Vcs22
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The document describes code for a calendar program in C. It includes functions to calculate the number of days between dates, store months and days in arrays, display the calendar for a given month and year with day names and dates, and allow the user to navigate to the next or previous month or year. Structures are used to store employee records with name, age, and salary fields. Arrays of structures hold multiple records. Pointers to structures are also demonstrated.
![main( )
{
......CCaalleennddaarr
{ 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 } ;
int days[ ] =
.i n.t i, s ; int firstday ; . .
totaldays = normaldays +
leapdays ;
s = 0 ;
for ( i = 0 ; i = ; i++ )
s = s +
totaldays += s ;
firstday = totaldays % 7 ;
. .
}
m - 2
days[ i ] ;
1/1/1 to 31/12/1998
totaldays
1/1/1999 to 31/7/1999
31 + 28 + 31 + 30
+ 31 + 30 + 31](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-2-320.jpg)
![......CCaalleennddaarr
( )
main( )
{
{ 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 } ;
int days[ ] =
.. ..
totaldays = normaldays + leapdays ;
if ( ( y % 400 == 0 ) || )
days[ 1 ] = 29 ;
s = 0 ;
for ( i = 0 ; i = ; i++ )
s = s +
. .
}
m - 2
days[ i ] ;
y % 100 != 0 y % 4 == 0
totaldays += s ;
firstday = totaldays % 7 ;](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-3-320.jpg)
![10
20 26 32
Mon - - -Tue - - -Wed---
......CCaalleennddaarr
main( )
{
.. ..
char *months[ ] = { ”January”, ”February”, … } ;
totaldays += s ;
firstday = totaldays % 7 ;
col = 20 + firstday * 6 ;
clrscr( ) ;
gotorc ( 8 , 3 5 ) ;
printf ( ”%s %d”, months [ m - 1 ], y ) ;
}
gotorc ( 10
, 3 5 ) ;
printf ( ”Mon Tue Wed Thu Fri Sat Sun”
) ;](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-5-320.jpg)
![......CCaalleennddaarr
main( )
{
i n t d a y s [ ] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, … } ;
/* print month year */
/* print days */
for ( i = 1 ; i = days[ m - 1 ] ; i++ )
{
gotorc ( r o w , c o l ) ;
printf ( ”%d”, i ) ;
col = col + 6 ;
if ( col 56 )
{
row++ ; col = 20 ;
} } }
10
20 26 32
Mon - - -Tue - - -Wed---
row = 12 ;](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-6-320.jpg)
![main( )
{
{ 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 } ;
int days[ ] =
scanf ( ”%d%d”, m, y ) ;
while ( 1 )
{
normaldays =
totaldays =
leapdays =
if ( y % 400 = = 0 ) || ( y % 100 ! = 0 y % 4 = = 0 ) )
days[ 1 ] = 29 ;
else
. . days[ 1 ] = 28 ;
} }
9, 1998
9, 2000
9, 1700
9, 1752
- 30
- 30
- 30
- 16
......CCaalleennddaarr](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-8-320.jpg)
![main( )
{
char n[ ] = { ’A’, ’X’, ’Y’, ’0’ } ;
int a[ ] = { 23, 27, 28 } ;
float s[ ] = { 4000.50, 5000.00, 6000.75 } ;
int i ;
for ( i = 0 ; i = 2 ; i++ )
printf ( ”%c %d %f”, n[ i ], a[ i ], s[ i ] ) ;
}
HHaannddlliinngg DDaattaa](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-10-320.jpg)
![main( )
{
AArrrraayy ooff SSttrruuccttuurreess
struct employee
{
char n ;
int a ;
float s ;
} ;
struct employee e[ ]
= {
{ ’A’, 23, 4000.50 } ,
{ ’X’, 27, 5000.00 } ,
{ ’Y’, 28, 6000.75 }
} ;
int i ;
for ( i = 0 ; i = 2 ; i++ )
printf ( ”%c %d %f”, e [ i ]
}
.n, e[ i ].a, e[ i ].s ) ;](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-12-320.jpg)
![Keyword eemmppllooyyeeee
Structure
name/
struct
{
tag
char n ;
int a ;
float s ;
} ;
struct employee e1, e2, e[ 10 ] ;
Structure elements/
members
Structure
variables Array of
structures](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-13-320.jpg)
![Conclusion
A structure is usually a collection of
dissimilar elements.
