GATE CS 2021 Set 2 Q10 — Programming in C

To find the output of the program, let's analyze how the 2D array is stored in memory and how the pointer arithmetic is evaluated.

1. Array Initialization

The array arr is declared as int arr[4][5], which has $4$ rows and $5$ columns. The elements are initialized using the formula:

$$\text{arr}[i][j] = 10 \times i + j$$

Let's write down the values for the relevant rows:

• Row 1 ($i = 1$):
  • $\text{arr}[1][0] = 10$, $\text{arr}[1][1] = 11$, $\text{arr}[1][2] = 12$, $\text{arr}[1][3] = 13$, $\text{arr}[1][4] = 14$
• Row 2 ($i = 2$):
  • $\text{arr}[2][0] = 20$, $\text{arr}[2][1] = 21$, $\text{arr}[2][2] = 22$, $\text{arr}[2][3] = 23$, $\text{arr}[2][4] = 24$

2. Pointer Arithmetic and Memory Layout

In C, a 2D array is stored in row-major order as a contiguous block of memory.

• The expression arr[1] decays to a pointer to the first element of the second row, i.e., &arr[1][0], and has the type int *.
• Adding an integer $k$ to an int * pointer shifts it forward by $k$ elements of type int.

Therefore, we can map the pointer offsets starting from arr[1]:

$$\begin{aligned} \text{arr}[1] + 0 &\to \&\text{arr}[1][0] \\ \text{arr}[1] + 1 &\to \&\text{arr}[1][1] \\ &\ \ \vdots \\ \text{arr}[1] + 4 &\to \&\text{arr}[1][4] \\ \text{arr}[1] + 5 &\to \&\text{arr}[2][0] \quad \text{(since each row has 5 elements)} \\ \text{arr}[1] + 6 &\to \&\text{arr}[2][1] \\ \text{arr}[1] + 7 &\to \&\text{arr}[2][2] \\ \text{arr}[1] + 8 &\to \&\text{arr}[2][3] \\ \text{arr}[1] + 9 &\to \&\text{arr}[2][4] \end{aligned}$$

3. Dereferencing

Dereferencing the pointer gives:

$$*(arr[1] + 9) = \text{arr}[2][4]$$

Substituting the values:

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

Thus, the program prints 24.