Java array giving "Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6"

Question

Here is the source code. I don't see any problems with this bubble sorting code:

int[] numbers = { 5, 8, 14, 1, 5678 };
int tempVar;
for (int i = 0; i < numbers.length; i++)
{
   for(int j = 0; j < numbers.length; j++)
   {
            if(numbers[i] > numbers[j + 1])
            {
                   tempVar = numbers [j + 1];
                   numbers [j + 1]= numbers [i];
                   numbers [i] = tempVar;
            }
   }
}
for (int i = 0; i < numbers.length; i++)
{
     System.out.println(numbers[i].toString());
}

Answer 1

Remove toString from the primitive array.Try the below code.

    int[] numbers = { 5, 8, 14, 1, 5678 };
    int tempVar;
    for (int i = 0; i < numbers.length; i++)
    {
       for(int j = i+1; j < numbers.length; j++)
       {
                if(numbers[i] > numbers[j])
                {
                       tempVar = numbers [j];
                       numbers [j]= numbers [i];
                       numbers [i] = tempVar;
                }
       }
    }
    for (int i = 0; i < numbers.length; i++)
    {
         System.out.println(numbers[i]);
    }

Answer 2

ArrayIndexOutOfBoundsException is thrown when you try to access the array over its defined length. You are exceeding the array bounds in your loop. And also the logic in the loop seems to be a bit wrong. Try the following :

    int[] numbers = { 5, 8, 14, 1,5678};
    int tempVar;
    for (int i = 0; i < (numbers.length - 1); i++)
    {
       for(int j = 0; j < (numbers.length-1); j++)
       {
                if(numbers[j] > numbers[j + 1])
                {
                       tempVar = numbers [j];
                       numbers [j]= numbers [j + 1];
                       numbers [j+1] = tempVar;
                }
       }
    }
    for (int i = 0; i < numbers.length; i++)
    {
         System.out.println(numbers[i]);
    }