window with the same name as the application object. This window is not very useful as it can’t receive or process any user input. It is therefore essential to override InitMainWindow to create a window object that does process user input. The following example creates a TFrameWindow object and makes it the main window.Program D.3: Creates and shows a maximized window entitled, “First MainWindow” #include <owlapplicat.h> #include <owlframewin.h> class TrialApp: public TApplication { public: TrialApp(char far *name): TApplication() {} void InitMainWindow(); }; // override InitMainWindow void TrialApp::InitMainWindow() { /* SetMainWindow takes a pointer to the TframeWindow object and returns a pointer to the old main window. The return value is zero for a new application that has not set a main window as yet. */ SetMainWindow(new TFrameWindow(0, "First Main Window")); /* value of nCmdShow controls the parameters of the window, e.g. when set to SW_SHOWMAXIMIZED will show a maximized window whenever you run this application. */ nCmdShow = SW_SHOWMAXIMIZED; }
- 570. 555Appendix D int OwlMain(int /*argc*/, char* /*argv*/ []) { return TrialApp("trial").Run(); /* The application name given in the TApplication constructor is used only if you do not provide a main window. As this program has used SetMainWindow function -- the caption bar is titled "First Main Window" and not "trial". */ } Doc/View Classes In Doc/View model data is contained in and accessed through a document object and displayed and manipulated through a view object. It comprises of TDocManager, TDocument and TView classes. Miscellaneous Classes ObjectWindows miscellaneous classes include menus and clipboard classes. Once an application is initialized, the application object’s MessageLoop starts running. All incoming messages from Windows are processed by the MessageLoop. One can override the default message handling functions and introduce additional message processing capability in the application. TAppliaction includes functions, such as MessageLoop, IdleAction, PreProcessMenu, and ProcessAppMsg which provide message handling functionality for any ObjectWindows application. TAppliaction also includes functions that enable one to load the Borland Custom Control Library (BWCC.DLL for 16- and 32- bit applications) and Microsoft 3-D Control Library (CTL3DV2.DLL and CTL3D32.DLL for16- and 32- bit applications, respectively. Function TApplication::EnableBWCC is used to open and close the Borland Custom Control Library. When one passes True to EnableBWCC, the function loads the DLL if it is not already loaded. Similarly, if one passes False to EnableBWCC it unloads the DLL. EnableCTL3D function works in the same way as EnableBWCC for loading and unloading Microsoft 3-D Control Library. Response tables are used to handle all events in an ObjectWindows application. These tables are used in the following way: Define the response table ObjectWindows provides DEFINE_RESPONSE_TABLEX macro to help a user to define response tables. DEFINE_RESPONSE_TABLE1( TrialFrame, TframeWindow) EV_WM_LBUTTONDOWN, EV_WM_LBUTTONUP, EV_WM_MOUSEMOVE, EV_WM_RBUTTONDOWN, END_RESPONSE_TABLE;
- 571. 556 Define the response table entries When a window or control receives a message, it checks the response table to see if there is entry for that message. If an entry is found, it is passed to the corresponding function else the message is passed to its parent. If parent is the main window, message is passed to the application object. If application object also does not have a response entry , the message is processed by the ObjectWindows default processing mechanism. Declare and define the response member functions ObjectWindows provides a large number of macros, called command message macros, that let a user assign command messages to any function. The different types of command message macros and the corresponding function are given Table D.D. Table D.3: Command Message Macros Macro Description EV_COMMAND (CMD, UserName) Calls function UserName when the CMD message is received. EV_COMMAND_AND-ID ( CMD, UserName) Calls function UserName when the CMD message is received. Passes the command’s ID(WPARAM parameter) to UserName. EV_COMMAND_ENABLE( CMD, UserName) Used to automatically enable and disable command controls such as menu items, tool bar buttons and so on. EV_MESSAGE(MSG, UserName) Calls UserName when the user- defined message MSG is received, MSG is passed to UserName without modification. EV_REGISTERED(MSG, UserName) Calls UserName when the registered message MSG is received. MSG is passed to UserName without modifications. Program D.4 It illustrates use of a response table to capture events such as pressing of the left and right buttons of the mouse. A message box with the caption “Message Sent” and message “Left mouse button pressed” appears when the left button of the mouse is pressed. Similar message will appear when the right button of the mouse is pressed. The program also shows how to override the CanClose member function of TWindow to ensure that everything has been cleaned up before saving. /*-------------------------------------------------------------------------------------------------------------- Program D.4 --------------------------------------------------------------------------------------------------------------*/
- 573. 558 and message "Left mouse button pressed" appears. Attribute MB_OK will put an "OK" button.*/ MessageBox(" Left mouse button pressed", "Message Sent", MB_OK); } void TrialWindow::EvRButtonDown(uint, TPoint&) { MessageBox(" Right mouse button pressed", "Message Dispatched", MB_OK); } class TrialApp : public TApplication { public: TrialApp() : TApplication() {} void InitMainWindow() { SetMainWindow(new TFrameWindow(0, " Trial program", new TrialWindow)); } }; int OwlMain(int /*argc*/, char* /*argv*/ []) { return TrialApp().Run(); } D.2.1 Graphics Device Interface It refers to the graphics output function of Windows and isolates a Windows program from the physical output device such as display or printer. In addition to producing output on the physical devices it also supports output on two pseudo devices bitmaps and metafiles. Bitmaps refer to a rectangular array of pixels and metafiles. Most of the PC output devices like video display adapters, dot-matrix printers and laser printers are raster devices and represent images as a pattern of dots, i.e., bitmaps. The important GDI functions can be grouped in following categories: Vector drawing functions to draw graphical objects such as lines, circles and rectangles; bitmap manipulation functions. Text output functions Palette management functions to display colors supported by a display adapter. Palette management in systems with super VGA & other enhanced graphics adapters, supporting more than 16 colors, is useful.
- 574. 559Appendix D D.2.2 Device Context (DC) DC can be considered as a generalized model of a graphics output device and is key to the GDI's support for device independent graphics in Windows. To perforn any sort of graphical operation in Windows, one needs a DC for the window or the area one wants to work with. DC, in fact, is a data structure which includes drawing objects , such as brush, pen and bitmap and graphics attributes, such as background color, text color, fill style, and fonts which control the appearance of the graphics and text. To generate graphics output using DC, one must call an appropriate GDI graphics function to access the DC, use this DC to draw and then release the DC. Windows permits only upto 5 DCs to be open at any given time for the entire Windows system. The handle to the DC (HDC) is a protected member variable of the TDC class. TDC is the root class for encapasulation of ObjectWindows GDI device contexts and is used for drawing in a variety of devices and screen areas, including the TClientDC for drawing in the client area of a window, TmemoryDC for drawing on to an in-memory bitmap and TPrintDC for output to a printer. All GDI functions are member functions of the TDC class. The names of the member functions are the same as the GDI functions but the member functions do not need HDC as an argument. Table D.2 gives an example below of the call pattern of functions to draw a line from (xleft, ytop) to (xright, ybottom) using GDI functions, OWL classes and TDC Class in OWL 2.0 with TPoint objects as argument, respectively. D.2.3 Contents of a DC A DC has the following graphics objects: A pen which controls the appearance of lines and borders of rectangles, ellipses, and polygons. Drawing mode specifies how to combine pen's color with the color that already exists on the drawing surface. A brush which defines a fill pattern. Font specifies the shape and size of textual output. Palette is an array of colors. Palette facilitates to pick the current selection of available colors out of the millions of colors that can be shown on the display. Bitmap is used to draw images. Region is defined by a combination of rectangles, ellipses and polygons used for drawing or for clipping. Table D.4: Call Pattern of Functions GDI Functions ObjectWindows Library (OWL) Classes TDC class with TPoint obj-ects (represents a point in the x-y plane) as arguments HDC hdc; // Assume that handle to DC is // properly set up TclientDC dc(HWindow); // Assume that TclientDC // is a member of the // TclientDC dc(HWindow); TPoint start, end;
- 575. 560 TWindow class // Assume that the //TPoint objects are //initialized MoveTo(hdc, xleft, ytop); LineTo(hdc, xright, ybottom); dc.MoveTo( xleft, ytop); dc.LineTo(xright,ybottom); dc.MoveTo( start); dc.LineTo( end); D.2.4 Acquiring a DC In order to draw on a graphics output device, it is necessory to first obtain a handle to a DC. Windows provides several methods for acquiring a DC handle. A DC can be acquired in response to the WM_PAINT message because this event causes redrawing of an application's window. UsingOWL classes and deriving application's main window from the TWindow class, one can generate all graphics output in a member function named Paint. The prototype of the Paint function is void Paint ( TDC& dc, // Device context for the display Boolerase, // TRUE = erase the background Trect&rect // Rectangle to be redrawn - defines area ); The Paint function is called by OWL whenever Windows sends a message to the window. To avoid waiting for the WM_PAINT message to update the window, one can either call the updateWindow function or create an instance of the TClientDC class to get a DC. When the TClientDC is destroyed, the DC is automatically released. Changes made the attributes from the handle to a DC are lost when the DC is released. One may also create a private DC for a window so that changes carried out in the attributes of the DC persist until the window is destroyed. The following example shows the procedure to create a persistent DC: A window class TrialWindow is derived from the OWL class TWindow override the GetWindowClass function add the CS_OWNDC flag to the class style as given below: // TrialWindow:public TWindow void TrialWindow :: GetWindowClass(WNDCLASS& wclass); { TWindow::GetWindowClass(wclass); wclass.style |= CS_OWNDC; } Now each window created using the window class will have its own private DC that exists until the window is destroyed. While using the CS_OWNDC style, the DC attributes need initialization
- 576. 561Appendix D only once and continue to be valid until they are changed. The CS_OWNDC style affects only the DC retrieved from GetDC and BeginPaint and not the DC obtained from other functions, such as GetWindowDC. One needs 800 bytes of storage for the DC for each window with the CS_OWNDC style. Thus, the window associated with each instance of TrialWindow class has its own private DC. One must still release the DC before exiting from the window function. SaveDC and RestoreDC functions let a user save the current DC, make temporary changes in the attributes and revert back to the original attributes. D.2.5 Graphics output using a DC A typical procedure to generate graphics output using DC is given below: Create the TDC object. For example, to draw in the client area of the window, one needs to create a TClientDC object. Define the graphics attributes. Invoke appropriate GDI drawing functions. If the new operator is used to create the TDC object, it must be destroyed by invoking the delete operator. DC can also be used to determine the capabilities of a device, such as number of color planes in a graphics display by calling the GetDeviceCaps function. Program D.5 This program makes it possible to create a freehand line drawing while keeping the left button of the mouse pressed. One can release the left button and move the mouse to a new position. Pressing the left button would let one draw freehand from the new position of the mouse. Pressing the right button of the mouse clears the entire client area of the window.
