A 2-D array A[4...7,-1...3] requires 2 bytes of storage space for each element. If the array is stored in row-major form having base address 100, then the address of A[6, 2] will be...................

Question

Accepted Answer

The correct option is (2):
The address of an element in a two-dimensional array stored in **Row-Major Order (RMO)** is calculated using a specific formula. RMO means that all elements of the first row are stored contiguously in memory, followed by all elements of the second row, and so on. The formula for RMO is: Address(A$$i, j$$) = Base Address + W \* $$ (i - L1) \* N2 + (j - L2) $$, where W is the size of each element, L1 and L2 are the lower bounds of the row and column indices, and N2 is the total number of columns. First, we calculate N2 (the number of columns): N2 = U2 - L2 + 1 = 3 - (-1) + 1 = 5. Now, for the element A$$6, 2$$, we substitute the values: Address = 100 + 2 \* $$ (6 - 4) \* 5 + (2 - (-1)) $$ = 100 + 2 \* $$ (2 \* 5) + (3) $$ = 100 + 2 \* $$10 + 3$$ = 100 + 2 \* 13 = 100 + 26 = 126. The address is 126.

The incorrect option is (1):
The address 150 would mean that the element A$$6, 2$$ is 50 bytes away from the base address, implying 25 elements precede it in memory. Our correct calculation shows only 13 elements precede it, requiring an offset of 26 bytes. This result is typically obtained if a calculation error is made, such as miscalculating the number of columns (N2) or using incorrect index arithmetic for the rows. For instance, calculating the offset as 25 instead of 13 would lead to 100 + 2 \* 25 = 150.

The incorrect option is (3):
The address 200 implies a 100-byte offset from the base address, meaning 50 elements precede the target element. The entire array has (7 - 4 + 1) \* (3 - (-1) + 1) = 4 \* 5 = 20 elements, taking up a total of 40 bytes (20 \* 2). Since the entire array ends at 100 + 40 - 2 = 138 (assuming 0-indexed offset), 200 is well outside the memory block allocated for the array, making it conceptually and mathematically incorrect.

The incorrect option is (4):
An address of 112 suggests an offset of only 12 bytes (6 elements). This is incorrect because, in row-major order, the rows before the target element A$$6, 2$$ (which are rows 4 and 5, two complete rows) already contain $2 \times 5 = 10$ elements. Along with the three elements of row 6 (indices -1, 0, 1) that precede A$$6, 2$$, the total count of preceding elements is $10 + 3 = 13$. The offset must be 26 bytes, not 12.