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

$6$
ip is an integer pointer and the initial assignment sets it to the element at array index $4$ i.e. $5$.(holds address of ar index $4$)
The next statement refers to the next integer after it which is $6 (ip[1]=*(ip+1))$.

Comments

The dereferencing part made it clear.

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I have taken size of one element as 2 (integer type) and that points to 8 and answer came as 9 but now i corrected thanks

Answer 2

Hence Ans is 6.

Answer 3

int arr[]={1,2,3,4,5,6,7,8,9,0,1,2,5} , *ip = arr+4;

printf("%d\n",ip[1]);

ip is an integer pointer that is currently holding the address where 5 is stored.(i.e. a[4])

ip[1] = *(ip+1) = value present at the next address i.e. 6. so it will print 6.

Answer 4

arr+0 points to the 1st element of the array(index 0) which is 1.

Hence arr+4 will point to the 5th element of the array(index 4) which is 5.

As per assignment $*ip=arr+4$ , $ip[0]=5$

Hence $ip[1]=6$

Answer 5

The program asks for what is the out put of printing ip[1] ,for that we need to understand what ip[1] represents here .it represents the second element of ip[ ] .What ip[ ] represents ? it represents arr[4] so what ip[1] will represent it will be arr[4+1] which is 6 .

Answer 6

After executing the line---- *ip=arr+4

ip will point to the element 5.

so ip[0]=5

ip[1]=6

ip[2]=7 ......

Answer-->6