Outputs or errors of given simple programs with explanation (1 to 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 base 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.