GO Classes 2024 | IIITH Mock Test 5 | Question: 16

Answer 1

$\textsf{p}$ points to $3.$
$\textsf{p += 2;}$ will make p point to $5$
$\textsf{*p += 2;}$ will make $\textsf{a[5]}$ as $7.$
$\textsf{*p++;}$ here postfix $\textsf{++}$ has higher precedence so we will do $\textsf{p++}$ first.

But since it is postfix hence result will not be reflected in same line.
Now we do $\textsf{*p}$ which is same as $7.$
In the next line, $\textsf{p}$ points to $\textsf{a[6]}.$

Comments

In the beginning p will point to 2 as &a[3] will give 2 as output I think so

Answer 2

//            FOR  simplicity 

#include <stdio.h>

int main()
{
   
    int a[7] = {0, 1, 2, 3, 44, 50, 64};
int *p = &a[3];
p += 2;
*p += 2;
printf("%d ", *p++);
printf("%d", *p);

    return 0;
}

 

o/p -->   52 64