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C Sample Question

  1. 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.
  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.
  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
  4. 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.

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