java.lang.ArrayIndexOutOfBoundsException: 10

Question

I can't figure out what the problem is someone please help. The program is supposed to find the index of the duplicate values in the array and print these out. Outputs java.lang.ArrayIndexOutOfBoundsException: 10.

private static String s = "";
private static int num = 0;

public static void main(String[] args) {
    int[] array = { 1, 5, 3, 8, 2, 3, 7, 1, 9, 3 };
    for (int i = 1; i <= array.length; ++i) {
        while (num <= array.length - 2 && array[num] == array[i]) {
            s += i + ","; 
            num += 1;
        }
    }
    System.out.println("index 0 are at positions" +s);
    System.out.println();
}

The time you spent writing this question would have been better spent typing "java array index out of bounds exception" in google. The top 10 hits have exactly the same problem you're having here. — Vincent van der Weele, Jun 13 '14 at 08:40
A length of ten means yourArray[0 - 9] position ZERO is the FRIST one, position 1 is the SECOND. — RossC, Jun 13 '14 at 08:57

score 5 · Answer 1 · edited Jun 13 '14 at 08:38

5

for (int i= 0; i<array.length; ++i) {
    while (num <=array.length-2 && array[num]==array[i]) {
        s += i+ ","; 
        num += 1;
    }
}

This should work for you. length of array is 10, so last index is 9, not 10! And start from 0!

edited Jun 13 '14 at 08:38

Mohammad Faisal

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answered Jun 13 '14 at 08:34

PKlumpp

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score 4 · Answer 2 · edited Jun 13 '14 at 08:37

4

array.length is 10, but last index in array is 9, because array index start from 0.

The stop condition in your for loop should be i<=array.length-1 if you want to loop through entire array.

edited Jun 13 '14 at 08:37

Mohammad Faisal

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answered Jun 13 '14 at 08:35

Kasper Ziemianek

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score 1 · Answer 3 · answered Jun 13 '14 at 08:36

1

First of all you should start from 0 instead of index 1. And secondly you should use < sign

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

answered Jun 13 '14 at 08:36

AurA

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score 0 · Answer 4 · answered Jun 13 '14 at 08:38

0

Only the max number of elements is set to 10 which means the index number would start at 0.

You can only get element until the 9th element.

java.lang.ArrayIndexOutOfBoundsException: 10

It is clear sign, that you're not having an 11th element.

for (int i = 1; i < array.length; ++i) {
    while (num <= array.length - 2 && array[num] == array[i]) {
        s += i + ","; 
        num += 1;
    }
}

This would do the job for you.

answered Jun 13 '14 at 08:38

Afzaal Ahmad Zeeshan

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thanks for the help but the above code doesn't print the index of the array at repeat positions. It prints :index 0 are at positions0,1,2,3,4,5,6,7,8,9, – Afshin Jun 13 '14 at 09:51

java.lang.ArrayIndexOutOfBoundsException: 10

4 Answers4