Consider the following ANSI C program.#include <stdio.h> int main(){ int arr[4][5]; int i, j; for (i=0; i<4; i++){ for (j=0; j<5; j++){ arr[i][j] = 10*i + j; } } printf("%d", *(arr[1] + 9)); return 0; }What is the output of the above program?

The 2D array arr[4][5] has 4 rows and 5 columns, stored sequentially in row-major order in memory.

The nested loops initialize each element arr[i][j] to $10 \times i + j$:

• Row 0: arr[0][0] to arr[0][4] have values $0, 1, 2, 3, 4$

• Row 1: arr[1][0] to arr[1][4] have values $10, 11, 12, 13, 14$

• Row 2: arr[2][0] to arr[2][4] have values $20, 21, 22, 23, 24$

• Row 3: arr[3][0] to arr[3][4] have values $30, 31, 32, 33, 34$

arr[1] decays to a pointer to the first element of the second row, i.e., &arr[1][0] (of type int *).

Pointer arithmetic arr[1] + 9 advances the pointer by $9$ int elements starting from arr[1][0]:

• $9$ elements ahead from arr[1][0] corresponds to index: $(1 \times 5 + 0) + 9 = 14$ in flat 1D indexing.

• Converting linear index $14$ back to 2D coordinates for an array with $5$ columns:

  $$\text{Row} = \lfloor 14 / 5 \rfloor = 2$$

  $$\text{Column} = 14 \pmod 5 = 4$$

Thus, *(arr[1] + 9) accesses arr[2][4].

The value stored at arr[2][4] is:

Hence, the correct option is C.