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Memory Address Calculation in an Array

Address Calculation in single (one) Dimension Array:

Array of an element of an array say “A[ I ]” is calculated using the following formula:

Address of A [ I ] = B + W * ( I – LB )

Where,
B = Base address
W = Storage Size of one element stored in the array (in byte)
I = Subscript of element whose address is to be found
LB = Lower limit / Lower Bound of subscript, if not specified assume 0 (zero)

Example:

Given the base address of an array B[1300…..1900] as 1020 and size of each element is 2 bytes in the memory. Find the address of B[1700].

Solution:

The given values are: B = 1020, LB = 1300, W = 2, I = 1700

Address of A [ I ] = B + W * ( I – LB )

= 1020 + 2 * (1700 – 1300)
= 1020 + 2 * 400
= 1020 + 800
= 1820 [Ans]

Address Calculation in Double (Two) Dimensional Array:

While storing the elements of a 2-D array in memory, these are allocated contiguous memory locations. Therefore, a 2-D array must be linearized so as to enable their storage. There are two alternatives to achieve linearization: Row-Major and Column-Major.

Address of an element of any array say “A[ I ][ J ]” is calculated in two forms as given:
(1) Row Major System (2) Column Major System

Row Major System:

The address of a location in Row Major System is calculated using the following formula:

Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]

Column Major System:

The address of a location in Column Major System is calculated using the following formula:

Address of A [ I ][ J ] Column Major Wise = B + W * [( I – Lr ) + M * ( J – Lc )]

Where,
B = Base address
I = Row subscript of element whose address is to be found
J = Column subscript of element whose address is to be found
W = Storage Size of one element stored in the array (in byte)
Lr = Lower limit of row/start row index of matrix, if not given assume 0 (zero)
Lc = Lower limit of column/start column index of matrix, if not given assume 0 (zero)
M = Number of row of the given matrix
N = Number of column of the given matrix

Important : Usually number of rows and columns of a matrix are given ( like A[20][30] or A[40][60] ) but if it is given as A[Lr- – – – – Ur, Lc- – – – – Uc]. In this case number of rows and columns are calculated using the following methods:

Number of rows (M) will be calculated as = (Ur – Lr) + 1
Number of columns (N) will be calculated as = (Uc – Lc) + 1

And rest of the process will remain same as per requirement (Row Major Wise or Column Major Wise).

Examples:

Q 1. An array X [-15……….10, 15……………40] requires one byte of storage. If beginning location is 1500 determine the location of X [15][20].

Solution:

As you see here the number of rows and columns are not given in the question. So they are calculated as:

Number or rows say M = (Ur – Lr) + 1 = [10 – (- 15)] +1 = 26
Number or columns say N = (Uc – Lc) + 1 = [40 – 15)] +1 = 26

(i) Column Major Wise Calculation of above equation

The given values are: B = 1500, W = 1 byte, I = 15, J = 20, Lr = -15, Lc = 15, M = 26

Address of A [ I ][ J ] = B + W * [ ( I – Lr ) + M * ( J – Lc ) ]

= 1500 + 1 * [(15 – (-15)) + 26 * (20 – 15)] = 1500 + 1 * [30 + 26 * 5] = 1500 + 1 * [160] = 1660 [Ans]

(ii) Row Major Wise Calculation of above equation

The given values are: B = 1500, W = 1 byte, I = 15, J = 20, Lr = -15, Lc = 15, N = 26

Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]

= 1500 + 1* [26 * (15 – (-15))) + (20 – 15)] = 1500 + 1 * [26 * 30 + 5] = 1500 + 1 * [780 + 5] = 1500 + 785
= 2285 [Ans]

More Questions to follow

[Question 2] ISC 2017 Computer Practical Paper Solved – Quiz Result

Solution of Program 2 of ISC 2017 Computer Science Paper 2 (Practical) Exam. Java program to input the answers of each participant row-wise and calculate their marks