- 577. 562 /*------------------------------------------------------------------------------------------------------------ Program 5.5: Draws freehand line drawing */------------------------------------------------------------------------------------------------------------ #include <owl/owlpch.h> #include <owl/applicat.h> #include <owl/framewin.h> #include <owl/dc.h> /* Definition of the TrialWindow class -- data member CommonDC, a pointer to the TClientDC object, is not destroyed automatically when TrialWindow object is destroyed. It is necessory to add a destructor to TrialWindow to ensure proper cleanup.*/ class TrialWindow : public TWindow { public: TrialWindow(TWindow* parent = 0); ~ TrialWindow() // Delete CommonDC { delete CommonDC; } protected: // Provides a valid device context across functions EvLButtonDown, // EvMouseMove, and EvLButtonUp. TDC* CommonDC; // Override member function of TWindow bool CanClose(); // Message response functions void EvLButtonDown(uint, TPoint&); void EvRButtonDown(uint, TPoint&); void EvMouseMove(uint, TPoint&); void EvLButtonUp(uint, TPoint&); DECLARE_RESPONSE_TABLE( TrialWindow); }; DEFINE_RESPONSE_TABLE1( TrialWindow, TWindow) EV_WM_LBUTTONDOWN, EV_WM_RBUTTONDOWN, EV_WM_MOUSEMOVE, EV_WM_LBUTTONUP, END_RESPONSE_TABLE; TrialWindow:: TrialWindow(TWindow* parent) { Init(parent, 0, 0); // Initialize TrialWindow base class
- 578. 563Appendix D CommonDC = 0; //Initialize CommonDC-- a pointer to the TClientDC--to zero. } bool TrialWindow::CanClose() { return MessageBox("Do you want to save?", " Not Implemented", MB_YESNO | MB_ICONQUESTION) == IDNO; } void TrialWindow::EvLButtonDown(uint, TPoint& point) { if (!CommonDC) { SetCapture(); CommonDC = new TClientDC(*this); CommonDC->MoveTo(point); } } void TrialWindow::EvRButtonDown(uint, TPoint&) { Invalidate(); // clears the entire client area of the window } void TrialWindow::EvMouseMove(uint, TPoint& point) { if (CommonDC) CommonDC->LineTo(point); } void TrialWindow::EvLButtonUp(uint, TPoint&) { if ( CommonDC) { ReleaseCapture(); delete CommonDC; CommonDC = 0; } } class TDrawApp : public Tapplication { public: TDrawApp() : TApplication() {} void InitMainWindow() { SetMainWindow(new TFrameWindow(0, "Freehand Draw", new TrialWindow)); } };
- 579. 564 int OwlMain(int /*argc*/, char* /*argv*/ []) { return TDrawApp().Run(); } D.2.6 GDI Coordinate System The GDI supports two coordinate systems referred to as physical(or device) and logicalcoordinates. The physical coordinate system is fixed for a device, e.g., for a window on the screen the origin is at the upper left corner of a window's client area, the positive x-axis extends to the right, and the positve y-axis extends downwards. In all device coordinate systems, units are in erms of pixels. Windows supports several logical coordinates systems and maps each one of them to the physical systems before displaying any graphics output. Figure D.2: Mapping of logical coordinates to physical coordinate system Windows supports eight mapping mode identifiers specifying logical units as 0.001 inch, 0.01 inch, 0.01 millimeter, or 0.1 millimeter and the positive y-axis extending downward or upwards. One
- 580. 565Appendix D can use the SetMapMode function to set a mapping mode, such as MM_ANISOTROPIC so that logical units along x- and y- axes can be set independently, or MM_TEXT in which logical coordinate is the same as the physical one. The GetMapMode function is available to determine the current mapping mode. The mapping modes are listed in table D.5. Table D.5: Mapping Modes Mapping Mode MM_TEXT MM_LOMETRIC MM_HIMETRIC MM_LOENGLISH MM_HIENGLISH MM_TWIPS MM_ISOTROPIC MM_ANISOTROPIC Logical Unit Pixel 0.1 mm 0.01 mm 0.01 in. 0.001 in. 1/1440 in. Arbitrary(x=y) Arbitrary(x!=y) X-axis Right Right Right Right Right Right Selectable Selectable Y-axis Down Up Up Up Up Up Selectable Selectable One can convert client area coordinates to screen coordinates or vice versa using the functions ClientToScreen and ScreenToClient. The position and the size of the whole window in terms of screen coordinates can be obtained using the GetWindowRect function D.2.7 Drawing with GDI Functions The GDI contains functions to draw individual pixels, lines, rectangles, polygons and ellipses. A summary of some important functions supported by the GDI is presented in the following sections. D.2.7.1 Drawing Points There are two ways of depicting a point on the screen. The main difference is in the way color is specified. One can specify the color either with a 32 bit value whose least significant three bytes represent the red (R), green (G), and blue (B) components of a color or using the RGB macro. The values of the red, green and blue components are between 0 and 255. dc.SetPixel(x, y, color); // Paints the point at the logical coordinates (x, y) with the // specified color. dc.SetPixel(x, y, RGB(r, g, b)); // Color is specified using a RGB macro. Table D.6: Functions to Draw Closed Figures
- 581. 566 Function Figure Rectangle(int xleft, int ytop, int xright, int ybottom); Draws a rectangle whose upper left corner is (xleft, ytop) and lower right corner is (xright, ybottom). RoundRect(int xleft, int ytop, int xright, int ybottom, int x_ellipse, int y_ellipse); Draws a rectangle with rounded corner. Each corner is rounded by a small ellipse whose width and height are x_ellipse and y_ellipse. Ellipse(int xleft, int ytop, int xright, int ybottom); Draws an ellipse bounded by rectangle whose opposite corners are given by two points. Pie(int xleft,int ytop,int xright, int ybottom, int xbegin, int ybegin, int xend, int yend); Draws a pie shaped wedge. Bounding rectangle is defined by the point (xleft, ytop) & point (xright, ybottom). The two straight edges of the pie are defined by the line joining the center of the ellipse and the points (xbegin, ybegin), and (xend, yend). Chord(int xleft, int ytop, int xright, int ybottom, int xbegin, int ybegin, int xend, int yend); Draws a segment of ellipse similar to the Pie but unlike Piejoins the end points by a straight line. Polygon(TPoint* pt, int numpt); Draws a polygon by joining the points in the array pt by straight lines. The first and last points in the array pt are joined by a line segment to give a closed figure. GDI provides a number of function to manipulate rectangles. OWL provides the RECT structure of Windows API, which is defined in the header file <windows.h) as follows: typedef struct tagRECT { int left; // upper left corner of the rectangle int top; int right; // lower left corner of the rectangle int bottom; } RECT; The functions included in GDI that manipulate rectangles are CopyRect Copyrect(&dest_rect, &src_rect); copy one rectangle structure to another
- 582. 567Appendix D EqualRect EqualRect(&rect1, &rect2); returns TRUE if two rectangle structures are equal Fill Rect dc.FillRect(rect, hbrush); fills the rectangle with the specified brush FrameRect dc.FrameRect(rect, brush); uses the brush to draw a rectangular frame InvertRect dc.InvertRect(rect); inverts all the pixels in the rectangles OffsetRect OffsetRect(&rect, x, y); move a rectangle a number of units along the x- and y-axes InflateRect InflateRect(dx,dy); increase or decrease the size of a rectangle SetRect SetRect(&rect, left, right, top, bottom); sets the fields of a RECT structure IntersectRect IntersectRect(&destrect, &srcrect1, &srcrect2); Obtains the intersection of two rectangles, UnionRect Un ionRect(&result_rect, &rect1, &rect2); Sets the fields of a RECT structure to be union of two other rectangles SetRectEmptySetRectEmpty(&rect); Sets the fields of a RECT structure to empty PtInRect PtInRect(&rect, point); Returns TRUE if a point is in a rectangle D.2.7.5 Creating and Painting Regions A region is a description of an area of the display that is a combination of rectangles, polygons, and ellipses. Regions are GDI objects like pens brushes and bitmaps. Regions are also useful in clipping by selecting the region in device context. When one creates a region, Windows returns a handle to the region of type HRGN. Two ways of creating rectangular regions are given below: hRgn = CreateRectRgn (xleft, ytop, xright, ybottom); or hRgn = CreateRectRgnIndirect(&rect); One can use the TRegion class to define a region in the device context and perform a number of operations, such as painting, filling, inverting and so on. Table D.7 describes the GDI functions available for manipulating regions. Table D.7:GDI functions that manipulate regions FUNCTION DESCRIPTION CreateRectRgn (xleft, ytop, xright, ybottom); Creates a rectangular region using specified coordinates for the opposite corners of the rectangle. CreateRoundRectRgn Creates a rectangular region wih rounded corners - similar to RoundRect function. CreateEllipticRgn (xleft, ytop, xright, ybottom); Creates an elliptic region. CreateEllipticRgnIndirect (&
- 583. 568 rect) Creates an elliptic region using the fields in a RECT structure CreatePolygonRgn (points, npoints, fill_mode); Creates a polygon region. Parameter fill_mode is ALTERNATE to WINDING CreatePolyPolygonRgn Creates a region out of multiple polygons. CombineRgn(hresult_rgn, hrgn1, hrgn2, combine_flag); Combines two regions into one according to a specified combining mode. EqualRgn (hrgn1, hrgn2); Returns TRUE if two regions are equal. dc.FillRgn (rgn, brush); It belongs to the TDC class that fills a region with the specified brush. FrameRgn (rgn, brush, pt); //pt is a TPoint with width and height of the frame It belongs to the TDC class and draws a frame around a region with the specified brush. rgn.GetRgnBox (rect); It belongs to the TRegion class that returns the bounding box of the region. InvalidateRgn (hrgn, erase_flag); It belongs to the TWindow class that marks the specified region for repainting. dc.InvertRgn (rgn); A member function of the TDC class that inverts the pixels in a region. OffsetRgn (hrgn, x, y); Moves a region by a specified x-and y-offset dc.PaintRgn (rgn); Fills the region with the current brush: PtInRegion(hrgn, x, y); Returns TRUE if the specified point is in the given region. RectInRegion(hrgn, & rectangle); Returns TRUE if any part of the specified rectangle is in the region. rgn.SetRectRgn (rect); It belongs to the TRegion class that changes the region to a rectangular shape specified by the TRect argument. ValidateRgn (hrgn); A member function of the TWindow class that removes the region from the area to be repainted: D.2.8 Drawing Mode GDI consists of number of line modes which can be used for combining the pen color with the existing color and draw lines and filled figures. The mode name starts with R2_ prefix. The default mode is R2_COPYPEN which overwrites whatever exists on the drawing surface. The
- 584. 569Appendix D drawing mode is also referred to as raster operation (ROP). The following statement can be used to set a new drawing mode and to save the previous mode: int old_ROP = dc.setROP2 (R2_BLACK); Table D.8 lists the drawing modes supported in GDI. Table D.8: Drawing modes in Windows GDI Mode Name Boolean Operation Comments R2_BLACK all bits zero Draws in black irrespective of pen color and existing color R2_COPYPEN pen Draws with the pen color R2_MASKNOTPEN (NOT pen) AND dest Inverts bits in the pen color, performs a bitwise-AND with existing color R2_MASKPEN pen AND dest Performs bitwise-AND of pen color and existing color R2_MASKPENNOT pen AND (NOT dest) Inverts existing color and does bitwise- AND with pen color R2_MERGENOTPEN (NOT pen) OR dest Inverts pen color and does bitwise-OR with existing color R2_MERGEPEN pen OR dest Performs bitwise-OR of pen color and existing color R2_MERGEPENNOT pen OR (NOT dest) Inverts existing color and does bitwise-OR with pen color R2_NOP dest Leaves existing color unchanged (hence the name NOP for ``no operation”) R2_NOT NOT dest Inverts existing color R2_NOTCOPYPEN NOT pen Draws with inverted pen color R2_NOTMASKPEN NOT (pen AND dest) Performs bitwise-AND of pen and
- 585. 570 existing color and inverts the result R2_NOTMERGEPEN NOT (pen OR dest) Does bitwise-OR of pen and existing color and inverts the result R2_NOTXORPEN NOT (pen XOR dest) Performs bitwise exclusive-OR of pen and existing color and inverts the result R2_WHITE all bits 1 Draws in white color R2_XORPEN pen XOR dest Does bitwise exclusive-OR of pen color and existing color D.2.9 Colors, Palettes and Bitmaps D.2.9.1 Colors Windows API defines a 32- bit integer called COLORREF. The low order bytes contain the intensities of the red, green and blue colors and the most significant byte contains the type which indicates whether to interpret the value as a color or a palette entry. The RGB macro represents the RGB color, e.g. RGB(0, 0, 255) denotes a full intensity blue color.Each color is represented by 8-bits, i.e. by intensities between 0 to 255. TColor class in OWL 2.0 maintains color as COLORREF value. Windows uses dithering (described in chapter 2) only when filling areas to display neighbouring pixels in different colors to create unique shades. It cannot use dithering when drawing points lines and text. D.2.9.2 PalettesThe number of colors that can be displayed at a time is determined by the number of bits of storage allocated for each pixel. In standard VGA, there is 4-bits storage to represent
- 586. 571Appendix D each color pixel, therefore, this adapter can display only 24 , i.e., 16 simultaneous colors. A super VGA card can have memory to support upto 8-bits per pixel and hence can display 28 (or 256) simultaneous colors. Display adapter represents each color by the RGB intensities. Given below is the number of distinct colors supported and number of simultaneous colors shown by a display adapter: Adapter Bits/pixel Intensities Distinct Colors Bits/Pixel Simultaneous Components R G B available Supported Colours VGA 6 26 26 26 26 x 26 x 26 = 262,144 4 24 = 16 SVGA 8 28 28 28 28 x 28 x 28 = 16,777,216 8 28 = 256 The display adapter converts each pixel’s contents into a color by interpreting the pixel’s value as an index into a table called color palette. The entries in this table are the RGB values. Windows defines a system palette which has 16 predefined colors for EGA and VGA. System palette has 20 predefined colors for display adapters which support more than 256 colors. The concept of logical palette in Windows enables a user to take advantage of the large number of colors available in the system. Windows supports an extended system palette that is the mirror image of the hardware palette. Windows assigns 20 values corresponding to the default colors in the extended pallete. Windows maps each color in the logical palette to the extended system palette as follows: • If color in logical palette already exists in the system palette, that color is mapped to the matching color index in the system palette. • If color in logical palette does not exist in the system palette, it is added in the system palette if there is space. • In case the system palette is full, a color in the logical palette is mapped to the closest matching color in the system palette. Windows supports only 20 default colors in the system palette. One needs to use a logical palette to develop applications with all 256 colors in the Super VGA mode. Table D.9 gives a brief description of the Windows functions to manipulate palette.
