Row Major

Question

An array A[10][20] starts at decimal address 1000. Each element of array occupies 1 B and array is stored in row major order. Compute address of A[5][6]

just give the output.No need very big derivation

Answer 1

For 1-indexed array

Addr(A[5][6])=1000+20*4+5

                    = 1085

For 0-indexed array

Addr(A[5][6])=1000+20*5+6

                    = 1106

Comments

@srivivek95

you are considering only 4 as index starts from 0 we should consider 0,1,2,3,4 rows so we have to cross 5*20 elements and in 5th row we should cross 0,1,2,3,4,5 column elements so 6

so totally A[5][6]=(5*20+6)*1b+1000=1106B

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@saipriyab This confusion came as the index was not mentioned. Now edited answer considering both the cases.

Answer 2

in row major order

add(A[i][j]=(i*n+j)*w+base      (where array is of order M*N and we assume that indexing starts with zero)

add[5][6]=(5*20+6)*1+1000

             = 1106

and when indexing is starts with 1

we use add(A[i][j])=((i-1)*n+(j-1))*w+base       (where w is the size of an element in byte)

add[5][6]=(4*20+5)*1+1000

             =1085