ArrayIndexOutOfBounds in sorting

Question

I've written my sorting problem as follows, but i am getting an ArrayIndexOutOfBounds exception. which i'm not able to figure it out. plz help.

System.out.println("Enter the total no of digits to sort:-  ");

n = Integer.parseInt(br.readLine());
x = new int[n];

System.out.println("Enter the elements:- ");

for(i = 0; i < n; i++)
    x[i] = Integer.parseInt(br.readLine());

for(i = 0; i < n; i++)
{
    for(j = 0; j < n; j++)
    {
        if(x[j] > x[j+1])  //ascending order
        {
            temp = x[j];
            x[j] = x[j+1];
            x[j+1] = temp;
        }
    }
}

Yep, my question is no longer relevant since your updated provided the answer. — Menno, May 01 '13 at 13:07

score 4 · Accepted Answer · answered May 01 '13 at 12:59

4

Since j goes up to n, j+1 is out of bound. You need to change it to

for(j=0;j<n-1;j++)

Doing so would make sure that x[j+1] is within bounds.

answered May 01 '13 at 12:59

Sergey Kalinichenko

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i did the same program in netbeans n gt an error saying..."java.lang.NoClassDefFoundError: javaapplication3/Main Exception in thread "main" Java Result: 1"........help on this plz. – jhe May 01 '13 at 13:11
@jhe This is likely to be an issue with your netbeans project configuration. Here is a [link to a relevant discussion](http://stackoverflow.com/q/2702185/335858). – Sergey Kalinichenko May 01 '13 at 13:23

hamid · Answer 2 · 2013-05-01T13:10:07.887

0

Error is here:

if(x[j] > x[j+1]) {
 ....

Because j+1 is equal ton

Make this change:

 for(j=0;j + 1<n;j++) {
  ...

edited May 01 '13 at 13:10

answered May 01 '13 at 13:00

hamid

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Nop, `j+1` IS `n` on the last loop. Though when setting `int[n]`, the last index is `n - 1`. So `int[n]` or its equivalent on the last loop `int[j+1]` will be outOfBound. *Example*: int[4] creates keys {0, 1, 2, 3} – Menno May 01 '13 at 13:08

ArrayIndexOutOfBounds in sorting

2 Answers2