Consider an array of size 3 by 4. What is the offset to reach the array[2][1] element?
array[2][1] with 4 columns, the calculation is \((2 \times 4) + 1 = 9\).To determine the memory offset for an element in a 2D array stored in row-major order, we calculate how many elements precede the target element in memory. The array has 3 rows (indices 0, 1, 2) and 4 columns (indices 0, 1, 2, 3). We are targeting the element at array[2][1]. In row-major order, elements are stored row by row, meaning all elements of row 0 are stored first, followed by all elements of row 1, and so on.
The offset for an element at array[i][j] in an array with C columns is calculated using the formula \(\text{Offset} = (i \times C) + j\). Here, the row index \(i = 2\), the column index \(j = 1\), and the total number of columns \(C = 4\). Applying these values, the offset is \((2 \times 4) + 1 = 8 + 1 = 9\). While the provided options (a. 8 and b. 10) do not include 9, the calculation based on the standard definition of row-major order offset is definitively 9. Assuming the question implies the count of preceding elements, we arrive at this result.
Based on the calculation, the number of elements preceding array[2][1] is 9. Therefore, the memory offset to reach the element array[2][1] is 9.
Understanding memory offsets is foundational in computer science, particularly in low-level programming and data structures. For example, when you access an image represented as a 2D array of pixels, the computer calculates the specific memory address of a pixel using the row-major offset formula to retrieve the color information quickly. This principle applies to optimizing memory usage in applications like gaming engines or large-scale data processing.