- 587. 572 Table D.9: Functions to manipulate palettes Function Description pal.AnimatePalette (start_indx, count, pal_entries); Belongs to the TPalette class - changes entries in a logical palette resulting in instant changes to colors on the display. CreatePalette Creates a logical palette and returns a handle (that can be used to create a TPalette) to the palette pal.GetNearestPaletteIndex (color); Belongs to the TPalette class - returns the index of the palette entry that most closely matches a specified TColor pal.GetPaletteEntries (start_indx, count, pal_entries); Belongs to the TPalette class - retrieves the color values for a specified number of entries in a logical palette dc.GetSystemPaletteEntries (start, num, pal_entries); Belongs to the TDC class - retrieves the color values for a specified number of entries in the system palette GetSystemPaletteUse (); Belongs to TDC class - returns a flag indicating whether an application can change the system palette dc.RealizePalette (); Belongs to the TDC class - maps the entries of the currently selected logical palette into the system palette dc.ResizePalette(); Belongs to the TPalette class - enlarges or reduces the size of a logical palette after it has been created pal.SetPaletteEntries (start, count, pal_entries); Belong to the TPalette class - changes the color values of a specified number of entries in the logical palette SetSysColors Sets one or more colors in the system palette, identified by constants such as COLOR_ACTIVEBORDER, COLOR_MENU and COLOR_WINDOW dc.SetSystemPaletteUse (flag); Belongs to the TDC class - allows the active application to change the entries in the system palette and flag is one of SYSPAL_NOSTATIC or SYSPAL_STATIC dc.UpdateColors (); Belongs to the class - updates the color of the pixels in the client area of a window to reflect the current entries in the system palette In case an application uses a logical palette, it should handle the following three palette specific Windows messages: • Whenever the application’s window becomes active, Windows sends a
- 588. 573Appendix D WM_QUERYNEWPALETTE message. This is the time when one can load the desired set of colors using RealizePalette function which maps logical palette to the system palette. • Windows sends a WM_PALETTECHANGING message to the top level windows of all applications when the system palette is about to change. • WM_PALETTECHANGED message is sent to the top level windows of all applications when the system palette has changed. D.2.9.3 Bitmaps Bitmap is a digital representation of the picture. Each pixel in the picture represents one or more bits in the bitmap. Monochrome bitmaps require only one bit per pixel but color bitmaps require additional number of bits to indicate the color of each pixel. Device Independent Bitmap (DIB) format is called device indepedent because it consists of the color table. The color table gives the pixel values corresponding to the RGB color values. The DIB format is an extension of bitmap format of OS/2 presentation manager. The WINDOWS.H header file contains a few structures that support OS/2 bitmaps. The GDI bitmap objects are called device dependent bitmap (DDB) objects because they must be compatible with a specific graphics output device. The DIB is represented by the TDib class and has no GDI handle like a regular bitmap as it is not a GDI object. The DIB is, therefore, kept outside GDI either as files or in the memory. One can convert a DIB into a GDI object but in such case the device independent color information in the DIB is lost. The DIB format is useful for storing images in a file. One can create a DIB using Microsoft Image Editor, Paintbrush, or Borland Resource Workshop. The files, thus, created have extension .BMP or in some cases extension .DIB. The internal representation of data, i.e., number of bits per pixel is : i) 1 for monochrome bitmaps; ii) 4 for 16- color bitmaps; iii) 16 for 256- color bitmaps and iv) 24 for 16,777,216 - color bitmaps. Given below is a constructor for creating a TDib object: Tdib( int width, int height, int ncolors, uint 16 mode=DIB_RGB_COLORS); ‘width’ and ‘height’ parameters give width and the height of the DIB in pixels; ncolors & mode refer to number of colors and RGB or PAL palette respectively. Program D.6 /*------------------------------------------------------------------------------------------------------------- Program D.6: This program shows how to a) Create a menu and use of SetMainWindow and GetMainWindow functions to activate the menu "COMMANDS" to set Pen Width & color
- 589. 574 b) Declare and define response table c) Process events caused by pressing and releasing of mouse buttons. d) Create a freehand line drawing while keeping the left button of the mouse pressed The left button can be released to move the mouse to a new position. Pressing the left button would allow freehand drawing from the new position.) e) Clear the entire client area of the window when the cleanup option is selected from the setup menu f) Output text (in the chosen color) on the window g) Show current mouse position by pressing the right button of the mouse h) Terminate a running program through use of menu Exit/Close option i) Select color using TChooseColorDialog j) Take value of a parameter using TInputDialog k) Use message box (MessageBox) -------------------------------------------------------------------------------------------------------------*/ #include <owl/owlpch.h> #include <owl/applicat.h> #include <owl/framewin.h> #include <owl/dc.h> #include <owl/inputdia.h> #include <stdlib.h> #include <string.h> #include "p3_6.rc" /* Definition of the TrialWindow class -- data member CommonDC, a pointer to the TClientDC object, is not destroyed automatically when TrialWindow object is destroyed. It is necessory to add a destructor to TrialWindow to ensure proper cleanup. */ class TrialWindow : public TWindow { public: TrialWindow(TWindow* parent = 0); ~ TrialWindow() // Delete CommonDC { delete CommonDC; delete Pen; } void GetPenSize(); void SetPenSize(int newSize); void SetPen(TColor& newColor, int penSize = 0); protected: // Provides a valid device context across functions EvLButtonDown,
- 590. 575Appendix D // EvMouseMove, and EvLButtonUp. TDC* CommonDC; TPen* Pen; int PenSize; TColor Color; // Override member function of TWindow bool CanClose(); // Message response functions void EvLButtonDown(uint, TPoint&); void EvRButtonDown(uint, TPoint&); void EvMouseMove(uint, TPoint&); void EvLButtonUp(uint, TPoint&); //Menu processing functions void CmPenSize(); void CmPenColor(); void CmAbout(); void CmClearWindow(); void CmClose(); DECLARE_RESPONSE_TABLE( TrialWindow); }; DEFINE_RESPONSE_TABLE1( TrialWindow, TWindow) EV_WM_LBUTTONDOWN, //Mouse left button down EV_WM_RBUTTONDOWN, //Mouse right button down EV_WM_MOUSEMOVE, EV_WM_LBUTTONUP, //Mouse left button up EV_COMMAND(CM_PENSIZE, CmPenSize), EV_COMMAND(CM_PENCOLOR, CmPenColor), EV_COMMAND(CM_ABOUT, CmAbout), EV_COMMAND(CM_CLEARWINDOW, CmClearWindow ), EV_COMMAND(CM_CLOSE,CmClose ), END_RESPONSE_TABLE; TrialWindow:: TrialWindow(TWindow* parent) { Init(parent, 0, 0); // Initialize TrialWindow base class CommonDC = 0; // Initialize CommonDC-- a pointer to the TClientDC--to zero. PenSize = 1; // Initialize pen size Color = TColor::Black; // Initialize color
- 591. 576 Pen = new TPen(Color, PenSize); //TColor::Black defined in owlcolor.h } void TrialWindow::SetPen(TColor& newColor, int penSize) { // If penSize isn't the default (0), set PenSize to the new size. if (penSize) PenSize = penSize; Color = newColor; delete Pen; Pen = new TPen( Color, PenSize); // pickup a pen of new color } bool TrialWindow::CanClose() // file save feature not implemented { return MessageBox("Do you want to save?", "Not implemented", MB_YESNO | MB_ICONQUESTION) == IDNO; // insert logic for saving the file } void TrialWindow::EvLButtonDown(uint, TPoint& point) { if (!CommonDC) { SetCapture(); CommonDC = new TClientDC(*this); //'this' is a local variable avialable // in the body of a nonstatic function CommonDC->SelectObject(*Pen); //select pen in device context CommonDC->MoveTo(point); } } /* Let us learn to output text on the window. Output mouse position when right button of mouse is pressed. */ void TrialWindow::EvRButtonDown(uint, TPoint& point) { char str[40]; short posx = point.x; short posy = point.y; TClientDC CommonDC(*this); wsprintf(str, "Mouse Position (%d,%d) ", posx, posy); CommonDC.SetTextColor(RGB(255,0,0)); // set text color to red CommonDC.SetTextAlign(TA_BOTTOM | TA_LEFT); //Text Alignment
- 592. 577Appendix D CommonDC.TextOut(20, 20, str, strlen(str)); } void TrialWindow::EvMouseMove(uint, TPoint& point) { if (CommonDC) CommonDC->LineTo(point); } void TrialWindow::EvLButtonUp(uint, TPoint&) { if ( CommonDC) { ReleaseCapture(); delete CommonDC; CommonDC = 0; } } void TrialWindow::CmPenSize() //Processes menu command Pen Size { GetPenSize(); } void TrialWindow::CmPenColor() { TChooseColorDialog::TData colors; //Processes menu command Pen Color static TColor custColors[16] = { 0x010101L, 0x101010L, 0x202020L, 0x303030L, 0x404040L, 0x505050L, 0x606060L, 0x707070L, 0x808080L, 0x909090L, 0xA0A0A0L, 0xB0B0B0L, 0xC0C0C0L, 0xD0D0D0L, 0xE0E0E0L, 0xF0F0F0L }; colors.Flags = CC_RGBINIT; colors.Color = TColor::Black; colors.CustColors = custColors; if (TChooseColorDialog(this, colors).Execute() == IDOK) SetPen(colors.Color); } void TrialWindow::GetPenSize()
- 593. 578 { char inputText[6]; wsprintf(inputText, "%d", PenSize); if ((TInputDialog(this, "Line Thickness","Input a new thickness:", inputText,sizeof(inputText))).Execute() == IDOK) { int newPenSize = atoi(inputText); if (newPenSize < 0) newPenSize=1; SetPenSize(newPenSize); } } void TrialWindow::SetPenSize(int newSize) { delete Pen; PenSize = newSize; Pen = new TPen(Color, PenSize); } void TrialWindow::CmClearWindow() { Invalidate(); // clears the entire client area of the window } void TrialWindow::CmAbout() { TDialog(this, IDD_ABOUT).Execute(); } void TrialWindow::CmClose() // terminate the program { EvClose(); // shuts down the main window and ends the application } class TDrawApp : public TApplication { public: TDrawApp() : TApplication() {} void InitMainWindow(); };
- 594. 579Appendix D void TDrawApp::InitMainWindow() { SetMainWindow(new TFrameWindow(0, "Freehand Drawing", new TrialWindow)); GetMainWindow()->AssignMenu("COMMANDS"); } int OwlMain(int /*argc*/, char* /*argv*/ []) { return TDrawApp().Run(); } Resource file for Program D.6 //---------------------------------------------------------------------------- // Resource file for program D.6 //---------------------------------------------------------------------------- #include <owl/inputdia.rc> #include <owl/except.rc> #define CM_ABOUT 205 #define CM_PENSIZE 206 #define CM_PENCOLOR 207 #define CM_CLEARWINDOW 208 #define IDD_ABOUT 211 #define CM_CLOSE 212 #ifdef RC_INVOKED #ifndef WORKSHOP_INVOKED # include <windows.h> #endif COMMANDS MENU { POPUP "&Setup" { MENUITEM "Pen &Size", CM_PENSIZE MENUITEM "Pen &Color", CM_PENCOLOR MENUITEM "&CleanUp", CM_CLEARWINDOW } POPUP "&Help" { MENUITEM "&About", CM_ABOUT } POPUP "&Exit" {
