Page 24 - DCAP605_ADVANCED_DATA_STRUCTURE_AND_ALGORITHMS

Unit 1: Introduction to Data Structure and Arrays

Notes
LB UB
2 2
LB
1
i th Row A(i, LB ) .......
2 A(l, j)

UB
1

If we want to calculate the number of elements till Ist Row then.
No. of elements = (UB – LB + 1) * (1 – 1 + 1)
2 2
or No. of elements = UB – LB + 1 …(6)
2 2
No. of elements in (j – 1) Rows = (j – 1) (UB – LB + 1). If s be the size of data types of Array
2
2
elements then memory space required for storing i-1 Rows will be.
Space Required = (UB – LB + 1) (i – 1)*s …(7)
2 2
If x be the address of A[LB , LB ] then Address of A(i, LB ) will be:
1 2 2
Add = x + (UB – LB + 1) (i-1)*s …(8)
2 2
Address of A[i, j] will be
Address of A[i, j] = x + [(UB – LB + 1) (i – 1) + (j – 1)]*s …(9)
2 2
This is Address Scheme for Row Major form. For Column Major form
Address of A[i, j] = x + [(UB – LB + 1) (j – 1) + (i – 1)]*s …(10)
1 1
1 2 3 4
1

2
A =
3

4

Consider a two-dimensional matrix of figure. Suppose address of A11 is 2000 and this two

dimensional array contains elements of 4-bytes each we want to calculate the address of A23,
then by the formula (9) we have
_ = 2000,
UB = 4, LB = 1
2 2
i = 2, j = 3,
and s = 4
then

add. of A23 = 2000 + [(4 – 1 + 1) (2 – 1) + (3 – 1)]*4 = 2024

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