Structure elements are always stored in
adjacent memory locations.
struct employee e[ 3 ] ;
A 23 400.50 X 27 500.00 Y 28
600.75
401 408 415](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-14-320.jpg)
![AArrrraayy ooff SSttrruuccttuurreess
struct employee e[ ] = { ... } ;
char *p ;
Ptr. to
structure
struct employee e[ 3 ] ;
A 23 400.50 X 27 500.00 Y 600.75
401 408 415
q = e ; r = e ;
p = e ;
p++ ;
printf ( ”%u”, q ) ;
28
struct employee *q ;
struct employee (*r )[3] ;
q++ ; r++ ;
printf ( ”%u”, p ) ;
printf ( ”%u”, r ) ;
440022
440088
442222
Array of
ptrs.](https://image.slidesharecdn.com/vcs22-140830105500-phpapp02/85/Vcs22-15-320.jpg)
Vcs22
- 1. main( ) { int m, y ; long int normaldays ; long int totaldays ; scanf ( ”%d%d”, &m, &y ) ; printf ( ”Enter month and year” ) ; normaldays = ( y - 1 ) * 365 L ; leapdays = ( y - 1 ) / 4 - ( y - 1 ) / 100 + ( y - 1 ) / 400 ; totaldays = normaldays + leapdays . . ; } 8 1999 1/8/1999 - ? 1/1/1 - Mon 1/1/1999 - ? 1/1/1 to 31/12/1998 x % 7 1/8/1999 - ? 1/1/1 to 31/7/1999 1/1/1 to 31/12/1998 +1/1/1999 to 31/7/1999 x % 7 CCaalleennddaarr int leapdays ;
- 2. main( ) { ......CCaalleennddaarr { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 } ; int days[ ] = .i n.t i, s ; int firstday ; . . totaldays = normaldays + leapdays ; s = 0 ; for ( i = 0 ; i = ; i++ ) s = s + totaldays += s ; firstday = totaldays % 7 ; . . } m - 2 days[ i ] ; 1/1/1 to 31/12/1998 totaldays 1/1/1999 to 31/7/1999 31 + 28 + 31 + 30 + 31 + 30 + 31
- 3. ......CCaalleennddaarr ( ) main( ) { { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 } ; int days[ ] = .. .. totaldays = normaldays + leapdays ; if ( ( y % 400 == 0 ) || ) days[ 1 ] = 29 ; s = 0 ; for ( i = 0 ; i = ; i++ ) s = s + . . } m - 2 days[ i ] ; y % 100 != 0 y % 4 == 0 totaldays += s ; firstday = totaldays % 7 ;
- 4. SSccrreeeenn August 1999 10 20 26 32 38 44 50 56 Mon - - -Tue - - -Wed - - -Thu- - -Fri - - -Sat - - -Sun
- 5. 10 20 26 32 Mon - - -Tue - - -Wed--- ......CCaalleennddaarr main( ) { .. .. char *months[ ] = { ”January”, ”February”, … } ; totaldays += s ; firstday = totaldays % 7 ; col = 20 + firstday * 6 ; clrscr( ) ; gotorc ( 8 , 3 5 ) ; printf ( ”%s %d”, months [ m - 1 ], y ) ; } gotorc ( 10 , 3 5 ) ; printf ( ”Mon Tue Wed Thu Fri Sat Sun” ) ;
- 6. ......CCaalleennddaarr main( ) { i n t d a y s [ ] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, … } ; /* print month year */ /* print days */ for ( i = 1 ; i = days[ m - 1 ] ; i++ ) { gotorc ( r o w , c o l ) ; printf ( ”%d”, i ) ; col = col + 6 ; if ( col 56 ) { row++ ; col = 20 ; } } } 10 20 26 32 Mon - - -Tue - - -Wed--- row = 12 ;
