Memory Storage and Array Address Computation, Summaries of Data Structures and Algorithms

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Array

Array ADT

• The simplest but useful data structure.

• Assign single name to a homogeneous collection of

instances of one abstract data type.

– All array elements are of same type, so that a

pre−defined equal amount of memory is

allocated to each one of them.

• Individual elements in the collection have an

associated index value that depends on array

dimension.

float marks[10];

Memory Storage

Memory Storage – One Dimensional Array

int student[4];

float marks[3];

student[1]

student[0] student[2]

student[3]

marks[1]

marks[0] marks[2]

Contd…

• Row-major order

Row0 Row1 Row

• Column-major order

Col0 Col1 Col2 Col3 Col

(0,0) (0,1) (0,2) (0,3) (0,4) (1,0) (1,1) (1,2) (1,3) (1,4) (2,0) (2,1) (2,2) (2,3) (2,4)

(0,0) (^) (1,0) (2,0) (0,1) (^) (1,1) (2,1) (0,2) (^) (1,2) (2,2) (0,3) (^) (1,3) (2,3) (0,4) (^) (1,4) (2,4)

Array Address Computation

1D array – address calculation

• Let A be a one dimensional array.

• Formula to compute the address of the

Ith^ element of an array (A[I]) is:

Address of A[I] = B + W * ( I – LB )

[0] 1

[1] 2

[2] 3

[3] 4

[4] 5

[5] 6

B

W

Given: B = 100, W = 4, and LB = 0

A[0] = 100 + 4 * (0 – 0) = 100

1D array – address calculation

• Let A be a one dimensional array.

• Formula to compute the address of the

Ith^ element of an array (A[I]) is:

Address of A[I] = B + W * ( I – LB )

[0] 1

[1] 2

[2] 3

[3] 4

[4] 5

[5] 6

B

Given:

B = 100, W = 4, and LB = 0 A[1] = 100 + 4 * (1 – 0) = 104 A[2] = 100 + 4 * (2 – 0) = 108 A[3] = 100 + 4 * (3 – 0) = 112 A[4] = 100 + 4 * (4 – 0) = 116 A[5] = 100 + 4 * (5 – 0) = 120

Example – 2

• If LB = 5, Loc(A[LB]) = 1200, and W = 4.

• Find Loc(A[8]).

Address of A[I] = B + W * ( I – LB )

Loc(A[8]) = Loc(A[5]) + 4 * (8 – 5)

Example – 3

• Base address of an array B[1300…..1900] is 1020

and size of each element is 2 bytes in the memory.

Find the address of B[1700].

Address of A[I] = B + W * ( I – LB )

• Given: B = 1020, W = 2, I = 1700, LB = 1300

Address of B[1700] = 1020 + 2 * (1700 – 1300)

2D Array – Address Calculation

• If A be a two dimensional array with M rows and N

columns. We can compute the address of an element at Ith

row and Jth^ column of an array ( A[I][J] ).

B = Base address/address of first element, i.e. A[LBR][LBC] I = Row subscript of element whose address is to be found J = Column subscript of element whose address is to be found W = Number of bytes used to store a single array element LBR = Lower limit of row/start row index of matrix, if not given assume 0 LBC = Lower limit of column/start column index of matrix, if not given assume 0 N = Number of column of the given matrix M = Number of row of the given matrix

[0] [1] [2] [3] [0] (^1 2 3 ) [1] (^5 6 7 ) [2] (^9 10 11 )

Row-major

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

Address of A[2][1] = B + W * (4 * (2 – 0) + (1 – 0))

M = 3 N = 4

Address of A[2][1] = B + W * (4 * (2 – 0) + (1 – 0))

Address of A[2][1] = B + W * (4 * (2 – 0) + (1 – 0))

Address of A[2][1] = B + W * (4 * (2 – 0) + (1 – 0))

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

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

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

Contd…

• Row Major

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

• Column Major

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

• Note: A[LBR…UBR, LBC…UBC]

M = (UBR – LBR) + 1

N = (UBC – LBC) + 1

Example – 4

• Suppose elements of array A[5][5] occupies 4 bytes, and

the address of the first element is 49. Find the address of the element A[4][3] when the storage is row major. Address of A[I][J] = B + W * ( N * ( I – LBR ) + ( J – LBC ))

• Given: B = 49, W = 4, M = 5, N = 5, I = 4, J = 3, LBR = 0,

LBC = 0. Address of A[4][3] = 49 + 4 * (5 * (4 – 0) + (3 - 0)) = 49 + 4 * (23) = 49 + 92 = 141