- 595. 580 MENUITEM "&Close", CM_CLOSE } } STRINGTABLE { CM_PENSIZE, "Change the pen width" CM_PENCOLOR, "Change the pen color" CM_ABOUT, "show an information dialog box" CM_CLEARWINDOW, "Clear window" CM_CLOSE, "Terminate the Program" } IDD_ABOUT DIALOG 37, 25, 170, 95 STYLE DS_MODALFRAME | WS_POPUP | WS_VISIBLE | WS_CAPTION | WS_SYSMENU CAPTION "About Freehand Drawing" FONT 8, "MS Sans Serif" { DEFPUSHBUTTON "Ok", IDOK, 60, 70, 50, 14 CTEXT "ObjectWindows 2.5 Program", -1, 23, 46, 127, 10 CTEXT "Freehand Drawing", -1, 23, 24, 126, 8 } #endif // RC_INVOKED D.2.10 Tool Bar and Status Bar A tool bar generally appears at the top of main window of application and consists of collection of push buttons. It enables the user to quickly initiate action as compared to the pull down menu. For example, in Word Perfect, the tool bar comprises of buttons, such as New Blank Document, Open, Save, Print, Cut, Copy, Paste, Undo, Redo, Bold, Italic, Underline, and Quick Format. Most packages provide buttons for special functions, like File open and Edit dialog box. A status bar is another common decoration that is included in many Windows applications. This is an area at the bottom of the main window where an application displays help messages as well as the status of some keys (Caps lock, Num lock and Insert). To insert status and control bars one has to change the main window from a TFrameWindow to a TDecoratedFrame and modify the function InitMainWindow() as follows: // Create the decorated frame window TDecoratedFrame* frame = new TDecoratedFrame(Name, "Caption of window", new TdrawWindow, true); //Create a status bar TStatusBar* sb = new TStatusBar(frame, Tgadget::Recessed, TstatusBar::CapsLock| TstatusBar::NumLock| TStatusBar::OverType); // Create a control bar
- 596. 581Appendix D TControlBar* cb = new TcontrolBar(frame); // Insert button gadgets into the TControlBar. Each button gadget has an associated bitmap // resource ID and a control ID that tie together a message response function to that button. cb->Insert(*new TButtonGadget(CM_FILENEW, CM_FILENEW, TButtonGadget:: Command)); cb->Insert(*new TButtonGadget(CM_FILEOPEN, CM_FILEOPEN, TButtonGadget:: Command)); cb->Insert(*new TButtonGadget(CM_FILESAVE, CM_FILESAVE, TButtonGadget:: Command)); cb->Insert(*new TButtonGadget(CM_FILESAVEAS, CM_FILESAVEAS, TbuttonGadget:: Command)); // Insert the status bar and control bar into the frame frame->Insert(*sb, TDecoratedFrame::Bottom); frame->Insert(*cb, TDecoratedFrame::Top); Source: Chapter 3 of the book ‘Computer Graphics for Engineers and Scientists’ by R.G.S. Asthana and N.K. Sinha, New Age International Publishers Pages 539, First Edition 1996, Second Edition: 1998, Reprint 2005.
- 598. APPENDIX E NUMBER SYSTEM Any use of numbers implies the use of a base value for the numbers. The simplest base value in a numbering scheme is '1'. In this scheme the number '2' is two things, or two groups of ones. The number '9' is nine things or nine groups of ones. The basic rules for a base numbering system entail ordering items, grouping ordered items and then expressing the groups and items in a consistent way. The way it represents the different groups gives the numbering system an order of magnitude. This can be expressed in several ways. A special symbol is used to represents a specified grouping value. For example, a picture of a hand may represent numeral 5. The Roman numerals and the Egyptian numbering represent each order of magnitude with a special symbol. This approach, however, limits how high a numbering system may count because a new symbol needs to be devised for each successive grouping. For example, the number 1,295,468 is represented in the Egyptian numbering system as follows: Each of the magnitudes of 10 was represented in the above number, for example the 4 frogs represent 4 hundred thousands and the 5 lotus flowers represent 5 thousands, etc. In this numbering system, only the magnitudes of 10 that are used are expressed in the written number. The number 6,060 is thus represented as:
- 599. 584 Grouping values can be represented by the place they hold in the representation of the number. Our Hindu/Arabic numbering system uses this method, as we have a place value for 1's, 10's, 100's, etc. Since no new symbols are used (0-9), the numbering system can continue to incredibly large values. From the use of this method a representation of 0 emerges so that there is a way to represent 0 of a group. The orders of magnitude may represent a consistent gradation of value where each successive order of magnitude will be 'n' times the last order of magnitude. In the decimal system, each successive place value is 10 times the last place value. Some examples of number base systems are: Binary (base 2), Hand (base 5), Octal (base 8), Decimal (base 10), Base 12 (duodecimal), Hexadecimal (base 16), Mayan (base 20) (Vigesimal), and Time and Ancient Sumerian (base 60) Base 60 (Sexagesimal). One can imagine that we would have evolved a strategy of using the letter A to represent ten; or we might have used IIIIIIIIII to represent that idea. The Romans used X. The Arabic system, which we use, makes use of position in conjunction with numerals to represent values. The first (right-most) column is used for "ones," and the next column is used for tens. Thus, the number fifteen is represented as 15 (read "one, five"); that is, 1 ten and 5 ones. Certain rules emerge, from which some generalizations can be made as given in Table E .1. Table E.1: Number systems Number System Base digits power Range of numbers Largest Decimal Smallest Decimal BaseRep- resentation Decimal 10 0-9 10 0 to 10n - 1 102 - 1 = 99 100 10010 Octal 8 0-7 8 0 to 8n - 1 82 -1 = 63 64 1008 Binary 2 0 & 1 2 0 to 2n - 1 22 -1 = 3 4 1002 Hexa- Decimal 16 0-9 & A-F 16 0 to 16n - 1 162 - 1 = 255 256 10016 To distinguish numbers written in each base, we write the base as a subscript next to the number. The number thirteen in base 10 would be written as 1310 and read as "One, three, base ten." Thus, to represent the number 1310 in base 8 you would write 158. This is read "one, five, base eight." Why 15? The 1 means 1 eight, and the 5 means 5 ones. One eight plus five ones equals fifteen. Around the Bases How do you generalize the process? To convert a base ten number to base 8, think about the columns: in base 8 they are ones, eight, sixty four, five-hundred-ten-two, and so on. Why these columns? They represent 1, 8, 64, 512 and so forth.
- 600. 585Appendix E To understand the procedure to convert from a decimal value to base 8, we take 202 as a decimal number and give below a step by step procedure: Step 1: Examine the number and decide which column to use first. The number is 202 so you know that column 4 (202/512) is 0 and you don’t have to worry about it. Step 2: To find out how many 64s there are, divide 202 by 64. The answer is 3, so put 3 in column 3. Step 3: Examine the remainder of step 2 which is 10 so divide remainder by 8 and u get 1, so put a one in the second column. Step 4: The remainder is now two. There are 2 ones in 2 so put a 2 in the1s column. Result: The resultant octal number is 0312. Let us take up another example. Say we want to convert 202 to a binary number. To convert the number to base 2, you follow the same procedure: There are no 256s, so column 8 is 0. There is one 128 in 202, so column 7 is 1 and 74 is the remainder. There is one 64s in 78 so column 6 is 1 and remainder is 10. There is no 16 in 10 so column 5 is 0. There is one 8 in 8, and so column 4 is 1 and remainder is 2. There is no 4 in remainder 2 so column 3 is 0 and the remainder remains 2. There is one 2 in 2, and so column 1 is 1 and remainder is zero. There is no remainder, so the column 1 is 0. The binary number, therefore, is 01101010. Why Base 2? Once you make this great leap of intuition, the power of binary becomes clear: With 1s and 0s you can represent the fundamental truth of every circuit (there is power or there
- 601. 586 isn't). All a computer ever knows is, "Is you is, or is you ain't?" Is you is = 1; is you ain't = 0. Bits, Bytes, and Nibbles Once the decision is made to represent true and false with 1s and 0s, binary digits (or bits) become very important. Since early computers could send 8 bits at a time, it was natural to start writing code using 8-bit numbers--called bytes. Half a byte (4 bits) is referred to as nibble. With 8 binary digits you can represent up to 256 different values. What’s a KB? It turns out that 1,024 is roughly equal to 1,000. This coincidence was too good to miss, so computer scientists started referring to 1024 bytes as 1KB or 1 kilobyte, based on the scientific prefix of kilo for thousand. Similarly, 1024 * 1024 (1,048,576) is close enough to one million to receive the design- nation 1MB or 1 megabyte, and 1,024 megabytes is called 1 gigabyte (giga implies thou- sand-million or billion). Hexadecimal Because binary numbers are difficult to read, a simpler way to represent the same values is sought. Translating from binary to base 10 involves a fair bit of manipulation of numbers; but it turns out that translating from base 2 to base 16 is very simple, because there is a very good shortcut. To understand this, you must first understand base 16, which is known as hexadecimal. In base 16 there are sixteen numerals: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. The last six are arbitrary; the letters A-F were chosen because they are easy to represent on a keyboard. To translate from hexadecimal to decimal, you can multiply. Thus, the number C8C represents: Hexadecimal C8C Result Step 1 C*256 12*256 = 3072 Step 2 8*16 = 128 Step 3 C*1 12*1 = 12 Decimal Sum of Step 1-3 3212
- 602. 587Appendix E Translating the number AB to binary is done by translating first to base 10, and then to binary: Conversion to base 10 Hexadecimal AB Result Step 1 A*16 10*16 = 160 Step 2 B*1 11* 1 = 11 Decimal 171 Sum of Step 1- 2 171 Conversion to binary Step 1 There is 1 128 Remainder 43 1 Step 2 There is 0 64 0 Step 3 There is 1 32 11 1 Step 4 There is 0 16 0 Step 5 There is 1 8 3 1 Step 6 There is 0 4 0 Step 7 There is 1 2 1 1 Step 8 There is 1 1 1 Binary 10101011 Thus, the answer in binary is 1010 1011. Let us look at this binary number as two sets of 4 digits and you are set to do a magical transformation. The right set is 1011, i.e., 11 decimal or in hexadecimal it is ‘B’. The left set is 1010, i.e., 10 in Decimal or ‘A’ in Hexadecimal. We replace each set with its equivalent hexa- decimal digit and we have: Number System Left Set Right Set Binary 1010 1011 Hexadecimal A B Putting the two hex numbers together we get the hex number ‘AB’ which is the real value of binary number ‘1010 1011’. This method always works irrespective of binary number of any length. To repeat the process, first reduce the given binary number in to sets of 4, translate each set of four to hex, and put the hex numbers together to get the result in hex. Similarly, you can easily convert any hex to binary.
