Q1: What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

#include<stdio.h>

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
    printf('%u, %u\n', a+1, &a+1);
    return 0;
}

ANS:A - 65474, 65476

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions. Step 2: printf('%u, %u\n', a+1, &a+1); The base address(also the address of the first element) of array is 65472. For a two-dimensional array like a reference to array has type 'pointer to array of 4 ints'. Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480 Then, &a has type 'pointer to array of 3 arrays of 4 ints', totally 12 ints. Therefore, &a+1 denotes '12 ints * 2 bytes * 1 = 24 bytes'. Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496 Hence the output of the program is 65480, 65496