C Technical Interview Questions and Answers

Predict the output or error(s) for the following:

43. #ifdef something
int some=0;
#endif
 
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
 
Answer:
Compiler error : undefined symbol some

Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration 
int some = 0;
effectively removed from the source code.

44.  #if something == 0
int some=0;
#endif
 
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
 
Answer
0 0 

Explanation
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.

45. What is the output for the following program
 
            main()
                            {
      int arr2D[3][3];
       printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
               }
Answer
1

46.    void main()
         {
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
         } 

 Answer
You can answer this if you know how values are represented in memory

Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

47. int swap(int *a,int *b)
{
 *a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
                        int x=10,y=20;
            swap(&x,&y);
                        printf("x= %d y = %d\n",x,y);
}

Answer
            x = 20 y = 10

Explanation
This is one way of swapping two values. Simple checking will help understand this.

48.    main()
{          
char *p = “ayqm”;
printf(“%c”,++*(p++));
}

Answer:
b

49.      main()
            {
             int i=5;
             printf("%d",++i++);
} 

Answer:
Compiler error: Lvalue required in function main
Explanation:
                        ++i yields an rvalue.  For postfix ++ to operate an lvalue is required.

50.     main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}

Answer:
b

Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.

51.
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
 
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}

Answer:
 bye

Explanation: 
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of

the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing  "bye".

52.
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
 
Answer:
1

Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.