Output of C programs | Set 64 (Pointers)

Prerequisite : Pointers in C Question 1 : What will be the output of following program?

C

#include "stdio.h"
int main()
{
    char a[] = { 'A', 'B', 'C', 'D' };
    char* ppp = &a[0];
    *ppp++; // Line 1
    printf("%c %c ", *++ppp, --*ppp); // Line 2
}

OPTIONS: a)C B b)B A c)B C d)C A

OUTPUT: (d) C A

Explanation: Line 1 : Now, ppp points to next memory location i.e., index 1 of the character array. Line 2 : Firstly, –*ppp= –(*ppp) is executed and hence the value ‘B’ (which is in the index 1 position of the char[] array) gets decremented by 1(i.e., it becomes ‘A’)and it is sent for printing. Then *++ppp= *(++ppp) is executed which initially increments the pointer to the next element of the array and prints the value in that index number 2 which is ‘C’. Although –*ppp is executed first compared to *++ppp, the display will be shown in the order as we mentioned in the printf() function in line 2. Hence we get output as C A. Question 2 : What will be the output of following program?

C

#include <stdio.h>
int main()
{
    int* ptr;
    *ptr = 5;
    printf("%d", *ptr);
    return 0;
}

OPTIONS: a) compilation error b) Runtime error c) 5 d) linker error

OUTPUT: c 5

Explanation: Pointer variable (*ptr) cannot be initialized. Question 3 : What will be the output of following program?

C

#include <stdio.h>
int main()
{
    int a = 36;
    int* ptr;
    ptr = &a;
    printf("%u %u", *&ptr, &*ptr);
    return 0;
}

OPTIONS: a) Address Value b) Value Address c) Address Address d) Compilation error

OUTPUT: (c)Address Address

Explanation: & and * cancelled each other and display the address stored in a pointer variable ptr i.e) the address of a. Question 4 : What will be the output of following program?

C

#include <stdio.h>
int main()
{
    int num = 10;
    printf("num = %d addresss of num = %u", num, &num);
    num++;
    printf("\n num = %d addresss of num = %u", num, &num);
    return 0;
}

OPTIONS: a) Compilation error b) num = 10 address of num = 2293436 num = 11 address of num = 2293438 c) num = 10 address of num = 2293436 num = 11 address of num = 2293440 d) num = 10 address of num = 2293436 num = 11 address of num = 2293436

OUTPUT: (d)

Explanation: Here a variable num holds the value 10 and get its address as 2293436, then increment is done as num++ which uses the same address space to store the incremented value. Question 5 : What will be the output of following program?

C

#include <stdio.h>
int main()
{
    int i = 25;
    int* j;
    int** k;
    j = &i;
    k = &j;
    printf("%u %u %u ", k, *k, **k);
    return 0;
}

OPTIONS: a) address address value b) address value value c) address address address d) compilation error

OUTPUT : (a)address address value

Explanation : Here a pointer variable *j hold the address of i and then another pointer variable *k hold the address of j. now k = address of j *k = address of i **k = value of i. This article is contributed by Abhishek kurmi.