java.lang.ArrayIndexOutOfBoundsException: -6

Question

Any solution for this i need to swap element x with y in unsorted array ,x and y are in array

I have this error for array lengths >= 6. The code works for tabX[] < 6 length.

java.lang.ArrayIndexOutOfBoundsException:-6.

int tabX[] = {
    20, 40, 50, 60, 80, 70
};
int elemA[];
int elemB[];
int sumA = 0;
int sumB = 0;
int tmp, lastButOne;
Arrays.sort(tabX);
while (sumA > sumB || sumA == 0 || sumB == 0) {
    for (int i = 0; i < tabX.length; i++) {
        sumA = 0;
        sumB = 0;
        elemA = new int[tabX.length - (i + 1)];
        elemB = new int[i + 1];
        for (int j = 0; j < tabX.length - (i + 1); j++) {
            elemA[j] = tabX[j];
            sumA += elemA[j];
        }
        for (int k = 0; k < elemB.length; k++) {
            elemB[k] = tabX[tabX.length - (k + 1)]; //i
            sumB += tabX[tabX.length - (k + 1)];
        }
        if (sumA == sumB) {
            System.out.println("Secventa A = " + Arrays.toString(elemA));
            System.out.println("Secventa B = " + Arrays.toString(elemB));
            break;
        } else if (sumA < sumB) {
            System.out.println("Not a solution");
            break;
        }
        /************ Swap element in array****************/
        lastButOne = tabX[tabX.length - (i + 2)];
        int indexOfLastButOne = Arrays.binarySearch(tabX, lastButOne);
        tmp = tabX[i];
        tabX[indexOfLastButOne] = tmp; //here is error
        tabX[i] = lastButOne;
    } //for
} //while

You haven't indicated which line is throwing the exception, and this is way more code than is required. Look at the exception - it looks like you're trying to access element -6. Work out why, using a debugger if necessary, then fix the problem. — Jon Skeet
– Jon Skeet, Commented Mar 3, 2015 at 14:33
Have you checked the size from array tabX[] before you started with for()? — Reporter
– Reporter, Commented Mar 3, 2015 at 14:34

Bram · Accepted Answer · 2015-03-03 14:48:40Z

2

Well, that's exactly how Arrays.binarySearch works: when the actual element you're asking for isn't there, you get the -(insertion point) - 1, which happens to be -6 in your case.

But also: binarySearch assumes the elements are sorted, so if they're not, it can indeed happen that it isn't found, even if the value is in the list.

edited Mar 3, 2015 at 14:48

answered Mar 3, 2015 at 14:37

Bram

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Axel · Accepted Answer · 2015-03-03 14:37:27Z

1

It would be helpfull if you pointed out where the exception is thrown (add a comment like // AIOOBE here. From looking at your code, I guess you get a negative return value from Arrays.binarySearch():

    int indexOfLastButOne = Arrays.binarySearch(tabX, lastButOne);
    ...
    tabX[indexOfLastButOne] = tmp; // ArrayOutOfBoundsException

From the documentation of Arrays.binarySearch: returns index of the search key, if it is contained in the array; otherwise, (-(insertion point) - 1).

answered Mar 3, 2015 at 14:37

Axel

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1 Comment

Ambra69 Over a year ago

Yes you are right about Arrays.binarySearch ,i debuged and the value is stored in tabX. P.S i am sorry about that , forget to show the error line

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