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Data Structures andData Structures andAlgorithms (Arrays)Algorithms (Arrays)

Jose Rizal University, Mandaluyong City, Philippines

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References:

Data Structures and Algorithms in JAVA

Goodrich and Tamassia (2004) John Wiley &

Sons, Inc. Data Structures & Algorithm Analysisin C Mark Allan Weiss. (1993) Benjamin

Cummings Publishing Company,

Inc./Jemma, Inc., Philippines

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QuestionsQuestions

How important are these primitive data

types in developing our programs?

What limitations regarding the use ofprimitive data types can you imagine?

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Introducing ArraysIntroducing Arrays

Arrays are a very useful data structure provided by Java and otherprogramming languages.

An array is a variable compose of a finite, ordered set of homogeneouselements, i.e. elements of the same data type

Individual elements of this sequence can be uniformly

referenced/updated/ etc. since it consumes a consecutive set ofmemory locations is a set of pairs - index and a value forms

one-dimensional array n-dimensional array

Question:

Based on the definition above, how can arrays address the limitationsof using primitive data types ?

0

1

2

3

4

5

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Creating ArraysCreating Arrays

Arrays in Java are handled as objects (hence allocatedon heap)

2 data types base type or component type

index type All data types (primitive and user-defined objects) canbe put into arrays

To create an array, use the newkeyword Java has special syntax for array type declaration:

int[] a = new int[5]; int[] indicates a is array of integer variables int[5] indicates a will have 5 elements

Each cells in the array are assigned default values (0 /null / etc.) when array created

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Array IndexingArray Indexing

Java provides a special syntax for uniformly accessing cells in anarray Assume previous declaration of a:

int[] a = new int[5];

This in effect creates five int variables named: a[0] a[1] a[2] a[3] a[4]

To modify contents of cell #2 to 6: a[2] = 6;

This access mechanism is called array indexing 2 basic operations

extraction and storing

In Java, the same as with C and C++, array cells are indexedbeginning at 0 and going up to n-1 (n is number of cells) Beware of index numbers that exceed the array structure. Also, beware of starting the indexing at 0!

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ExampleExample

If we use the following codes:

int[] arr1 = new int[5];

arr1[2] = 6;

Which cell in arr1 would contain the value

6?

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0 1 2 3 4

6

Cell Index numbers

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Handling arrays usingHandling arrays using

loopsloops

Loops Are Useful for Processing Arrays

Loop counter can be used to iteratethrough index values

Each iteration of loop thus processes adifferent array element

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Examples of handlingExamples of handling

arraysarrays

int i;

int[] arr1 = new int[5];

arr1[2] = 6;for (i=0; i

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Exercise QuestionsExercise Questions

In the above coding, how would you changethe size of the array to accommodate 10integer values instead of 5?

What would happen if: instead of i++, i was used in the for loop

statement? if instead of i=0, i=1 was used? If instead of i

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If an array is declared to be A[n], then:

n = number of elements

If an array is declared to be A[n][m], then:

n*m = number of elements

If given n-dimensional array declaration,

A[b][c][d][e]..[n] = b,c,..n

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To determine the ith element of a single-dimensionarray: A[i] = + (i) * esize

where:

- base or starting addressi - element

esize - element size in bytes

2-dimensional arrays:A[i][j] = + [(i) * (UB2) + (j)] * esize

where: UB2 = upper bound of the 2nd dimension.

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3-dimensional arrays:

A[i][j][k] = + [(i) * (UB2) *(UB3) + (j)*(UB3)

+ (k)] * esize

where:UB2 = upper bound of the 2nd dimension;

UB3 = upper bound of the 3rd dimension

esize = element size in bytes

Typical esize: integer - 4 byteschar - 1 byte

float - 4 bytes

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ExamplesExamples

1. Given A[10][3][3][6], = 2000, esize=4

bytes:

a. find the formula to represent anelement in a 4-dimensional array.

b. find the total number of elements

c. find the address of A[2][2][0][4]

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ExamplesExamples

2. Given X[8][3][6][2][3], = 3000, esize = 3

bytes

a. find the total number of elementsb. find the address of X[0][2][5][1][2]

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ExamplesExamples

3. Consider the following declaration:class Myclass {

int A;char [] B = new char[10];

float C;char D;}Myclass class1 = new matrix [121][4][5];matrix A1;

A. Compute the address of element A1[120][3][3] given that thebase address is 2000.

B. Assume that we do not the size of Myclass in number of bytesbut we know that the address of A1[20][2][3] is 2160. Give thesize of Myclass in bytes. Assume the base address is 2000.