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.

16. 228822-- 228888
CChhaapptteerr 99