3 Function Overloading

WEL COME
PRAVEEN M JIGAJINNI
PGT (Computer Science)
MTech[IT],MPhil (Comp.Sci), MCA, MSc[IT], PGDCA, ADCA,
Dc. Sc. & Engg.
email: praveenkumarjigajinni@yahoo.co.in

CHAPTER 3
Function
Overloading

3
Introduction
The polymorphism refers to ‘one name having many
forms’ ‘different behaviour of an instance
depending upon the situation’. C++ implements
polymorphism through overloaded functions and
overloaded operators. The term ‘overloading’
means a name having two or more distinct
meanings. Thus, an ‘overloaded function’ refers
to a function having (one name and) more than
one distinct meanings. Similarly, when two or
more distinct meanings are defined for an
operator, it is said to be an ‘overloaded operator’.

4
Function Overloading
A function name having several definitions that are
differentiable by the number or types of their arguments.
OR
Function Overloading not only implements
polymorphism but also reduces number of comparisons
in a program and thereby makes the program run faster.
For example;
float divide (int a, int b);
float divide (float x, float y);

5
Declaration and Definition
The key to function overloading is a function’s
argument list which is also known as the
function signature. It is the signature, not the
function type that enables function overloading.
Note:
A function’s argument list (i.e., number and type of
argument) is known as the function’s signature.

6
If two functions are having same number and types
of arguments in the same order, they are said to
have the same signature. Even if they are using
distinct variable names, it doesn’t matter. For
instance, following two functions have same
signature.
void squar (int a, float b); //function 1
void squar (int x, float y); //same function as
that of function 1

7
To overload a function name, all you need to do is,
declare and define all the functions with the
same name but different signatures, separately.
For instance, following code fragment overloads
a function name prnsqr( ).
Void prnsqr (int i);
Void prnsqr (char c);
Void prnsqr (float f);
Void prnsqr (double d);
//overloaded for floats #3
//overloaded for double floats #4
//overloaded for character #2
//overloaded for floats #3
//overloaded for integer #1

8
void prnsqr (int i)
{cout<<“Integer”<<i<<“’s square is”<<i*i<<“n”;
}
void prnsqr (char c);
{cout<<c<<“is a character”<<“Thus No Square for
it”<<“n”;
}
Void prnsqr (float f)
{cout<<“float”<<f <<“’s square is”<<f *f<<“n”;
}
void prnsqr (double d)
{cout <<“Double float”<<d<<“’s square
is”<<d*d<<“n’;
}
After declaring overloading functions, you must
define them separately, as it is shown below for
above given declarations.

9
Thus, we see that is not too much difficulty
in declaring overloaded functions; they are
declared as other functions are. Just one thing is
to be kept in mind that the arguments are
sufficiently different to allow the functions to be
differentiated in use.
The argument types are said to be part of
function’s extended name. For instance, the
name of above specified functions might be
prnsqr() but their extended names are
different. That is they have prnsqr(int),
prnsqr(char), prnsqr(float), and prnsqr(double)
extended names respectively.

10
When a function name is declared more than once in a
program, the compiler will interpret the second (and
subsequent) declaration(s) as follows:
1) If the signatures of subsequent functions match the
previous function’s, then the second is treated as a re-
declaration of the first.
2) If the signatures of two functions match exactly but
the return type differ, the second declaration is treated
as an erroneous re-declaration of the first and is
flagged at compile time as an error.
For example,
float square (float f);
double square (float x); //error

11
Functions with the same signature and same name
but different return types are not allowed in C++.
You can have different return types, but only if
the signatures are also different:
float square (float f);
double square (double d);
//different signatures, hence
//allowed

12
3) If the signature of the two functions differ in
either the number or type of their arguments,
the two functions are considered to be
overloaded.
Use function overloading only when a function is
required to work for alternative argument
types and there is a definite way of optimizing
the function for the argument type.

13
Restrictions on Overloaded
Functions
Several restrictions governs an acceptable set of
overloaded functions:
Any two functions in a set of overloaded
functions must have different argument lists.
Overloading functions with argument lists of the
same types, based on return type alone, is an
error.
Member functions cannot be overloaded solely
on the basis of one being static and the other
nonstatic.

