C Pointer Basics

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Question 1

Consider the following declaration:

int a, *b=&a, **c=&b;

The following program fragment

a=4;
**c=5;

A
does not change the value of a
B
assigns address of c to a
C
assigns the value of b to a
D
assigns 5 to a

Question 2

Consider the following C program.

#include<stdio.h>
void mystery(int *ptra, int *ptrb) 
{
   int *temp;
   temp = ptrb;
   ptrb = ptra;
   ptra = temp;
}
int main() 
{
    int a=2016, b=0, c=4, d=42;
    mystery(&a, &b);
    if (a < c)
       mystery(&c, &a);
    mystery(&a, &d);
    printf("%d", a);
}

The output of the program _____________ Note : This question was asked as Numerical Answer Type.

A
2016
B
0
C
4
D
8

Question 3

What will be the output produced by the following C code:

int main()
{
    int array[5][5];
    printf("%d",( (array == *array) && (*array == array[0]) ));
    return 0;    
}

A
1
B
0
C
2
D
-1

Question 4

Consider the following C code

int main()
{
   int a = 300;    
   char *b = (char *)&a;
   *++b = 2;
   printf("%d ",a);
   return 0;
}

Consider the size of int as two bytes and size of char as one byte. Predict the output of the following code . Assume that the machine is little-endian.

A
556
B
300
C
Runtime Error
D
Compile Time Error

Question 5

Consider the following function implemented in C:

void printxy(int x, int y)
{
    int *ptr;
    x = 0;
    ptr = &x;
    y = *ptr;
    *ptr = 1;
    printf("%d,%d", x, y);
}

The output of the printxy(1,1) is

A
0,0
B
0,1
C
1,0
D
1,1

Question 6

Consider the following snippet of a C program. Assume that swap(&x, &y) exchanges the contents of x and y.

int main()
{
    int array[] = {3, 5, 1, 4, 6, 2};
    int done = 0;
    int i;

    while (done == 0)
    {
        done  = 1;
        for (i = 0; i <= 4; i++)
        {
            if (array[i] < array[i+1])
            {
                swap(&array[i], &array[i+1]);
                done = 0;
            }
        }
        for (i = 5; i >= 1; i--)
        {
            if (array[i] > array[i-1])
            {
                swap(&array[i], &array[i-1]);
                done = 0;
            }
        }
    }

    printf("%d", array[3]);
}

The output of the program is _____. Note: This question appeared as Numerical Answer Type.

Question 7

Faster access to non-local variables is achieved using an array of pointers to activation records, called a

A
stack
B
heap
C
display
D
activation tree

Question 8

‘ptrdata’ is a pointer to a data type. The expression *ptrdata++ is evaluated as (in C++) :

A
Depends on compiler
B
(*ptrdata)++
C
*(ptrdata)++
D
*(ptrdata++)

Question 9

Consider the following table

A. Activation record	p. Linking loader
B. Location counter	q. Garbage collection
C. Reference counts	r. Subroutine call
D. Address relocation	s. Assembler

Matching A, B, C, D in the same order gives :

A
p, q, r, s
B
q, r, s, p
C
r, s, q, p
D
r, s, p, q

Question 10

What does the following C-statement declare? int (*f) (int*);

A
A function that takes an integer pointer as argument and returns an integer
B
A function that takes an integer as argument and returns an integer pointer
C
A pointer to a function that takes an integer pointer as argument and returns an integer
D
A function that takes an integer pointer as argument and returns a function pointer

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