Exception in array element swapping

Question

I have this code to swap pairs of array elements:

int[] a= new int[]{1,2,3,4};
for(int i=0; i<a.length ;i++)
{
    int temp= a[i];
    a[i] = a[i+1];
    a[i+1] = temp;
}

However, I am getting the following exception:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 7
    at com.B.main(B.java:14)

Why am I getting this exception? How can I fix it?

this code can throw two ArrayIndexOutOfBoundsException 1) From for loop condition i < a.length while accessing a[i] 2) when condition i < a.length -1 while accessing a[i+1) — swapy
– swapy, Commented Jul 28, 2013 at 7:36

Maroun · Accepted Answer · 2013-07-28 07:21:47Z

4

Lets draw a table:

 i | a[i]
---+------
 0 |  1 :)
 1 |  2 :)
 2 |  3 :)
 3 |  4 :)
 4 |  ? :_(

Note that arrays are zero-based in Java, that means, if you have an array of size N (4 in your case), then the indexes are from 0 to N - 1 (0 to 3 in your case).

So when you try to access a[a.length - 1 + 1] (a[i+1] in the last iteration) you're getting ArrayIndexOutOfBoundsException.

answered Jul 28, 2013 at 7:21

Maroun

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Comments

kakabali · Accepted Answer · 2013-07-28 17:01:14Z

1

The error is due to the reason that you are accessing the element at a.length which is not available, so the code throws ArrayIndexOutOfBoundsException so please use a.length - 1 in the for loop. The problem in your case was at last iteration. You were trying to use a[4], but the elements in array a[ ] started from a[0] and ended at a[3].

edited Jul 28, 2013 at 17:01

answered Jul 28, 2013 at 7:19

kakabali

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Comments

Jigar Joshi · Accepted Answer · 2013-07-28 07:14:34Z

0

You are going upto a.length - 1

    for(int i=0; i<a.length ;i++)

and you are trying to access element at a.length which is out of bound

answered Jul 28, 2013 at 7:14

Jigar Joshi

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Comments

SheetJS · Accepted Answer · 2013-07-28 07:15:41Z

0

You are starting from i=0. So it should be a.length-1

edited Jul 28, 2013 at 7:15

SheetJS

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answered Jul 28, 2013 at 7:14

Vikas V

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