C Aptitude - P2 Solved

1.      main()

{

            extern int i;

            i=20;

printf("%d",i);

}

Answer: 

Linker Error : Undefined symbol '_i'

Explanation:

                        extern storage class in the following declaration,

                                    extern int i;

specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

2.      main()

{

            int i=-1,j=-1,k=0,l=2,m;

            m=i++&&j++&&k++||l++;

            printf("%d %d %d %d %d",i,j,k,l,m);

}

Answer:

                        0 0 1 3 1

Explanation :

Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression  ‘i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

3.      main()

{

            char *p;

            printf("%d %d ",sizeof(*p),sizeof(p));

}

Answer:

                        1 2

Explanation:

The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

4.      main()

{

            int i=3;

            switch(i)

             {

                default:printf("zero");

                case 1: printf("one");

                           break;

               case 2:printf("two");

                          break;

              case 3: printf("three");

                          break;

              } 

}

Answer :

three

Explanation :

The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

5.      main()

{

              printf("%x",-1<<4);

}

Answer:

fff0

Explanation :

-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

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