MCQ on Pointers in C/C++(Pointers Multiple choice questions)

What does the following C-statement declare?

//Declaration

int ( * f) (int * ) ;

Which of the following is true with respect to Reference?

What is the output of the below code?

#include <string.h>
#include <stdio.h>
#include <stdlib.h>
void fun(char** ptr)
{
    ptr++;
}
int main()
{
    char *str = malloc(20*sizeof(char));
    if(str == NULL)
        return -1;

    strcpy(str, "Aticleworld");

    fun(&str);

    puts(str);

    free(str);

    return 0;
}

What is the output of the below code?

#include <stdio.h>

int main()
{
    int data = 24;
    int *ptr = NULL;
    int **ptr1 = NULL;
    ptr = &data;
    ptr1 = &ptr;
    printf("%d\n", *ptr );
    printf("%d\n", **ptr1);

    return 0;
}

What is the output of the below code?

#include<stdio.h>
int main()
{
    register int data = 10;
    int *ptr = NULL;
    ptr = &data;
    printf("%d ",*ptr);
    return 0;
}

What is the output of the below code?

int main()
{
    int aiData[5] = {100,200,300,400,500};
    
    int *piData = aiData;
    
    ++*piData;
    
    printf("aiData[0] = %d, aiData[1] = %d, *piData = %d", aiData[0], aiData[1], *piData);
    
    return 0;
}

What is the output of the below code?

#include <stdio.h>

int main()
{
    char *pcData="aticleworld";
    printf("%c ",6[pcData]);

    return 0;
}

Which is an indirection operator among the following?

Below declaration means,

//Declaration

int (*p) [5];

What is the output of the below code?

#include<stdio.h>

void fun(int *p)
{
  int q = 40;
  p = &q;
}

int main()
{
  int data = 27;

  int *ptr = &data;

  fun(ptr);

  printf("%d", *ptr);

  return 0;
}

What is the output of the below code?

#include "stdio.h"

typedef void (*fPtr)(int);

void display(int a)
{
    printf("%d\n",a);
}

int main()
{
    fPtr fPtr1 = NULL, fPtr2 = NULL;

    fPtr1 = &display;

    fPtr2 = display;

    (*fPtr1)(10);

    fPtr2(10);

    return 0;
}

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

What is the output of the below code?

#include<stdio.h>

int main()
{
    int data = 27;
    int *ptr = &data;
    
    printf("%u %u", *&ptr, &*ptr);
    
    return 0;
}

What is the output of the below code?

#include <stdio.h>

int f(int x, int *py, int **ppz)
{
    int y, z;
    **ppz += 1;
    z  = **ppz;
    *py += 2;
    y = *py;
    x += 3;
    return x + y + z;
}

void main()
{
    int c, *b, **a;
    c = 4;
    b = &c;
    a = &b;
    printf( "%d", f(c,b,a));
    return 0;
}

What is the output of the below code?

#include<stdio.h>

void fun(int **p)
{
  static int q = 40;
  *p = &q;
}

int main()
{
  int data = 27;

  int *ptr = &data;

  fun(&ptr);

  printf("%d", *ptr);

  return 0;
}

In the below statement, ptr1 and ptr2 are uninitialized pointers to int i.e. they are pointing to some random address that may or may not be a valid address.

//Declaration

int* ptr1, ptr2;

What is the output of the below code?

#include <string.h>
#include <stdio.h>
#include <stdlib.h>

void fun(char** ptr)
{
    (*ptr)++;
}

int main()
{
    char *str = malloc(20*sizeof(char));
    if(str == NULL)
        return -1;
        
    strcpy(str, "Aticleworld");
    
    fun(&str);
    puts(str);
    free(str);
    
    return 0;
}

What does the following ‘C’ statement declare?

//Syntax

int * f [ ] ( );

What is the output of the below code?

