Question

Consider a 2 - D array A[m][n] where m are the number of rows and n are the number of columns. Assuming that each of the element takes 's' memory locations. The array A is stored in a row-major order, then the address of the element A[i][j] can be given as : (Assuming the location of $1^{st}$ element as b and starting of the array index from 1)

• A. $\left[\right(i-j)+\frac{j(i-1)}{2}]$
• B. $b+(j-1)\times s+(i-1)\times s\times n$
• C. $b+(i-1)\times s+1(j-1)\times m\times s$
• D. $(j-i)+\left[\right(i-j)n-\frac{(i-1)(i-2)}{2}]$

Answer 1

The correct option is B $b+(j-1)\times s+(i-1)\times s\times n$

The address of an element A[i][j] in row-major order is given as :
A[i][j] = Base (A) + (j - start index) $\times$ size of element) + (i - start index) $\times$ size of element $\times$ n
Here :
Base (A) = b
Size of element =S
Start index = 1
m : no of rows, n : no. of colums
$\therefore$ we get A[i][j] = b + (j - 1) $\times$ s + (i - 1) $\times$ s $\times$ n