c question set 1

(Q-1)
#include<stdio.h>
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.

(Q-2)

#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.

(Q-3)

#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);

}

Note:

100; is an executable statement but with no action. So it doesn't give any problem

(Q-4)

main()

{

 int i=400,j=300;

 printf("%d..%d");

}

Answer:  400..300

Explanation:

printf takes the values of the first two assignments of the program. Any number of printf's may be given.

 All of them take only the first two values. If more number of assignments given in the program,then printf 

will take garbage values.

(Q-5)

main()

{

    int i=1;

    while (i<=5)

    {

       printf("%d",i);

       if (i>2)

 goto here;

       i++;

    }

}

fun()

{

   here:

     printf("PP");

}

Answer: Compiler error: Undefined label 'here' in function main

Explanation:

Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is

available in function fun() Hence it is not visible in function main.