GATE CS 2021 Set 2 Q.20 (CS Q.10): Pointers, Arrays and Strings

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?

Official answer and solution

Official answer: C
From: Official GATE 2021 answer key

Solution

Trace row-major memory addressing and pointer arithmetic for a two-dimensional array in C.

The array is declared as int arr[4][5], which contains 4 rows of 5 integers each. In row-major order, elements are laid out contiguously in memory row by row:

Row 0 contains arr[0][0] to arr[0][4], with values $0, 1, 2, 3, 4$ .
Row 1 contains arr[1][0] to arr[1][4], with values $10, 11, 12, 13, 14$ .
Row 2 contains arr[2][0] to arr[2][4], with values $20, 21, 22, 23, 24$ .
Row 3 contains arr[3][0] to arr[3][4], with values $30, 31, 32, 33, 34$ .

The expression arr[1] evaluates to a pointer of type int* pointing to the first element of row 1, &arr[1][0].

Adding an offset of 9 using pointer arithmetic advances the pointer by 9 int elements:

arr[1] + 0 points to arr[1][0] (index 5 overall)
arr[1] + 4 points to arr[1][4] (index 9 overall)
arr[1] + 5 points to arr[2][0] (index 10 overall)
arr[1] + 9 points to arr[2][4] (index 14 overall)

Dereferencing *(arr[1] + 9) accesses arr[2][4]. The value assigned in the loop was:

\text{arr}[2][4] = 10 \times 2 + 4 = 24

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology - Session 2 (CS2)
Question: 20 of 65, printed as CS Q.10
Section: Computer Science and Information Technology

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