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C Aptitude Questions and Answers

Predict the output or error(s) for the following:
1.    void main()
            int  const * p=5;

Compiler error: Cannot modify a constant value.
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]);

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].

3.      main()
            float me = 1.1;
            double you = 1.1;
printf("I love U");
                        printf("I hate U");

I hate U

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--);

5 4 3 2 1

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;     }
printf(" %d ",*p);
++p;     }
                        2 2 2 2 2 2 3 4 6 5

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.

6.      main()
            extern int i;
Linker Error : Undefined symbol '_i'
                        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 .

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