- 7. main( ) { . . scanf ( ”%d%d”, m, y ) ; while ( 1 ) { normaldays ____ = ( y - 1 ) * 365L ; c__a__lender gotorc ( 20, 35 ) ; printf ( ”Rt-Next mth…” ) ; ch = getkey( ) ; switch ( ch ) { case 77 : m++ ; if ( m 12 ) { y++ ; m = 1 ; __}__ } } } ......CCaalleennddaarr Next year Prev. mth Next Prev. year mth
- 8. main( ) { { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 } ; int days[ ] = scanf ( ”%d%d”, m, y ) ; while ( 1 ) { normaldays = totaldays = leapdays = if ( y % 400 = = 0 ) || ( y % 100 ! = 0 y % 4 = = 0 ) ) days[ 1 ] = 29 ; else . . days[ 1 ] = 28 ; } } 9, 1998 9, 2000 9, 1700 9, 1752 - 30 - 30 - 30 - 16 ......CCaalleennddaarr
- 9. TTeerrmmiinnoollooggyy Fields Record Database Name Age Salary ’’AA’’ 2233 44000000..5500 ’X’ 27 5000.00 ’Y’ 28 6000.75
- 10. main( ) { char n[ ] = { ’A’, ’X’, ’Y’, ’0’ } ; int a[ ] = { 23, 27, 28 } ; float s[ ] = { 4000.50, 5000.00, 6000.75 } ; int i ; for ( i = 0 ; i = 2 ; i++ ) printf ( ”%c %d %f”, n[ i ], a[ i ], s[ i ] ) ; } HHaannddlliinngg DDaattaa
- 11. main( ) { struct employee { char n ; int a ; float s ; } ; struct employee e 1 = { ’A’, 23, 4000.50 } ; struct employee e2 = { ’X’, 27, 5000.00 } ; struct employee e3 = { ’Y’, 28, 6000.75 } ; printf ( ”%c %d %f”, n, a, s ) ; e1. e1. e1. printf ( ”%c %d %f”, e2.n, e2.a, e2.s ) ; printf ( ”%c %d %f”, e3.n, e3.a, e3.s ) ; } . structure operators SSttrruuccttuurreess
- 12. main( ) { AArrrraayy ooff SSttrruuccttuurreess struct employee { char n ; int a ; float s ; } ; struct employee e[ ] = { { ’A’, 23, 4000.50 } , { ’X’, 27, 5000.00 } , { ’Y’, 28, 6000.75 } } ; int i ; for ( i = 0 ; i = 2 ; i++ ) printf ( ”%c %d %f”, e [ i ] } .n, e[ i ].a, e[ i ].s ) ;
- 13. Keyword eemmppllooyyeeee Structure name/ struct { tag char n ; int a ; float s ; } ; struct employee e1, e2, e[ 10 ] ; Structure elements/ members Structure variables Array of structures
- 14. Conclusion A structure is usually a collection of dissimilar elements. Structure elements are always stored in adjacent memory locations. struct employee e[ 3 ] ; A 23 400.50 X 27 500.00 Y 28 600.75 401 408 415
- 15. AArrrraayy ooff SSttrruuccttuurreess struct employee e[ ] = { ... } ; char *p ; Ptr. to structure struct employee e[ 3 ] ; A 23 400.50 X 27 500.00 Y 600.75 401 408 415 q = e ; r = e ; p = e ; p++ ; printf ( ”%u”, q ) ; 28 struct employee *q ; struct employee (*r )[3] ; q++ ; r++ ; printf ( ”%u”, p ) ; printf ( ”%u”, r ) ; 440022 440088 442222 Array of ptrs.