Predict the output (or) error(s) for the following 6-10

6.     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 output.

7.     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.

8.    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.her program with memory space allocated for it. Hence a linker error has occurred

9.     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.

10.    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.