C Program Samples-5

1.#define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:

64

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

2.main()

{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s   %s",p,p1);
}

Answer:

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

• *p that is value at the location currently pointed by p will be taken
• ++*p the retrieved value will be incremented
• when ; is encountered the location will be incremented that is p++ will be executed

Hence, in the while loop initial value pointed by p is 'h', which is changed to 'i' by executing ++*p and pointer moves to point, 'a' which is similarly changed to 'b' and so on. Similarly blank space is converted to '!'. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches '\0' and p1 points to p thus p1doesnot print anything.

3.#include‹stdio.h›

#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:



50
Explanation:



The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

4.#define clrscr() 100

main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :

main()
  {
       100;
       printf("%d\n",100);
  }

5.main()

{
printf("%p",main);
}

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.