Java Error ArrayIndexOutOfBoundsException

[Rohit Rane]

class selection
{
	public static void sort(int arr[])
	{
		for(int i= arr.length;i>0;i++)
		{
			int m=0;
			for(int j=1;j<=i;j++)
			{
				if(arr[j]>arr[m])
				{
					m=j;
				}
			}
			int temp= arr[i];
			arr[i]= arr[m];
			arr[m]= temp;
		}
		System.out.println("Array After Sort : ");
		System.out.print("{");
		for(int i=0;i<arr.length;i++)
			System.out.println(arr[i]+",");
		System.out.print("}");
	}
	public static void main(String args[])
	{
		int arr[] = {11,55,13,5,3};
		System.out.print("Array before sort : ");
		System.out.print("{");
		for(int i=0;i<arr.length;i++)
			System.out.print(arr[i]+",");
		System.out.println("}");
		sort(arr);
	}
}

Output :
Array before sort : {11,55,13,5,3,}
Excepion in thread "main" java.lang.Array IndexOutOfBound sException: 5
at selection.sort( selection.java: 10)
at selection.main( selection.java: 33)

[Rabbit]

You're going out of bounds in the first for loop in the sort function.

[Rohit Rane]

so what changes in code should i do

[Rabbit]

Well you should probably initialize i to 0 and the continue condition should be i < arr.length. But what you actually want to put in there is dependent on your business logic. You haven't explained what that is so I can't say what is correct. With the information you've given me, I can only tell you where it's wrong, not how to fix it.

[Rohit Rane]

well this is part of data structure in java
topic named is Selection sorting
n tnx for replying u r answer will be well appretiated

[Rabbit]

It sounds like you're just doing a bubble sort. In which case you should change those values to the ones I listed in the fourth post.

[neeraj0708]

hi,

As array are "0 based index" and you are running the loop more than its length in first for loop. and incrementing the value of i.

try this

for(int i= arr.length-1;i>0;i--)

Hope it will work fine.