GATE CSE 2019 | Question: 24

Question

Consider the following C program:

#include <stdio.h>
int main() {
        int arr[]={1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 5}, *ip=arr+4;
        printf(“%d\n”, ip[1]);
        return 0;
}

The number that will be displayed on execution of the program is _______

Comments

arr is base address of element $1$. 

arr+$4$ is base address of element $5$. Let it ip.

ip[$1$]=*(ip+$1$)= Base address of element $6$. 

6 is the answer. 

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*p = arr + 4  = arr[4]

*(p+0) = arr[4]

*(p+1) = arr[4+1] = arr[5]

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Easy

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ip[1] is *(ip+1) i.e 6

Answer 1

given :

*ip = (a + 4) = &(a[4])

ip = (a+4)

ip[1] = *(ip+1) = *( (a+4) +1 )

ip[1] = *(a+5) = a[5] = 6

 

correct me if i am wrong

Answer 2

This is how I've solved

*ip=arr+4 (It is saying move 4 elements ahead and stop)

Techically: ip=arr(base address) + 4*size(each element)

From the perspective of ip: {5, 6, 7, 8, 9, 0, 1, 2, 5} ip is pointing 5

ip[1] Asked ip[1]=ip(address)+1*size(element)   6 is the answer

 

Answer 3

This is how I've solved

*ip=arr+4 (It is saying move 4 elements ahead and stop)

Techically: ip=arr(base address) + 4*size(each element)

From the perspective of ip: {5, 6, 7, 8, 9, 0, 1, 2, 5} ip is pointing 5

ip[1] Asked ip[1]=ip(address)+1*size(element)   answer 6

 

Answer 4

Correct answer: 6