Jun 12, 2012

C Aptitude Questions and answers with explanation

C Aptitude 20
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

The C aptitude questions and answers for those questions along with explanation for the interview related queries.

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.




Predict the output or error(s) for the following:


C aptitude 20.1
  main()
            {
                int *j;
                {
                 int i=10;
                 j=&i;
                 }
                 printf("%d",*j);
}


Answer: 10

Explanation: The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives up to the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

C aptitude 20.2
   main()
            {
            int i=-1;
            -i;
            printf("i = %d, -i = %d \n",i,-i);
            }


Answer: i = -1, -i = 1

Explanation: -i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

C aptitude 20.3
     main()
 {
   const int i=4;
   float j;
   j = ++i;
   printf("%d  %f", i,++j);
 }

Answer: Compiler error

Explanation: i is a constant. you cannot change the value of constant

C aptitude 20.4
      main()
{
  int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
  int *p,*q;
  p=&a[2][2][2];
  *q=***a;
  printf("%d..%d",*p,*q);
}

Answer: garbagevalue..1

Explanation: p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer.

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