1. What would be the output
main( )
{
int i = 10 ; float a = 3.14 ; char ch = ’z’ ;
j = &i ; b = &a ; dh = &ch ;
k = &j ; c = &b ; eh = &dh ;
printf ( ” %d %d %d
” , j , b , dh ) ;
printf ( ” ” , *k ,*c , *eh ) ;
%d %d %d
i
a ch
j b dh
4000 5000 6000
4000 5000 6000
500 1500 2500
1000 2000 3000
printf ( ” ” , sizeof ( i ), sizeof ( a ), sizeof ( ch ) ) ;
%d %d %d
10
3.14 z
1000 2000 3000
1000 2000 3000
k c eh
1000 2000 3000
2 4 1
CCoonnttiinnuueedd......
int *j, **k ; float *b, **c ; char *dh, **eh ;

2. i
a ch
j b dh
4000 5000 6000
4000 5000 6000
500 1500 2500
2 2 2
printf ( ”%d %d %d ”, sizeof ( j ), sizeof ( b ), sizeof ( dh ) ) ;
printf ( ”%d %d %d”, sizeof ( k ), sizeof ( c ), sizeof ( eh ) ) ;
printf ( ”%d %f %c” , **k, **c, **eh ) ;
}
10
3.14 z
1000 2000 3000
1000 2000 3000
k c eh
2 2 2
10 3.14 z
......CCoonnttiinnuueedd

3. Truly Speaking
main( )
{
int i = 25 ;
float a = 3.14 ;
j = &i ;
int *j ;
float *b ;
b = &a ;
}
printf ( ”%d%f”, *j, *b ) ;
i
Binary of 25 401
401 402
a
501 502 503 504
j
b
Binary of 3.14 501

4. Accessing Individual Bytes
a
main( )
{
float a = 3.14 ;
501
502
float *b ; int *c ; char *d ;
b c d
b = &a ;
c = &a ;
d = &a ;
printf ( ” %f %d %d
”, *b, *c, *d ) ;
printf ( ”%d %d %d %d”, *d, *(d+1), *(d+2), *(d+3) ) ;
printf ( ”%d %d”, *( c+1 ), *( c+2 ) ) ;
}
503 504
501
0’s &
1’s
0’s &
1’s
0’s &
1’s
0’s &
1’s
501
501

5. main( )
{
int i = 25 ; int *j ;
float a = 3.14 ;
float *b ;
char far *kb ;
kb = 0x417 ;
j = &i ;
b = &a ;
}
Why far and near

6. i a j b kb
25 3.14 200 300
200 300
0x417
Code Segment
Turbo C
DOS/Windows
BDA 0x417
IVT
0xB8000000
Screen
M A
0xB8000000
M
A B C D E F
384 KB
High Memory
640 KB
Base or
Conventional
Memory
Data
Segment
Free Memory
(640 + 64 ) * 1024
HEX
64 * 6

7. main( )
{
Pointer Sizes
int i = 25 ; int *j ;
float a = 3.14 ;
float *b ;
char far *kb ;
kb = 0x417 ;
j = i ;
b = a ;
}
2 2
printf ( ”%d %d %d”, sizeof ( j ), sizeof ( b ),
sizeof ( kb ) ) ;
4

8. How Much Memory
main( )
{
int far *m ;
m = 0x413 ;
printf ( ”%d” ,*m ) ;
} 0x413 0x414
640
640 KB
BDA
Ideally
632
634
636
?

9. main( )
{
int a = 10, b = 20 ;
printf ( ”%d%d”, a, b ) ;
swapv ( a , b ) ;
printf ( ”%d%d”, a, b ) ;
}
swapv ( int x, int y )
{
}
int t ;
t = x ; x = y ; y = t ;
printf ( ”%d%d”, x, y ) ;
10 20
10 20
a b
x y t
x y t
x y t
x y t
10 20 g
10 20 10
20 20 10
20 10 10
20 10
10 20

10. swapr ( a , b ) ;
printf ( ”%d%d”, a, b ) ;
swapr ( )
{
}
int t ;
t = *x ;
*x = *y ;
*y = t ;
10 20
20 10
a b
20
20 10
300
x y t
main( )
{
int a = 10, b = 20 ;
printf ( ”%d%d”, a, b ) ;
int *x , int *y 200 } 10
200 300 10

11. main( )
{
int a = 10, b = 20, c = 30
sumprod ( a, b, c, ) ;
printf ( ”%d%d”, s, p ) ;
}
, s, p ;
60 6000
sumprod ( int x, int y, int z, )
{
}
x + y + z ;
x * y * z ;
s, p
int *ss, int*pp
ss =
pp =
*
*
a b c s p
6G0 60G00
100 200
10 20 30
x y z ss pp
10 20 30 100 200

12. Advanced Features of
Functions
• Returning Non-integer Value
• Call By Value / Reference
• Recursion

13. To Be A Successful Software
Developer
Creative Intelligence
Analytical Ability
Patience
Ability To Think