- 603. 588 Notes
- 604. APPENDIX F Recursion F.1 Introduction In computer science, recursion specifies (or constructs) a class of objects or methods (or an object from a certain class) by defining a few very simple base cases or methods (often just one), and then defining rules to break down complex cases into simpler cases. F.2 Recursive Functions A useful way to think of recursive functions is to imagine them as a process being performed where one of the instructions is to "repeat the process". This makes it sound very similar to a loop because it repeats the same code, and in some ways it is similar to looping. On the other hand, recursion makes it easier to express ideas in which the result of the recursive call is necessary to complete the task. Of course, it must be possible for the "process" to sometimes be completed without the recursive call. One simple example is the idea of building a wall that is ten feet high; if one wants to build a ten foot high wall, and then one may first build one foot high wall, and then add an extra foot of bricks repeatedly till the height become 10 feet. Conceptually, this is like saying the "build wall" function builds wall of one feet height and repeats itself till the height is equal to 10 feet.. Each recursive function should have the following properties. Base case: Must have at least one case which can be solved without recursion. Making progress: At each step there must be some progress toward the base case so that the base case condition is eventually met. Design rule: during the design of the recursive algorithm, assume that all simpler recursive calls work. Let us look at an example to further understand the concept of recursion. /* Example F.1: Factorial function n! = n · (n − 1) · (n − 2) · · · 3 · 2 · 1 n! = n · (n − 1)! n! = n · (n − 1) · (n − 2)! A recursive algorithm to calculate a factorial of a function */
- 605. 590 #include <iostream.h> int factorial(int n) { if (n > 1) { return n * factorial(n - 1); //‘Line R’ ‘Recursive call to the function factorial } else { return 1; // ‘Line B’ Base Case } } int main() { int n; int fact; cout << "Enter an integer for which the factorial is to be calculated: "; cin >> n; fact = factorial(n); // Line ‘C’ cout << n << "! = " << fact << endl; return 0; } Example F.1 illustrates a recursive algorithm to calculate the factorial of an integer entered by the user. So lets suppose the user enters the integer ‘n’ as the input to the program. The statement ‘fact = factorial (n);’ marks the first call to the factorial function. Once this function is called as it might already be evident from the function structure that the function does indeed have two decision cases. The first being if the value of the inputted integer ‘n’ is greater than ‘1’ in which case the function calls itself again (recursive call). Where as in the second case if the value of ‘n’ is lesser than or equal to ‘1’ in which case it returns the value ‘1’. In order to really understand this example, let us trace it say with value of n = 4. The program execution reaches ‘Line C’ in the main function, as a result of which the factorial function is called with ‘n = 4’. As ‘4 > 1’ so the ‘Line R’ in the factorial function is called with ‘n = 3’ (factorial (n - 1)). Again as ‘3’ is greater than ‘1’ the ‘Line R’ is executed in the factorial function and it is called again with ‘n = 2’. Similarly, factorial (2) results in calling the factorial function again with ‘n = 1’. When this happens the ‘Line B’ gets executed and the integer value of ‘1’ is returned. As a result of which
- 606. 591Appendix F factorial (2) returns (2 * factorial (1)) = 2 (‘return n * factorial(n - 1)’ Line C). Similarly factorial (3) returns (3 * factorial (2)) = 6 and finally returning factorial (4) as 24 (4 * factorial(3)). The calling sequence of the factorial function is also illustrated in figure F.1 given below. Figure F.1: Function Factorial a) Calling Sequence b) Returning Sequence You should now be able to understand better how a recursive function solves a problem by representing it as an instance of the same problem with a smaller input size in each iteration of the function. Let us look at another example which demonstrates a recursive function to calculate the greatest common divisor (GCD) of two numbers. // GCD recursive function # include <iostream> using namespace std; //Function prototype int gcd(int, int); int main() { int number1, number2; cout<<" Enter the two integers "; cin>>number1>>number2;
- 607. 592 cout<<" Enter the greatest common divisor of " << number1; cout << " and " << number2 << " is:"<<gcd(number1, number2)<<endl; // Line R return 0; } /*********************************************************************** GCD function This function uses recursion to calculate the greatest common divisor of two integers passed into the parameters num1 and num2 ***********************************************************************/ int gcd(int num1, int num2) { if (num1 % num2 == 0) return num2; else return gcd(num2, num1 % num2); } To understand this program let us again trace it for a simple example say ‘number1 = 4’ and ‘number2 = 14’. When the program execution reaches ‘Line R’ in the main function the ‘GCD’ function is called with ‘num1 = 4’ and ‘num2 = 14’. As (4%14) is not equal to zero the flow of the program enters the else structure and the function is recursively called with ‘num1 = 14’ and ‘num2 = 4’. As again in this iteration (14%4) is not equal to zero the flow of program enters the else structure and the function is recursively called with ‘num2 = 4’ and ‘num1=2’ and then with ‘num1 = 4’ and ‘num2 = 2’. During this last iteration of the function as the condition specifies the base case (4%2 == 0). The integer ‘num2=2’ is returned as the GCD for the two integers. F.3 Practical Applications of Recursion In this section we will look at some of the practical applications of recursion. F.3.1 Tower of Hanoi The Tower of Hanoi puzzle was invented by the French mathematician Eduard Lucas in 1883. We are given a tower of eight disks (initially four in the applet below), initially stacked in increasing size on one
- 608. 593Appendix F of three pegs. The objective is to transfer the entire tower to one of the other pegs exercising the following set of rules (fig F.2): Move the disks from pile 1 to pile 3 using pile 2 as a temporary holding location. Move one disc at a time. A disk may not be replaced by a smaller disks Figure F.2: Tower of Hanoi The recursive algorithm for solving this puzzle is given below: Algorithm: Tower Of Hanoi 1. Move n − 1 disks from pile 1 to pile 2, using pile 3 are temporary holding location. 2. Move 1 disk from pile 1 to pile 3. 3. Move n − 1 disks from pile 2 to pile 3, using pile 1 as temporary holding location. Exercising this algorithm we can now develop the recursive function for doing so. The recursive function implementing this algorithm is given in program F.1.
- 609. 594 Program F.1: Tower Of Hanoi int towers (int n, int start, int goal, int temp) { if ( n==0) return 0 ; // Base Case towers ( n-1, start, temp, goal ); cout << "Move from "<<start << " to " << goal ; towers ( n-1,temp , goal , start) ; } F.3.2 Binary Search Recall the binary search algorithm covered in ‘Part I Chapter 5, page 155’. We implemented the binary search algorithm using an iterative procedure. The same algorithm can also be implemented using an recursive algorithm as shown below. Algorithm: o The element to be searched for is taken from the user say the ‘Key’. o The given data array is taken. Any arbitrary lower limit or the lower bound for the data values is taken. Similarly one defines the upper limit or the upper bound. Then we obtain the middle value using. Middle = (lower bound + upper bound)/2; o The search ‘key’ is compared to the Middle value if ‘Key=Middle’ then the search is complete. If Key<Middle, the program continues and the search is confined to the first half only. On the other hand if Key>Middle then also the program continues and the search is confined to the latter half. This process is continued till the time ‘Key = Middle’. Program F.2 implements the algorithm given above using an recursive function. Program F.2: Binary Search int binsrch(list x[],int srch ,int left ,int right) { int mid = (left + right)/2 ;
- 610. 595Appendix F if (x [ mid]== srch) return (mid) ; else if(x(mid)>srch && mid>left) return binsrch (x , srch , left ,mid-1); else if (x(mid)<srch && right>mid) return binsrch (x , srch , mid+1 , r i g h t ) else return UNSUCCESSFUL; } F.4 Recursion Vs Iteration It is a popular computer science belief that any algorithm that can be coded with recursion can also be coded with an iterative control structures. Both approaches achieve the same goal, but the question arises which one is the best to use among them? Although a recursive algorithm might be easier to understand but there are several reasons why use of recursion in curbed. Recursive algorithms are certainly less efficient than corresponding iterative algorithms. Each time a function is called, the system incurs an overhead that is not necessary with a loop. Also, in some cases it is possible the algorithm design for an iterative procedure might be simpler. The speed and amount of memory available to modern computer today however decreases the impact of recursion so much that its inefficiency is no longer a strong argument against it. Today the choice for a recursive or an iterative algorithm is just a matter of design.
- 611. 596 Notes
- 612. APPENDIX G XML & C++ G.1 Overview of XML Extended Markup Language (XML) is a subset of the Standard Generalized Markup Language (SGML), a complex standard for describing document structure and content. XML is a language for organizing - not merely presenting – data. XML is intended to be a metalanguage, i.e., a language in which you can describe another language. XML, therefore, empowers you to design your own markup. Hyper Text Markup Language (HTML) most popular in making web pages has following limitations: • It’s only a presentation format and does not describe contents. • HTML search engines can carry out key word searches and yield high number of irrelevant results as they do not have capacity to search based on context. • Links are hard coded into a document when it is created. • HTML does not permit link to an element or to multiple locations; XML does. XML defines structure of data in an open and self describing manner facilitating transfer over a network and consistence processing by the receiver. G.1.1 Data, Information and Knowledge XML has potential to tag data so that it can be interchanged among applications, such as business-to-business e-commerce, directory services, financial transactions and wireless devices. The tags are context related and instantly turn data into information. Here context refers to the identity of, and relationships between a data set’s entities. A set of numbers [B 75 120 R] alone may not make any sense but if we associate it with say, type, passenger capacity, maximum permissible speed and color respectively of a railway coach – it becomes information. Information, in fact is union of data and context whereas knowledge is union of information with specific issues. XML has potential to play a key role in knowledge management as it is designed to do just this, i.e., to make information self-describing. G.1.2 XML – Syntax and Organization XML defines both document syntax and organization. This aspect is exploited to create data markup tags, schemas (Collection of tags and data structuring rules), and document type definitions (DTDs).
- 613. 598 Schemas and DTDs are transmitted with XML documents to identify how receiving application should interpret the tags, and developers may create style sheets to customize output. XML uses markup tags as well, but, unlike HTML (Hyper-text markup language), XML tags describe the content, rather than the presentation of that content. By avoiding formatting tags in the data, but marking the meaning of the data itself with custom user definable tags, it is easier to search various documents for a tag and view documents tailored to the preferences of the user. The XML processor can exist on the server, the client, or both. XML is the would-be de-facto standard of the web. While HTML has been around for some time without actually giving the users any great flexibility, the advent of XML threatens to dwarf its existence. XML, as the name implies, gives you the freedom to name and define your own tags. If you have ever had a look at any HTML file, you would have seen tags like <title> </title> spread around. That’s the way to write a tag, then input whatever you want to write in it and then close (using / ) the tag. For example: <title> My Home Page </title> In HTML, you thus had a fixed number of tags like title, table, body that could be used to generate web pages. It severely undermined the potential of the World Wide Web. The entry of XML changes all that. In this new paradigm, you can define your own tags and then use them, as you want them to be used. Thus the number of tags that can be used becomes unlimited. For example: <my_anniversary> 05/20/77 </my_anniversary> Of course, there are several restrictions on how they have to be written and named and all that. For instance, your element name can’t start with an underscore; you can’t have spaces in the name, etc. Thus XML not only allows you to define your own tags, but also allows you to describe what they are. This approach is markedly different from HTML where you can only display your data but can’t describe it in a nice way. For example, if you had seen a date in an HTML document, you wouldn’t have known if it were someone’s birthday or someone’s anniversary or whatever. You would have to go to the document and actually read the text to know anything about it. To illustrate the advantages of XML, we are presenting below a code segment in HTML and XML.. Example – G.1: A code segment in HTML and XML HTML XML <Table> <Employee> <TR> <EmployeeID> 1001 </EmployeeID> <TD> Name </TD> <TD> Ankit </TD> <DOB> 25/12/1986 </DOB> </TR> <Name> Ankit </Name> <TR> </Employee> <TD> DOB </TD> <TD> 25/12/1986 </TD> </TR> <TR> <TD> EmployeeID </TD> <TD> 1001 </TD> </TR> </Table> G.1.3 Terminology associated with any XML document Element: Basically the names of the tags you use.
