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Sample Technical Questions

C Sample Question

49. main( )
    {
     char  *q;
     int  j;
     for (j=0; j<3; j++) scanf("%s" ,(q+j));
     for (j=0; j<3; j++) printf("%c" ,*(q+j));
     for (j=0; j<3; j++) printf("%s" ,(q+j));
    }

    Explanation:

    Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

    M

    O

    U

    S

    E

    \0

    When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

    M

    T

    R

    A

    C

    K

    \0

    The third input starts filling from the location 102

    M

    T

    V

    I

    R

    T

    U

    A

    L

    \0

    This is the final value stored .
    The first printf prints the values at the position q, q+1 and q+2 = M T V
    The second printf prints three strings starting from locations q, q+1, q+2
    i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

50. 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'.

51. 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'.

52. main()
    {
     int  i, n;
     char *x = "girl";
     n = strlen(x);
     *x = x[n];
     for(i=0; i‹n; ++i)
       {
          printf("%s\n",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 ('\0') to the first location. Now the string becomes "\0irl" . 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] = '\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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