سوال ۶۸ کنکور ارشد علوم کامپیوتر ۱۳۹۷

We are given two cell addresses that share the same column index ($j=1$) and differ in the row index ($i=0$ vs $i=1$):

$\text{addr}(A[0,1]) = \mathtt{7A00}, \qquad \text{addr}(A[1,1]) = \mathtt{7A10}.$

The step taken when the row index increases by one (column fixed) is

$\mathtt{7A10} - \mathtt{7A00} = \mathtt{10}_{16} = 16 \text{ bytes}.$

Two storage orders for a 2-D matrix. Let the element size be $s$ and the number of columns be $c$.

Moving from $A[0,1]$ to $A[1,1]$ advances by a full row width of $16$ bytes — i.e. one entire row ($c$ elements of a small word size, e.g. $c\cdot s = 16$) is jumped over to reach the next row. Crossing one whole row when the first index changes is exactly the signature of row-major storage, where rows are laid out one after another and the first (row) index is the slow / outer index.

The round $16$-byte boundaries ($\mathtt{7A00}\to\mathtt{7A10}$) confirm that one row occupies $16$ bytes, so the matrix is stored row by row — i.e. row major.

Options 1 (hash) and 2 (random) are not contiguous addressing schemes for matrices and are ruled out immediately.

Correct answer: option 4, row major.

Pitfall: With only two cells sharing the column index, the $16$-byte step is the row stride. A large stride when the first (row) index changes signals row-major; a one-element stride would have signaled column-major. Always ask which index moved before deciding the order.