GATE CS 2002 Q2.8 — Programming in C

In C, two-dimensional arrays are stored in row-major order. This means elements are stored row by row in contiguous memory locations.

For a 2D array declared as char a[M][N], the memory address of the element at index [i][j] is calculated using the formula:

$$\text{Address}(a[i][j]) = \text{Base Address} + (i \times N + j) \times \text{sizeof}(\text{element})$$

Given the following values from the question:

• Array declaration: char a[100][100], so the number of columns $N = 100$.
• Base Address = $0$
• Target element: a[40][50], so $i = 40$ and $j = 50$.
• Since the array is of type char, the size of each element is $1$ byte (under standard byte-addressable memory).

Substituting these values into the formula:

$$\text{Address}(a[40][50]) = 0 + (40 \times 100 + 50) \times 1$$

$$\text{Address}(a[40][50]) = 4000 + 50 = 4050$$

Therefore, the address of a[40][50] is 4050.