#include<stdio.h>

int main()
{
    int i = 3;
    int *j;
    int **k;
    j = &i;
    k = &j;
    k++;
    printf("%d ",**k);
    return 0;
}

What is the output of the below code?

#include <stdio.h>

int main(void)
{
    int aiData[5] = {100,200,30,40,50};
    
    int *piData = aiData;
    
    *++piData;
    
    printf("aiData[0] = %d, aiData[1] = %d, *piData = %d", aiData[0], aiData[1], *piData);
    
    return 0;
}

The reason for using pointers in a C program is

Consider the following declaration:

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

The following program fragment

a=4;

**c=5;

What is the output?

#include<stdio.h>

void fun(int *p, int *q)
{
    p = q;
    *p = 2;
}

int main()
{
    int i = 0, j = 1;

    fun(&i, &j);

    printf("%d %d", i, j);

    return 0;
}

What is the output of the below program? Assume the sizeof an integer and a pointer is 4 bytes.

#include <stdio.h>

int main()
{
    int (*ptr)[5][10];

    printf("%d",  sizeof(*ptr));

    return 0;
}

What is the output of the below program?

#include<stdio.h>

void fun(int arr[])
{
    int i;
    for (i = 0; i < (sizeof(arr)/sizeof(arr[0])); ++i)
    {
        printf("%d ", arr[i]);
    }
}

int main()
{
    int arr[4] = {10, 20,30, 40};

    fun(arr);

    return 0;
}

What is the output of the following program?

#include <stdio.h>

void fun(int x)
{
    x = 30;
}

int main()
{
    int y = 20;

    fun(y);

    printf("%d", y);

    return 0;
}

Is there any issue with the below program?

#include "stdlib.h"

int main()
{
    int *ptr1;
    int **ptr2;
    int **ptr3;
    
    ptr1 = (int*)malloc(sizeof(int));
    ptr2 = (int**)malloc(10*sizeof(int*));
    ptr3 = (int**)malloc(10*sizeof(int*));
    
    free(ptr1);
    free(ptr2);
    free(*ptr3);
    
    return 0;
}

What is the output of the below code?

#include<stdio.h>

int main()
{
    printf("%u",sizeof(NULL));
    
    return 0;
}

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

What is the output of the below code?

#include 

int main()
{
    void *ptr;

    ptr = (void*)0;

    printf("%u",sizeof(ptr));

    return 0;
}

The following statement in ‘C’ declares?

int (*f())[ ];

The output of the following program?

#include <stdio.h>

int main()
{
    int *ptr;
    int x;

    ptr = &x;
    *ptr = 0;

    printf(" x = %d\n", x);
    printf(" *ptr = %d\n", *ptr);

    *ptr += 5;
    printf(" x  = %d\n", x);
    printf(" *ptr = %d\n", *ptr);

    (*ptr)++;
    printf(" x = %d\n", x);
    printf(" *ptr = %d\n", *ptr);

    return 0;
}

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

Assume that an int variable takes 4 bytes and a char variable takes 1 byte

#include<stdio.h>

int main()
{
    int arr[] = {10, 20, 30, 40, 50, 60};

    int *ptr1 = arr;
    int *ptr2 = arr + 5;

    printf("%d\n",(ptr2 - ptr1));

    printf("%d\n",(char*)ptr2 - (char*) ptr1);

    return 0;
}

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]);
}

GATE-CS-2017 (Set 2)

The output of the program is _____.

What is the output of the below code?

#include <stdio.h>

int main(void)
{
    int * piData;
    {
        //block
        int Data = 27;
        piData = &Data;
    }
    printf("piData = %d\n", *piData);

    return 0;
}

What are the different ways to initialize an array with all elements as zero?

Can we combine the two statements?

char *ptr;

ptr = (char*) malloc(100);

What is the output of the below program?

#include<stdio.h>

int main()
{
    int * ptr = NULL;
    
    printf("%d",*ptr);
    
    return 0;
}

What is the output of the above program?

