Aptitude in C language

1.          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.

2.        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".

3.      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, indirecting it with * is same as s[i]. i[s] may be surprising. But in the  case of  C  it is same as s[i].

4.      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.

5.      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 .