14
Typedef declaration do not define new types;
they introduces synonyms for existing types.
They do not affect the overloading mechanism.
Consider the following code:
typedef char* PSTR;
void Print (char * szToPrint);
void Print (PSTR szToPrint);

15
CALLING OVERLOADED
FUNCTIONS
Overloaded functions are called just like other
functions. The number and type of arguments
determine which function should be invoked.
For instance consider the following code fragment:
prnsqr (‘z’);
prnsqr (13);
prnsqr (134.520000012);
prnsqr (12.5F);

16
Steps Involved in Finding the
Best Match
A call to an overloaded function is resolved to a particular
instance of the function through a process known as
argument matching, which can be termed as a process of
disambiguation. Argument matching involves
comparing the actual arguments of the call with the
formal arguments of each declared instance of the
function. There are three possible cases, a function call
may result in:
a) A match. A match is found for the function call.
b) No match. No match is found for the function call.
c) Ambiguous Match. More than one defined
instance for the function call.

17
1. Search for an Exact Match
If the type of the actual argument exactly matches
the type of one defined instance, the compiler
invokes that particular instance. For example,
void afunc(int);
void afunc(double);
afunc(0);
0 (zero) is of type int , thus the call exactly matches
afunc(int).
//overloaded functions
//exactly match. Matches afunc(int)

18
2. A match through promotion
If no exact match is found, an attempt is made to
achieve a match through promotion of the actual
argument.
Recall that the conversion of integer types (char,
short, enumerator, int) into int (if all values of
the type can be represented by int) or into
unsigned int (if all values can’t be represented
by int) is called integral promotion.

19
For example, consider the following code
fragment:
void afunc (int);
void afunc (float);
afunc (‘c’); //match through the promotion;
matches afunc (int)

20
3. A match through application of
standard C++ conversion rules
If no exact match or match through a promotion is
found, an attempt is made to achieve a match
through a standard conversion of the actual
argument. Consider the following example,
void afunc (char);
void afunc (double);
afunc (471); //match through standard
conversion matches afunc
(double)

21
The int argument 471 can be converted to a double value
471 using C++ standard conversion rules and thus the
function call matches (through standard conversion)
func(double).
But if the actual argument may be converted to multiple
formal argument types, the compiler wil generate an
error message as it will be ambiguous match. For
example,
void afunc (long);
void afunc (double);
afunc(15);
Here the int argument 15 can be converted either long or
double, thereby creating an ambiguous situation as to
which afunc() should be used.
//Error !! Ambiguous match

22
4. A match through application of a
user-defined conversion.
If all the above mentioned steps fail, then the
compiler will try the user-defined conversion in
the combinations to find a unique match.
Any function, whether it is a class member or just
an ordinary function can be overloaded in C++,
provided it is required to work for distinct
argument types, numbers and combinations.

23
Default Arguments Versus
Overloading
Using default argument gives the appearance of
overloading, because the function may be called
with an optional number of arguments. For
instance, consider the following function
prototype:
float amount (float principal, int time=2, float
rate=0.08);

24
Now this function may be called by providing just one
or two or all three argument values. A function call
like as follows:
cout<<amount (3000);
will invoke the function amount() with argument
values 3000, 2, and 0.08 respectively. Similarly a
function call like
cout <<amount (3000,4);
Will invoke amount() with argument values 2500, 5,
and 0.12 respectively. That is if argument values are
provided with the function call, then the function is
invoked with the given values. But if any value is
missing and there has been default values specified
for it, then the function is invoked using the default
value.

25
However if you skip the middle argument time
but C++ makes no attempt at this type of
interpretation. C++ will take 0.13 to be the
argument value for time and hence invoke
amount() with values 2000, 0 (0.13 converted to
int, thus 0) and 0.08 (the default rate). That is,
with default arguments C++ expects that only
the arguments on the right side can be defaulted.
If you want to default a middle argument, then
all the arguments on its right must also be
defaulted.

3 Function Overloading

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