Why does Arrays.binarySearch returning that value?

[Mike Savvy]

Hello!

    int [] numberes = new int[] {3,2,1};
    System.out.println(Arrays.binarySearch( numberes, 3 ));

Why does returned value equals -4 ?
If I add 0 at the end of the array,  it will be -5, but if I add a 3 instead of 0, it will be 3. Then I add 3 and 45 ( {3,2,1,0,3,45} ) and it will 4...

I know, the returned value is not defined for not sorted array, but I think it is important to understand why this values and not other.

P.S. Explanation in the book OCAJ SE8 from Selikoff&Boyarsky on page 126 is not enough.

Best regards
Mike Savvy

[Stephan van Hulst]

Mike Savvy wrote:I know, the returned value is not defined for not sorted array, but I think it is important to understand why this values and not other.

It's not important. Undefined means undefined.

While it's not important, it might be interesting.

The reason is that it first compares 3 to the value at the middle of the array, which is 2. It sees that 3 is greater, so it then continues to the right half of the array. It sees that 3 is greater than 1, so it continues to the right of 1. There are no more elements there, and it hasn't found the value 3, so instead of returning the index of the found element, it returns (-(insertion point) - 1).

For an insertion point of 3 (the index to the right of value 1), the return value is (-(3) - 1) = -4.

[Mike Savvy]

I see, thank you Stephan van Hulst!