- 614. 599Appendix G Attribute: As the name implies, properties of the elements that you want to use. For instance, if <person> is an element in your file, his nationality can be an attribute. You state this in the following way. <person nationality="Any country"> pretty vague</person>. DTD: This is the Document Type Definition. This is a set of rules that you specify for your elements in the XML file. Living in a practical world, you would not want your elements to go astray and accommodate everything possible. You would want to write some rules that they adhere to. All this is done in a DTD. CSS: This is the Cascading Style Sheet. In order to display your XML file the way you want you would like to write some styles for the elements of the XML file. CSS is the file where you can specify the stylistic rules for your elements. Well-formed document: Unlike HTML, XML is strict about nesting the tags and writing the end tag. For example, if you had a tag like <body> in your HTML document and forgot to end the content with a </body>, it is still fine. But you can’t do the same thing with XML. Any tag that starts has to end. Also the nesting has to make sense. For instance to use <city> Delhi City <state-province> Delhi </city> </state-province> would be wrong. As you can see, the nesting is not appropriate. To say <city> Delhi City <state-province> Delhi</state-province> </city> makes more sense and is correct. Any XML document that adheres to these rules is a well-formed document. Valid Document: In addition to the well-formed ness, if an XML document also adheres to the rules specified in a DTD, it is supposed to be valid. G.1.4 XML Repositories Repositories would provide XML documents templates based on schemas for domain specific data, such as say within the financial industry. Two such repositories, which offer public access or pointers to XML schemas for application development and processing of XML data, are Microsoft’s BizTalk.org and Oasis’ XML.org. G.1.5 XML Summary Humans have successfully done business for long by exchanging standardized documents, such as, purchase orders, invoices, manifests, receipts and so on. XML was designed for document exchange, and it is becoming evident that electronic commerce will heavily rely on flow of agreements, expressed in millions of XML documents pulsing around the Internet. Thus, for its users, the XML-powered Web will be faster, friendlier and a better place to do business. G.2 C++ and XML C++ is a popular programming language for which many XML related efforts already exist. Several toolkits and libraries have been produced for C++ based manipulation. Those toolkits mainly fall into two categories: event-driven processors and object model construction processors.
- 615. 600 G.2.1 Event-driven approaches In an event-driven approach, a parser reads the XML data and notifies specialized handlers to carry out desired actions. In fact, an event-driven process calls specific handlers when the contents of a XML document are encountered. For instance, calling endDocument() when the end of the XML document is found. XML parser implementations differ in their application program interfaces, e.g., • A parser could report to a handler of the start of an element, passing it only the name of the element and then attribute information in another call or • A parser reports to a handler when it encounters the same start-element tag and passes it name of the element along with a list of the attributes and values of that element. • A parser could use an STL list of strings, while another could use a specially made class to hold attributes and values. virtual void HypotheticalHandler::startElementHandler (const String name,const list<String> attributes) = 0; // IBM's xml4c2 parser. virtual void DocumentHandler::startElement (const XMLCh* const name, AttributeList& attrs) = 0; As you can see, the way processors notify applications about elements, attributes, character data, processing instructions and entities is parser-specific and can greatly influence the programming style behind the XML-related modules of your system. SAX (the Simple API for XML) is a standard event-driven XML processing API. Microsoft C++ provides support for SAX. G.2.2 Non SAX event-driven approaches Expat is an XML 1.0 parser written in C by James Clark. It is event-driven, in the sense that it calls handlers as parts of the document are encountered by the parser. User-defined functions can be registered as handlers. C++ wrappers for expat Parser are available. C++ wrapper will encapsulate the C details and provide you with a clean list of methods that you can override to suit your particular needs. G.2.3 ‘expatpp’ Interface The expatpp interface defines wrappers for all the methods in expat and includes the following members: virtual void startElement (const XML_Char* name, const XML_Char** atts); virtual void endElement(const XML_Char* name); virtual void charData(const XML_Char *s, int len); virtual void processingInstruction(const XML_Char* target, const XML_Char* data); virtual void defaultHandler(const XML_Char *s, int len);
- 616. 601Appendix G virtual void unparsedEntityDecl(const XML_Char *entityName, const XML_Char* base, const XML_Char* systemId, const XML_Char* publicId, const XML_Char* notationName); virtual void notationDecl(const XML_Char* notationName, const XML_Char* base, const XML_Char* systemId, const XML_Char* publicId); // XML interfaces int XMLPARSEAPI XML_Parse (const char *s, int len, int isFinal); XML_Error XMLPARSEAPI XML_GetErrorCode(); int XMLPARSEAPI XML_GetCurrentLineNumber(); This interface defines a handler base for expatpp. Expat is under the GPL (General Public License), is very fast and very portable. But it is just plain C, so you need to find a way to integrate it with your OO C++ project. A good option would be to wrap expat using a C++ class that will encapsulate the C details and provide you with a clean list of methods that you can override to suit your particular needs. This is how wrappers like expatpp work. Expatpp is a C++ wrapper for expat. It was developed by Andy Dent with this basic idea: the constructor of expatpp creates an instance of an expat parser, and registers dummy functions as handlers that call the corresponding expatpp override-able methods. G.2.4 DOM The Document Object Model is a language- and platform-independent interface that allows programs and scripts to dynamically access and update the content structure and style of documents. There is a core set of interfaces that every DOM 1.0-compliant implementation must provide. Expressing a document as a structure and making it available to the application is not new: all major browsers have done so for years in their own proprietary way. The important idea behind the XML DOM is that it standardizes the model to use when representing any XML document in memory. DOM-based C++ parsers produce a DOM representation of the document instead of informing the application when they encounter elements, attributes etc. The DOM, as the name implies, is an object model as opposed to a data model. The object- oriented interfaces define the semantics of a structural model, independently of the implementation chosen for it. That means that DOM parser implementations are free to choose whatever internal representation they like, as long as they comply with the DOM interfaces. G.2.5 DOM Interfaces The DOM level 1 core defines a basic set of interfaces that allow the manipulation of XML documents. It provides methods for the access and population of the document. These methods are encapsulated in two sets of interfaces: the fundamental core interfaces and the extended interfaces. For a complete description and all the methods, you will need to download a DOM library. Again, xml4c2 is a good choice because of its excellent documentation.
- 617. 602 G.2.6 XSD ‘XSD’, is a cross-platform, open source W3C XML Schema to C++ translator. Given an XML instance description (XML Schema), it generates C++ classes that represent the given vocabulary as well as parsing and serialization code (collectively called a mapping or binding). Compared to APIs such as DOM and SAX, the generated code allows you to access the information in XML instance documents using your domain vocabulary instead of generic elements, attributes, and text. Static typing helps catch errors at compile-time rather than at run- time. Automatic code generation frees you for more interesting tasks while minimizing the effort needed to adopt your applications to changes in the document structure. XSD supports two C++ mappings: in-memory C++/Tree and event-driven C++/Parser. The C++/Tree mapping consists of C++ classes that represent data types defined in XML Schema, a set of parsing functions that convert XML instance documents to a tree-like in-memory data structure, and a set of serialization functions that convert the in-memory representation back to XML. The C++/Parser mapping provides parser templates for data types defined in XML Schema. Using these parser templates you can build your own in-memory representations or perform immediate processing of XML instance documents. XSD features C++ standard library-based language mappings, configurable base character type (char/wchar_t), platform-independent generated code, comprehensive documentation, and more. XSD runs on a range of platforms, including GNU/Linux, Windows, Mac OS X, and Solaris. Supported C++ compilers include GNU g++ (3.3, 3.4, 4.0), Intel C++ 8.1, Sun C++ 5.7 (Studio 10) and 5.8 (Studio 11), Microsoft Visual C++ 7.1 (Visual Studio .NET 2003) and 8 (Visual Studio 2005).
- 618. APPENDIX H Exceptions and Exception Handling H.1 Exceptions Exceptions provide a way of handling “less than ideal” situations, such as • Action your program should take if it runs out of memory when trying to dynamically allocate storage • what should your program do if the file you’re trying to open is locked or does not exist When writing reusable code, there are times when your classes will encounter situations where they don’t know what to do, and can’t continue. One trivial solution is to just terminate the program. However, if the program is controlling a chemical process in real time then this can not be a desirable solution. The ideal situation will be that your program informs that it could not continue execution for whatever reason and the application code has necessary provision to resolve this condition suitably. Exceptions are a brilliant programmatic concept, but implementation of them in C++ language leaves lot to be done. Even ANSI C code is also somewhat lax in how programmers deal with exceptions and exception handling. H.2 Conventional Error Handling Methods Methods of handling errors usually involve returning an error code from every function, either through a return value, an extra reference parameter, or a global variable such as ‘errno’. These methods of error reporting, however, have several limitations: • Checking for an error condition after every function call practically doubles the amount of code you have to write. • Programmer may forget to check for an error condition, or otherwise choose to ignore them. • Propagating errors upward from function to function can become untidy because of the constraint of returning one type of error condition from a function.
- 619. 604 H.3 Exceptions in C++ C++ provides a way to minimize these problems through the use of exceptions. You need not deal with exceptions on each function call. If you call a series of functions dealing with one aspect of a program, you need to look for the error conditions only in one location. Suppose you have three functions A, B and C. You call a function A which calls function B which in turn calls function C, and C reports an error. If you handle the exception in A then instead of reporting error, C will return control to A if B does not handle that error type. If you do not handle an exception and ignore it, your program will terminate. You can also report different types of exceptions from a single function. H.4 Types of Exception You can define a separate exception handler for each of exception types. The type of an exception can be any type in C++, e.g., it can be a built-in type, such as an int or a char *, or it can be a user-defined type. You may define a class which encapsulates the information relevant to the exception, e.g., if you had an out-of-range exception when subscripting an instance of an ‘IntArray’, you may want to have an exception class which encapsulates the valid range of the array as well as the index requested. H.5 Throwing Exceptions Reporting an exception in C++ is called ‘throwing’ an exception. You do this by using the throw keyword along with an object or variable you want to throw. You can throw an exception of any type. The code below can be used to throw a string exception, int IntArray::operator[](int index) const { if (index < 0 || index >= numElements) throw "Out of Bounds"; return this->elements[index]; } Further you can also create a special ‘IndexOutOfBounds’ class and throw a temporary instance of it. The advantage is that you have a type specific to the nature of the exception in this case.
- 620. 605Appendix H int IntArray::operator[](int index) const { if (index < 0 || index >= numElements) { throw IndexOutOfBounds(index, numElements); } return this->elements[index]; } H.6 Catching Exceptions Catching exceptions is the C++ lingo for handling exceptions. If you don’t catch an exception that has been thrown, your program will terminate. The basic control structure for exception handling in C++ is the try/catch block. A ‘try/catch’ block that handles ‘IndexOutOfBounds’ exceptions might look like this: try { // code that might throw an IndexOutOfBounds exception } catch (IndexOutOfBounds iobe) { // code designed to handle any exception of type IndexOutOfBounds } Any exceptions thrown within the ‘try’ block will be handled by a matching exception handler in the ‘catch’ block. You can catch different types of exceptions within one ‘try/catch’ block, and you can catch exceptions of any type by using an ellipsis (...) as the exception type. This is also called a catch-all handler, and is demonstrated below: try { // Call some code here... } catch (char* str) { // Handle character string exceptions here... } catch (IndexOutOfBounds& boundsError) { // Handle IndexOutOfBounds exceptions here... } catch (...) { // Handle all other exceptions here... } The order in which catch blocks are placed is important. The first matching Exception handler will be invoked, so if you placed the catch-all exception handler first, no other exception handler would be called. One should never throw and catch an actual object— always use a reference, a pointer, or a built-in as the exception type. You can choose to filter out some exceptions by handling them in a particular ‘try/catch’ block and allowing an outer ‘try/catch’ block to handle the ones you don’t want to handle. This can be done by re-‘throw’ing the exceptions you don’t want to handle from the ‘catch’ block.
