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

Answer 1

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.

Answer 2

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) .