Predict the output (or) error(s) for the following 1 - 5

1.      void main()

            {

            int const * p=5;

            printf("%d",++(*p));

            }

Answer:

            Compiler error: Cannot modify a constant value.

Explanation:

            p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.    main()

            {

            char s[ ]="man";

            int i;

            for(i=0;s[ i ];i++)

            printf("\n%c%c%c%c",

            s[ i ],*(s+i),*(i+s),i[s]);

            }

Answer:

mmmm

aaaa

nnnn

Explanation:

            s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address.    I is the index number/displacement from the  as address. So, in directing it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3.     main()

            {

            float me = 1.1;

            double you = 1.1;

            if(me==you)

            printf("I love U");

            else

            printf("I hate U");

            }

Answer:

            I hate U

Explanation:

            For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Rule of Thumb:

            Never compare or at-least be cautious when using floating point numbers with relational operators.                           (== , >, <, <=, >=,!= ) .

4.     main()

            {

            static int var = 5;

            printf("%d ",var--);

            if(var)

            main();

            }

Answer:

            5 4 3 2 1

Explanation:

            When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like

any other ordinary function, which can be called recursively.

5.     main()

            {

            int c[ ]={2.8,3.4,4,6.7,5};

            int j,*p=c,*q=c;

            for(j=0;j<5;j++)

             {

            printf(" %d ",*c);

            ++q;

             }

            for(j=0;j<5;j++)

            {

            printf(" %d ",*p);

            ++p;

            }

            }

Answer:

            2 2 2 2 2 2 3 4 6 5

Explanation:

            Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.