- 621. 606 H.7 Resource Management Between the point where an exception is thrown and an exception is caught, all fully initialized direct objects that are local variables or parameters to functions are destroyed while the stack is unwound to the point where the exception is caught. For instance, consider the following code: bool TrainList::contains(Train *f) const throw(char*) { Train Train; Station *a = new Station(“SFO”, “San Francisco”); ... throw “Out of Bounds”; ... } try { if (TrainList->contains(Station)) ... } catch(char* str) { // Handle the error } In this case, the destructor for the ‘Train’ object is called. However, the destructor for the ‘Station’ object is not, because it is not a direct object. Because of this, it is often wise to declare objects directly in the presence of exceptions. H.8 Exception Specifications C++ allows you to specify which exception types a function will throw, or to specify that a function will throw no exceptions at all. This specification is provided as part of a function interface. For example, we might have a function ‘ReadStationsFromFile’ that might throw a ‘FileNotFoundException’ exception. The prototype for this function would then need to be: void ReadStationFromFile(const char *filename) throw (FileNotFoundException); Exception specifications can have any number of exception types, or even no exception types. If we select no exception types in an exception specifier then the function will not throw any exceptions. Class methods can have exception specifications: class Stack { ... Stack(); // can throw any exception type ~Stack() throw (); // can't throw any types whatsoever
- 622. 607Appendix H int pop(void) throw(EmptyStackException); int peek(void) const throw (EmptyStackException); void push(int value) throw (FullStackException, bad_alloc); ... }; The exception specification must be duplicated word for word when implementing a function or method or there will be a compile-time error according to the C++ standard. The compiler won’t check the transitivity of exception specifications. For instance, the ‘pop’ method above can call a function that throws exceptions other than those of type ‘EmptyStackException’. If you provide no exception specification, it means that the function can throw any type of exception. The compiler can’t always check to see that a function doesn’t throw an exception type that isn’t specified, but it can catch many cases of this. H.9 Efficiency Concerns Throwing exceptions is costly in terms of performance. It takes minimal overhead to use ‘try/catch’ blocks, but the actual throwing of exceptions is very expensive. Because of this, you should only be throwing exceptions in situations where an unexpected event is occurring. Unexpected events are presumably rare enough that the performance penalty isn't what's important. H.10 Exceptions in Constructors It is recommended that exceptions should be thrown within constructors. This way it is known that the constructor could not do its job. However, as a constructor has no explicit return type, there is no easy way to report error conditions except using an exception. A lot of care should be taken to make sure that a constructor never leaves an object in a half constructed state. It is generally unsafe to throw exceptions from within a copy constructor. This is because copy constructors often acquire and release resources. If an exception is thrown in the middle of the copy constructor, it could leave the copied object in an unpredictable state. H.11 Throwing Exceptions in Destructors Destructors should be exception-safe. This is because a destructor will be automatically called when an exception is thrown to de-allocate all objects that are on the stack between the exception throw and catch. If a destructor ends up throwing an exception, either directly or indirectly, the program will terminate. To avoid this problem, your destructor really needs to catch any exceptions that might be thrown by calls to helper methods and functions.
- 623. 608 The lesson is that never allow an exception of any sort to be thrown from a destructor, the alternative opens up the possibility that a program with bugs simply terminates without presenting any easily identifiable source for the bug.
- 624. APPENDIX I C++ Namespaces I.1 Introduction As a programmer you often face the problem that global names of some functions, variables, types, and enumerators--declared in one library clash with names found in another library or in your own application. Namespaces are a recent addition to C++, accepted in July 1993 at the Munich meeting of X3J16. Namespace is a named declarative region and resolves such problems in C++. For example, a function name in a graphics library from a source say ‘x’ and a math library from other source say, ‘y’ can both have a method called Circle::draw(int radius, a, b). You are writing a program that needs to integrate classes from Library ‘x’ as well as library ‘y’, but the compiler generates an error because the two class objects have the same name. You cannot modify the class libraries because the software vendor does not provide the source code. What you have is only library of object code. The solution to such problem is that the software vendor places each set of classes in its own namespace with its own name, which must be used to qualify the name of its members. For instance, the Circle object that is part of the graphics library might exist in a namespace entitled ‘x’, while the object that is part of the Maths library might exist in a namespace say, ‘y’. Thus, circle function from source ‘x’ and ‘y’ can be invoked by x_graphics::Circle::draw, and y_math::Circle::draw respectively. Even if you are using a library that predates namespaces, you can use namespaces to encapsulate names in your own code that clash with those of the library. A namespace definition is syntactically a declaration, but it may only appear at the global scope or within another namespace definition. See the declarations given below: • Namespace definition namespace identifier { ... List of zero or more declarations } • Extending a namespace namespace x { // Declare ‘x’ as a namespace. int f() { ... } typedef int T; }
- 625. 610 ... namespace y { // Add more to namespace ‘x’. int g(); } If the namespace identifier is not declared in the current scope, it is declared as a namespace. If it has already been declared, the definition "continues" the namespace. In this way, a namespace can be extended to allow a single large namespace to be defined over different header files. • Use of long names for namespaces to minimize potential clashes namespace Distributed_System_Object_Model_Library { ... } • using a shorthand name or alias namespace DSOML = Distributed_System_Object_Model_Library; A namespace-aliasing definition allows you to use a shorthand name ‘DSOML’ instead of long name as shown above. I.2 Using the Names The names declared in a namespace can be denoted explicitly using the existing qualified name syntax for classes. For instance, Colors::invert denotes a member invert of namespace Colors. namespace Colors { enum color {red, green, blue}; typedef int color; color invert(color); extern color c; }; ... func() { Colors::color C = ...; Colors::invert(C); } A using declaration declares a name N from a namespace as a current declaration in the scope in which the using declaration appears. The declaration of N is an alias of its declaration in the namespace. For example, a using declaration uses a qualified name to refer to the used name.
- 626. 611Appendix I func() { using Colors::red, Colors::green, Colors::blue; Colors::color a = red; using Colors::invert; a = invert(a); } A using declaration can introduce a function that overloads other functions in the same scope. And just as with normal declarations, a using declaration can cause a duplicate declaration if a nonfunction of the same name has already been declared; func() { extern int invert(int); using Colors::invert; // invert is overloaded. using Other_namespace::invert;// OK if invert is a function. Colors::color C = ...; invert(C); //1 Selects from 3 inverts. typedef int T; using Other_namespace::T; // Error. } Unlike normal declarations, using declarations can overload functions with the same argument types. The ambiguity is detected at the point of use, rather than the point of declaration. Suppose Other_namespace::invert had the same signature as Colors::invert; then for //1 invert(C), an error message would result. If ambiguity exists between two functions with identical argument types, where one function was introduced by a using declaration and the other by a non-using declaration, the non-using declaration is preferred over the using declaration. extern int invert(Colors::color); using Colors::invert; // invert is now overloaded. Colors::color C = ...; invert(C); Comments Function invert (C) calls the extern int function rather than the function introduced by // the using declaration and no warning is generated. I.3 The using Directive Namespace can be defined by placing a using namespace statement in the source file that references the namespace’s member object. For example, the following statement instructs the compiler that the file uses members of the ‘x’ namespace.
- 627. 612 using namespace x; A using directive is quite different from a using declaration. The directive imports an entire namespace at once, but not as declarations in the scope containing the directive. A using directive begins with using namespace followed by the name denoting the namespace, e.g. func() { using namespace Colors; color C = red; invert(C); } The names are treated as though they were declared globally, so local declarations of the same name, hide the global declarations. The local declaration //1 hides the global; //2 invokes //1Other_namespace::invert. func() { using namespace Colors; using Other_namespace::invert; //1 local declaration color C = red; invert(C); //2 gets //1. } Case 3: Code below introduces two Global namespace members (see //1 and //2). At //3 there is the choice of two globally declared invert functions. As stated above, ambiguity between two functions with identical argument types is resolved in favor of a function that was not introduced by a using directive. Similarly, ambiguity that cannot be resolved via overload resolution is reported at the use of a name, not at the occurrence of the using directive. func() { using namespace Colors; //1 invert declared globally using namespace Other_namespace;//2 another global invert color C = red; invert(C); //3 chooses between 1 and 2 } Case 4: Names introduced by using directives are ignored when an expression uses explicit qualification. Thus, even though the names are treated as being declared globally, a reference ::global_name, as in code given below, does not refer to any names introduced by using directives.
- 628. 613Appendix I namespace A { int g; } using namespace A; int g; // OK, now we have two of them globally. func() { g++; // Ambiguous: A::glob or ::glob. ::g++; // Refers to global, non-namespace glob. A::g++; // Refers to namespace A's glob. } ANSI C++ and Namespace std All the identifiers in the ANSI standard header files are part of the namespace. In ANSI C++, cin and cout are written as std::cin and std::cout . If you do not wish to specify std:: with cin or cout (or any of the ANSI standard identifiers), you must write the following statement in your program: using namespace std; Use "using namespace std" at the top of each file, outside of any function - this will make the standard library facilities available throughout the program. However, it does make globally visible a lot of function and variable names that you're probably not going to use. For example, we use a graphics include file that defines a value RED. Unfortunately RED is also defined in the std namespace, thus causing a potential name-clash when a global "using namespace std" is used. Program I.1 #include <iostream> using namespace std; namespace x { int a, b, c; } int main() { x::a = 1; x::b = 2; x::c = 3; cout << "The values are:n"; cout << x::a << " " << x::b << " " << x::c << endl; return 0; } Program Output The values are: 1 2 3
- 629. 614 Comments In Program 1, the variables a, b, and c are defined in the scope of ‘x’ namespace. Each time the program accesses one of these variables, x:: must precede the variable name. Otherwise, a compiler error will occur. I.4 Example of Name clashes If you have a local integer with the same name as a global integer, then the local variable simply masks the global one - it's not an error. Neither is it an error to have 2 sufficiently different types of things with the same name - you can't have a string and an integer with the same name in the same function, but the following (which uses i for 2 different purposes) is legal, neither usage interfering with the other. int main() { class i {int j;}; enum letter {i, j}; letter l=i; return 0; } There are cases where the compiler gives an error message when it can't resolve the situation. The following program is written to illustrate some potential sources of conflict. Though it uses many i symbols it compiles on our system without error. The comments show some conflicts which would cause trouble. It might be worth trying to compile the code with and without the changes to see what your compiler says. #include <iostream> using namespace std; namespace extra { int i; } void i(){ // Compilation would fail were the next line "using extra::i" using namespace extra; int i; // this i masks the i in the 'extra' namespace i=9; cout << i << endl; }
- 630. 615Appendix I int main() { enum letter { i, j}; // Compilation would fail were the next line "class i { letter i; };" class i { letter j; }; // Can you create an object called 'j' of type i by doing "i j;" ? // Compilation would fail were the next line "i();" ::i(); return 0; } I.5 Friend and extern Declarations Friends first declared in a class are declared in the smallest enclosing nonclass, nonfunction - prototype scope. With the addition of namespaces, this means a friend declaration like that in code given below can be introduced into the namespace. The same holds for an extern declaration first declared within a function in a namespace. Rather than being a global extern, the declaration is owned by the namespace. namespace x { class C { // This is x::func. friend func(); }; } int x::func() { ... } I.6 Unnamed Namespaces Identifier in a namespace definition can be omitted, in which case you are then referring to that "unnamed" namespace that exists for every compilation unit e.g. ‘std’ in C++. The unnamed namespace is unique for that compilation unit, and a using directive for it is automatically assumed. The resultant effect is that you can enclose your local data in the unnamed namespace without fear of it clashing with local data in other compilation units. Code segment (i) below is equivalent to having written to (ii), as in C++ ‘std’ is a compiler-chosen name unique to each compilation unit.
