Consider the following code snippet. What is the output?#include <stdio.h>int main() {    int arr[4] = {10, 20, 30, 40};    int *p = &arr[1];    printf("%d\n", *(p + 2));    return 0;}

The output of the code will be 40.

Here's the step by step explanation:

1. An array arr of 4 integers is declared and initialized with the values 10, 20, 30, 40.

2. A pointer p is declared and it is initialized to point to the second element of the array (i.e., arr[1] which is 20).

3. In the printf statement, *(p + 2) is used. This is pointer arithmetic, where p + 2 means the address of the third element after p (including p). Since p is pointing to arr[1], p + 2 will point to arr[3].

4. The * operator is used to dereference the pointer, i.e., get the value stored at the address the pointer is pointing to. So, *(p + 2) will give the value stored at arr[3], which is 40.

So, the output of the code will be 40.