introduction of array

INTRODUCTION

OF

ARRAY

Topics To Be Discussed……………………….

Introduction

Types of array

One Dimensional Array

Internal representation of one-dimensional array

Example 1

Example 2

Array

An array is a fine, ordered set of homogeneous elements. By homogeneous

we mean that all the elements of the set of same type. By ordered set, we

mean that all the element of the set has unique position and can be

accesses by referring to its position within the set. So basic properties of

array are:

1. Homogeneity of its elements

2. Ordering imposed on elements

3. Finite number of elements.

An array is a set of elements of the same data type represented by a single

name.

Each individual array element can be referred to by specifying the array name

followed by a index or subscript enclosed in brackets. For example marks is A

name of array containing n elements, then the individual array element will be

Marks[1],marks[2],……………m, MARKS[N]

Types of Arrays

There are two type of arrays :

Linear Array or One-Dimensional Array : An array in which each elements

can be referred by one subscript of index is known as one-dimensional array

or linear arrays. One Dimensional arrays are also known as vectors

Multi Dimensional Array: An array in which each element can be referenced

by more than one subscripts is known as multi-dimensional array. In two

dimensional array, each element of the array can be referenced by two

subscripts

One Dimensional Array

One Dimensional array is a linear data structure in which the position of an

element within array can be given by just one index or subscript. For

example, if we have 5 students with roll no 1,2,3,4,5 and their marks are

70,80,90,85,75 then with one dimensional array named MARKS is shown

in following figure:

INDEX CONTENTS

LB 1

2

3

4

UB 5

In this example, the individual elements are identified by

MARKS[1],MARKS[2],……… MARKS[5]. The index or the subscript,

which identifies the position has values from 1 to 5.

70

75

809085

Lower Bound(LB)

Lower bound is the smallest number that an index can have. In this example

LB = 1.

Upper Bound(UB)

Upper bound is the smallest number that an index can have. In this example

UB = 5.

Length and Size of the Array

Length of the array can be obtained by the following formula :Length = UB – LB +

1

In this example

Index Set

The set of all possible values of index is known as Index Set. Index Set is

denoted by I. in this example: I={1,2,3,4,5}

Value Set

All array elements have some values. The set of values of array elements is

known as value set and is denoted by T. In this example

T = {70,80,90,85,75}

Length = 5 – 1 +1 = 6

Example

Consider an array A[105 : 112]. Find the number of elements in

this array.

Length = UB – LB + 1

= 112 – 105 + 1

= 8

Internal Representation of One-Dimensional array

With in computer’s memory array are represent using sequential representation.

Address of first element is known as base address and address of other element

can be obtained by adding world length to base address.

Lo A [1]

Lo + C A [2]

Lo + 2C A [3]

Lo + (I-1)C A [ I ]

Since the elements of an array are stored in successive memory locations, the

address of Kth element of an array can be obtained if we know :

1.Address of the first element of the array. The address of the first element of

the array is known as base address denoted by Lo

2.Number of memory locations required to store one element denoted by C

Formula to calculate the address of the Kth element is :

Lo + (K - 1)*C If Lower Bound = 1

Lo + (K - LB)*C If lower Bound ≠1

EXAMPLE - 1

Consider A linear array A[16:30]. If Base = 100 and world length = 4 then

calculate the address of A[27].

Solution :

A [27] = Lo+(K-LB)*C

= 100+27-16)*4

= 144

EXAMPLE - 2

Assume that a linear array with LB=1. if address of A{25]=375 and address

of A[30]=390, then find the address of A[16].

Solution: Formula for the address of Kth element is

A[K] = Lo+(K-LB)*W

Given A[25] = Lo+(25-1)*C=375 Lo+24C = 375

A[30] = Lo+(30-1)*C=290 Lo+29C = 290

+5C= -15

C= 3

Lo+24C =375

Lo+24*3=375

Lo+72=375

Lo=375-72

Lo=303

A[16] = ? A[16] = Lo+(K-LB)*W

= 303+(16-1)*3

= 303 + 45

= 348

THANKS