C Interview Questions And Answers

C Objective Interview Questions And Answers(10)

C Subjective Interview Questions And Answers(100)     

C syntax, semantics and simple programming questions(61)    

C Aptitude Questions(179) Objectives 

C Aptitude Questions

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C Aptitude Questions  

Questions 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. 

Questions 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.

Questions 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 other program with memory space allocated for it. Hence a linker error has occurred .

Questions 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.

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