UGC NET CSE | August 2024 | Part 2 | Question: 21

Question

What will be the output of the following $\text{C}$ code?

#include<stdio.h>
void main()
{
    int arr[5]={10,20,30,40,50};
    int *p = (int*) (&arr+1);
    printf("\%d\%d", *(arr + 1), *(p -1));
}

• A. $10$ $50$
• B. $20$ $50$
• C. $30$ $40$
• D. $20$ $40$

Comments

20 50

correct me if any ;)

Answer 1

int arr[5]={10,20,30,40,50}; //for suppose base address is 1000 then address of arr+4 (last element

 in the array) is 1016.Assuming int=4B.

int *p = (int*) (&arr+1);

&arr represents address of whole array,so &arr+1 =1000+size(&arr)=1020.

p=1020.

And "arr" represents first element address which is 1000.

In the printf statement *(arr + 1)=*(1000+sizeof(int))=*(1004)=second element in the array=20

And *(p-1)=(1020-size(int))=*(1016)=last element in the array=50.

The Output is 20 50