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.

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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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.

C aptitude 16.1

 main( )
{
 void *vp;
 char ch = ‘g’, *cp = “goofy”;
 int j = 20;
 vp = &ch;
 printf(“%c”, *(char *)vp);
 vp = &j;
 printf(“%d”,*(int *)vp);
 vp = cp;
 printf(“%s”,(char *)vp + 3);
}

Answer: g20fy

Explanation: Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C aptitude 16.2

  main ( )
{
 static char *s[ ]  = {“black”, “white”, “yellow”, “violet”};
 char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
 p = ptr;
 **++p;
 printf(“%s”,*--*++p + 3);
}

Answer: ck

Explanation:In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

C aptitude 16.3

    main()
{
 int  i, n;
 char *x = “girl”;
 n = strlen(x);
 *x = x[n];
 for(i=0; i   {
printf(“%sn”,x);
x++;
   }
 }

Answer:(blank space)
irl
rl
l

Explanation: Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘’) to the first location. Now the string becomes “irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

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