#include<stdio.h>

int main()
{
    int a;
    char *x;

    x = (char *) &a;

    a = 512;

    x[0] = 1;

    x[1] = 2;

    printf("%d\n",a);

    return 0;
}

A function q that accepts a pointer to a character as an argument and returns a pointer to an array of 5 integers can be declared as

What is the output of the below code?

#include<stdio.h>

int main()
{
    int *ptr;
    
    *ptr = 5;
    
    printf("%d", *ptr);
    
    return 0;
}

What does the following expression mean? 

//pointer Quiz

char ∗(∗(∗ a[T]) ( )) ( );

What is the output of the below code?

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int i;
    int *ptr = (int *) malloc(5 * sizeof(int));
    if(ptr == NULL)
    {
        return -1;
    }
    for (i=0; i<5; i++)
    {
        *(ptr + i) = i;
    }
    printf("%d ", *ptr++);
    printf("%d ", (*ptr)++);
    printf("%d ", *ptr);
    printf("%d ", *++ptr);
    printf("%d ", ++*ptr);

    return 0;
}

Assume int is 4 bytes, char is 1 byte and float is 4 bytes. Also, assume that pointer size is 4 bytes (i.e. typical case)

char *pChar;
int *pInt;
float *pFloat;

sizeof(pChar);
sizeof(pInt);
sizeof(pFloat);

What’s the size returned for each of the sizeof() operator?

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 :

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;
}

What is the output of the below program?

#include<stdio.h>

int main()
{
    char *ptr = "Aticleworld.com";

    printf("%c\n", *&*&*ptr);

    return 0;
}

What is the output of the below code?

#include<stdio.h>

void fun(int **p)
{
  int q = 40;
  *p = &q;
}

int main()
{
  int data = 27;

  int *ptr = &data;

  fun(&ptr);

  printf("%d", *ptr);

  return 0;
}

Consider this C code to swap two integers and these five statements after it:

void swap(int *px, int *py)
{
*px = *px - *py;
*py = *px + *py;
*px = *py - *px;
}

P1: will generate a compilation error P2: may generate a segmentation fault at runtime depending on the arguments passed P3: correctly implements the swap procedure for all input pointers referring to integers stored in memory locations accessible to the process P4: implements the swap procedure correctly for some but not all valid input pointers P5: may add or subtract integers and pointers.

Consider a compiler where int takes 4 bytes, char takes 1 byte and pointer takes 4 bytes.

#include <stdio.h>

int main()
{
    int arr1[] = {1, 2,3};
    int *ptr1 = arr1;

    char arr2[] = {1, 2,3};
    char *ptr2 = arr2;

    printf("sizeof arr1[] = %d ", sizeof(arr1));

    printf("sizeof ptr1 = %d ", sizeof(ptr1));

    printf("sizeof arr2[] = %d ", sizeof(arr2));

    printf("sizeof ptr2 = %d ", sizeof(ptr2));

    return 0;
}

What is the output of the below program? The assumed size of char, int, and double is 1,4,8.

#include<stdio.h>

int main()
{
    int a, b, c;
    char *p = 0;
    int *q = 0;
    double *r = 0;
    
    a = (int)(p + 1);
    b = (int)(q + 1);
    c = (int)(r + 1);
    
    printf("%d %d  %d",a, b, c);
    
    return 0;
}

Suppose that in a C program snippet, the following statements are used.

i) sizeof(int);

ii) sizeof(int*);

iii) sizeof(int**);

Assuming the size of the pointer is 4 bytes and the size of the int is also 4 bytes, pick the most correct answer from the given options.

calloc(m, n); is equivalent to

Prior to using a pointer variable, it should be

Consider the following C function

#include <stdio.h>
int main()
{
char c[ ] = "ICRBCSIT17";
char *p=c;
printf("%s", c+2[p] – 6[p] – 1);
return 0;
}

The output of the program is?