Array Index out of bound 2D array

Question

I have a 2D array where no. of rows is 1 and no. of columns is > 1.

double[][] T = new double[1][24];
System.out.println(T[1].length);

But when i print the length of the columns, its says the index is out of bounds.

but when i print the following,

System.out.println(T[0].length);

I get the result as 24. But shouldn't T[0] should be equal to 1 and T[1] be equal to 24? Why am I getting this error? I suppose, java considers the above array as 1D array since it has only one row. but I need it to be a 2D array for further processes. Could anyone please help?

Mena · Accepted Answer · 2015-11-12 12:48:17Z

6

Array indices are 0-based.

If your array length is 1 (for the 1st dimension here), then you can only reference element 0.

In other words:

The declaration states the desired size (1 here)
The element reference states the desired 0-based index (0 here)

answered Nov 12, 2015 at 12:48

Mena

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thegauravmahawar · Accepted Answer · 2015-11-12 12:49:29Z

2

Array indices start from 0.

Your array length is 1, so T[0] is a valid index but T[1] is not.

answered Nov 12, 2015 at 12:49

thegauravmahawar

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Jr. Captain · Accepted Answer · 2015-11-12 12:52:55Z

2

System.out.println(T[0].length);

is correct. Because Array index starts from 0

answered Nov 12, 2015 at 12:52

Jr. Captain

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karim mohsen · Accepted Answer · 2015-11-12 12:58:33Z

1

Array is zero based so when double[][] T = new double[1][24]; you created a 2d array first dimension is 1 and the second is 24 so that's 1*24 Array

So System.out.println(T[1].length); gave you java.lang.arrayindexoutofboundsexception As arrays are zero based

if you wrote this double[][] T = new double[3][24]; you got 3*24 Array :-

System.out.println(T[0].length);//24
System.out.println(T[1].length);//24
System.out.println(T[2].length);//24
System.out.println(T[3].length);//java.lang.arrayindexoutofboundsexception

answered Nov 12, 2015 at 12:58

karim mohsen

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Rakesh KR · Accepted Answer · 2015-11-12 13:04:30Z

1

Array index starts from 0.

double[][] T = new double[1][24];

which can be also declared as follows:

double[][] T = new double[1][];
T[0] = new double[24];

answered Nov 12, 2015 at 13:04

Rakesh KR

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Mark Ola · Accepted Answer · 2015-11-12 13:05:47Z

1

imagine matrix of 1X24. you don't have 2nd column so obvious it will through nullpointerException.

answered Nov 12, 2015 at 13:05

Mark Ola

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Mauricio Gracia Gutierrez · Accepted Answer · 2015-11-12 13:09:40Z

1

In Java/Android and many other languages this is how is declared

matrix[numRows][numColumns]

And since Array indices start from 0 you need to be carefull to be sure that you are accessing a single value

x = matrix[0][1]

Or if you are accesing the first row (as a single dimension array)

x = matrix[0]

answered Nov 12, 2015 at 13:09

Mauricio Gracia Gutierrez

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punya · Accepted Answer · 2015-11-12 13:13:05Z

1

Array index start with 0 not 1.

If your array length is 1 then you can start from 0. so T[0] is a valid index but T[1] is not

answered Nov 12, 2015 at 13:13

punya

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