- 631. 616 (i) namespace { int a, b, c; void f() { ... } } (ii) namespace std { int a, b, c; void f() { ... } }; using namespace std;
- 632. IINNDDEEXX SYMBOLS # (number sign) 16 Operators && (AND) logical operator 33 !(NOT) logical operator 33 || (OR) logical operator 33 & (address of) operator 231 + (arithmetic) operator 28 = (assignment) operator 20 -- (decrement) operator 34 / (division) operator 28, 29 . (dot) operator 192 == (equal) operator 32, 33 >> (insertion) operator 27 ++ (increment) operator 34 << (extraction) operator 27 * (multiplication) operator 28, 29 != (not equal to) operator 32 % (remainder) operator 28, 29 - (subtraction) operator 28, 29 -A- abstract classes 274 abstraction 281, 334 access specifiers default 22, 24, 83 public access specifier 336,353 protected access specifier 336,353 private access specifier 336,353 access modifier ‘const’ 25 accessing array elements 146 functions 118 pointers 234 structures 192 class members 275 characters in string objects 162 access modifiers 25 ‘Ada’ 7 aggregate 509 ‘Algol’ 6 algorithm 45, 46 ampersand(&) 231, ANSI 531 AND logical operator 33 append 448, 460 API 549 ‘APL’ 6 arguments default arguments 120 constant arguments 121 arithmetic operators 28 array array declaration 144 array of pointers 252 arrow operator 192
- 633. 618 addition operator(+) 28 address of operators (&) 231 ASCII (American standard code for information exchange) 531 asterix (*) 28 automatic type conversion 38 assembler 5 attribute 272 -B- base 583 base class 334 ‘BASIC’ 6 Bifurcation of control loops 81 binary files 458-470 binary operator 34 binary number 584 binary search 155, 594 bit-maps 574 block of instructions 16,70 boolean boolean logical AND(&) 33 boolean logical OR operator 33 borland C++ 545 borland C++ builder 545 braces 15, 71 break keyword 23 break instruction 81 bubble sort 159 buffer 446 built in data types 18 bytes 18 -C- C programming language 8 C++ 10 C# 9 C++ keywords 23 C++ standard libraries 535 call by reference 122-126 call by value 122 cascading of operators 27 cascading style sheets(css) 599 casting 37 char 18 character character array 161 character set 17 character string 161 cin operator 14 circular/ring linked list 387 class abstract classes 274 base classes 334 conditional control structures 70 conditional expression 35 conditional operator (?:) 35 <conio.h> header file 37 const keyword 23 constant 24, 25 constant variable 24, 25 container 535 continue expression 82 continue keyword 23 control structures if then else 70-73 while loop 74-76 do while loop 76-78 for loop 78-81 switch case construct 83 control characters 531 constructor 290-301 default constructor 292 parameterized constructor 293
- 634. 619 INDEX class hierarchy 335 class keyword 23,24 class members 281 stream classes 442 close a file 448 COBOL 5 comments 16 compile a program 545-548 concepts and modeling 538 concrete class 274 copy constructor 294 constructor overloading 295 count 510 compiler 528-529 Borland compilers 545 Visual C++ 546 cout (<<) 16, 27, 28 creating link lists 378 text files 443 binary files 460 -D- Data data abstraction 281 data hiding 281 data in abstract class 274 data member 274,280 data structure 357-374 data type hierarchy 38 data types 17, 18 databases 489 decrement 34 decrement operator (--) 34 default case (switch case construct) 84 default keyword 24 # define 24 delete delete operators 250, 251, 252 deleting data in SQL 513 delete nodes from a link list 386 deque (dynamic queue) 409 deque/dequeue operation 412 dereferencing a pointer 235 devised context 559 difference engine 4 derived class 334-336 destructor 290, 301 directives 16 division 28, 29 division by zero 37 do…while 76 document type definition (DTD) 599 DOM 601 dot operator(.) 192 double 18 double backslash() 16-17 double keyword 24 drawing mode 570 drawing with GDI functions 566 dynamic data structure 376,379 -E- ENIAC 4 encapsulation 273, 280 endl manipulator 36 end of file (EOF) 447, 448 entity 493 egyptian numbering 583 else keyword 23 exception in constructors 607 exceptions in destructors 607 exception throwing 604 exception specifications 606 exception types 604 exit function 109 extended character set 533
- 635. 620 equal to (==) relational operator 32 Errors 37 escape characters 25, 26 escape sequence 25 exceptions 603 exception catching 605 extension 574 extraction operator 27 -F- fibonacci series 141 files 443-485 file object 448 closing of files 448 opening of files text files 443-449 binary files 458-466 float keyword 23 float variable 18 floating point variables 22 flow charts 46 for keyword 23 for loops 73 formal parameters 124 FORTRAN 5 Friend function 319 fstream class 443 Functions Function body 117 Function call 119 Function prototype 118 Return type of a function 118 Function name 115 Function overloading 282 Function header 115 Function ambiguity 283 -G- GDI coordinate system 564 get member function 162 getch member function 37 getline member function 446 global variable 127 goto statement 83 goto keyword 23 Graphics device interface 559 GUI 549 H-K hand (base ‘5’) 584 header files 16 hexadecimal 584 high level language 8, 12 identifier 17 if selection statement 70 ifstream 446 # include preprocessor directive 16 increment 34 increment operator(++) 34 inline functions 126 int 18 int keyword 23 integer 18 integer constant 25 integer arithmetic 28 <iomanip.h> header file 35 ios base class 442 iostream library 442 character functions
- 636. 621 INDEX indirection operator (*) 235 infix expression 405 infix to postfix 407 inheritance Single 343 Multiple 343 inheritance hierarchy 335 input stream 442 I/O header file ‘iostream.h’ 16 isalnum 131 isalpha 131 isdigit 131 isupper 131 toupper 131 iterator 535,536 ‘Java’ 8 jump statements 70,71,108 Keyword 23 -L- Languages 4 Last In First Out (LIFO) 390 Last In Last out (LILO) 409 Less than operator ‘<’ 32 Less than or equal to operator ‘<=’ 32 LIFO data structure 391 linear search 152 link lists 375-440 ‘LISP’4 logic errors 37 logical operators 32 and operator (&&) 33 or operator (||) 33 not operator (!) 33 local variables 127 long long data type 18 long keyword 23 M-N machine language 4 <math.h> header file 132 main keyword 23 main function 15 mapping 174 mathematical functions fabs 132 abs 133 log 133 log10(y) 133 pow(x,y) 133 member functions 274 memory address 230, 245, 17 memory allocation 229,17 merging of arrays 157 Microsoft visual C++ 546, 547 multiple inheritance 343 multi list 388 namespaces 609 namespaces ‘using directive’ 611 ‘using names’ 610 nested ‘if’ then ‘else’ structure 73 Structures 208 new keyword 23 new operator 250 node 376 normalization 496 not equal to (!=) 32,33 logical not operator (!) 32,33 Null keyword 23 number system 58
- 637. 622 O-Q object 274 object data model 492,493 object oriented programming 271-330 object oriented design 272 object oriented language 11 octal 584 ofstream class 441,442 opening a binary file 458-470 a text file 443-458 operator operator keyword 23 operator precedence 29 ‘Or’ logical operator 33 ostream class 442,443 overloading 282, 295 palletes 572 ‘Pascal’ 7 parameters 120,121,122,119,118 parameterized constructor 293 pass by value 119 pass by reference 122 passing an array 149 passing structures 203 structure members 205 pointers to structures 207 ‘Perl’ 9 ‘PL/I’7 pointer pointer arithmetic 241 pointer expression 241 arrow operator 192 pointer variables 230 pointer diagrams 233,234 pointer to an array 240 pointer to pointers 245 pointers with functions 248 array of pointers 252 pointer to a structure 253 polymorphism 281 pop 394 postfix operators 42 postfix expression/notation 42 ‘pow’ member function 133 precedence precedence of operators 29 precedence of datatypes 38 preprocessor directives 16 printable character set 532,533 private 336 prefix operators 42 procedural programming 12, 272 program statements 16 protected 336 psuedocode 46 public 336 push 391 ‘put’ member function 175 queue deque/dequeue 409 dynamic queue 417 enqueue 409 front 409 rear 409 static queue 410
- 638. 623 INDEX R-S read from a file 446, 459 read member function 461 recursion 589 recursive functions 589 reference parameters 122 refinement 539 register keyword 23 remainder/modulus operator 28 repetitive statements 50 return statement 119 relational database terminology 493 relational operators less than operator (<) 506 greater than (>) 506 less than or equal to (<=) 506 greater than equal to (>=) 506 not equal to (!=) 506 runtime error 37 scope of variables 126 scope resolution operator 276.277 setw manipulator 35 selection sort 160 sexagesimal 584 shared program technique 4 ‘Short Code’ 4 short keyword 23 signed keyword 23 ‘Simula’6 sorting 158 ‘SQL’ 503-522 stack dynamic stack 395 push 394 pop 394 STL 535-544 <stdio.h> header file 162 streams 442 string functions strlen 163 strcat 166 strcmp 165 strcpy 164 <string.h> header file 163 structures structure arrays 198 structure definition 188 structure declaration 188 structure variables 189 struct keyword 23 subroutines 4,9 subtraction operator 28 switch keyword 23 switch-case statement 83 syntax errors 37 T-Z ternary operators 34 text files 443-458 then keyword 23 tokens 17 tower of Hanoi 592 type type casting 39, 40 type conversion 37,38 typedef keyword 23 structure variables 189 pointer variables 230 void keyword 23 volatile keyword 23 windows graphics programming 551 while loop 74 XML 597 XSD 602
- 639. 624 two dimensional arrays 170 unary operator 34 union keyword 23 unsigned keyword 23 unsigned data types 41-42 unsigned int 42 unsigned long 42 unsigned char 42 variables integer variables 18 character variables 20 floating point variables 22
- 640. CD’s with the Book This book contains following CD’s: 1. MicrosoftVisual C++ 2005 Express Edition and 2. CD entitled, ‘CD accompanying Book’, containing write-up on Borland C++ 5.5 and MSVisual C++ 2005, Data StructureTutorial, source code from book and additional resources on the web. VVVVVisual C++ 2005 Express Editionisual C++ 2005 Express Editionisual C++ 2005 Express Editionisual C++ 2005 Express Editionisual C++ 2005 Express Edition This write-up has two aims — guide you step at a time in the installation process of Microsoft Visual C++ 2005 Express Edition and show you how to build your first program in it. Installation of Visual C++ 2005 Express Edition 1. WhenyouinserttheCDintothedrive,theinstallationprogramshouldrunautomatically. If it does not run, open ‘My Computer’or ‘Windows Explorer’, go to the CD and double click the ‘setup.exe’ file to start the installation. 2. Click ‘Next’ button to begin the installation. 3. On the next screen you should accept the License agreement and click ‘Next’ button to proceed with the installation. 4. The next screen will show you the installation options. a. It is recommended that you install the MSDN Express Library because that will provide you the help files for the Visual C++ Express compiler. b. SQL Server 2005 Express Edition is a basic version of Microsoft SQL Server Database. Install it if you are going to write programs that require a database system. 5. The next screen will show you the folder where Visual C++ 2005 Express Edition will be installed. Click on ‘next’ button to continue. The installation will now begin and you shall see a progress indicator on the screen.
- 641. 6. After the installation is finished the screen will inform you about completion of installation. Now click the ‘Exit’ button to come out of the setup program. Writing your first program in Visual C++ 2005 Express Edition Writing your first C or C++ program with Visual C++ 2005 Express Edition is very simple. This program will write Hello World on the screen using ‘cout’. Here are the steps that you need to follow: 1. Start Visual C++ 2005 Express Edition by clicking on Start -> Programs ->Visual C++ 2005 Express Edition -> Microsoft Visual C++ 2005 Express Edition. 2. Click on File -> New Project. On the left hand under ‘Project Types’choose ‘Win32’. Ontherighthandchoose‘Win32ConsoleApplication’.Atthebottomyoushouldgivea name to your project so type it in the ‘Name’ box. Click on OK. 3. TheWin32ApplicationWizard will come up. Click on ‘Next’. Make sure you click on ‘EmptyProject’inthenextscreenandthenclickon‘Finish’. 4. Now Visual C++ Express is ready with your project settings and you can add programming code to it. Right click on ‘Source Files’in the ‘Solution Explorer’that is showing on the left and click on ‘Add -> New Item’. 5. In the next screen, under ‘Categories’ on the left choose ‘Code’ and choose ‘C++ File (.cpp)’ on the right. 6. You need to give a name to the C++ file that you are adding to your project. So type it in the‘Name’boxatthebottomandclickon‘Add’.Thenamecanbethesameasthename for your project. 7. Now your C++ file will be open in the code window and you can start writing your program.TypethefollowingcodeintotheC++file. #include<iostream> #include<conio.h> using namespace std ; voidmain() { cout << “Hello Worldn”; getch (); }
- 642. 8. Press ‘F5’to compile, link and run the program.You should see the output ‘HelloWorld’ in a console screen. It should be noted thatVisual C++ Express cannot compile single .C or .CPPfiles. Source code files must be contained within a project. So there are 2 steps to create a program. 1. Create a project 2. Create your .C or .CPP file “inside” the project. Moreinformationavailableathttp://msdn.microsoft.com/vstudio/express/visualc/ PleasedoregisteryourinstallationofVisualC++2005ExpressEdition.Registrationcanbedone online and it is free. To register, start Visual C++ 2005 Express Edition and click on ‘Help -> RegisterProduct’. There are a number of benefits for a registered user and they can be seen at http:// msdn.microsoft.com/vstudio/express/register/