What does this boolean "(number & 1) == 0" mean?
On CodeReview I posted a working piece of code and asked for tips to improve it. One I got was to use a boolean method to check if an ArrayList had an even number of indices (which was required). This was the code that was suggested:
private static boolean isEven(int number)
{
return (number & 1) == 0;
}
As I've already pestered that particular user for a lot of help, I've decided it's time I pestered the SO community! I don't really understand how this works. The method is called and takes the size of the ArrayList as a parameter (i.e. ArrayList has ten elements, number = 10).
I know a single & runs the comparison of both number and 1, but I got lost after that.
The way I read it, it is saying return true if number == 0 and 1 == 0. I know the first isn't true and the latter obviously doesn't make sense. Could anybody help me out?
Edit: I should probably add that the code does work, in case anyone is wondering.
Answer
Keep in mind that "&" is a bitwise operation. You are probably aware of this, but it's not totally clear to me based on the way you posed the question.
That being said, the theoretical idea is that you have some int, which can be expressed in bits by some series of 1s and 0s. For example:
...10110110
In binary, because it is base 2, whenever the bitwise version of the number ends in 0, it is even, and when it ends in 1 it is odd.
Therefore, doing a bitwise & with 1 for the above is:
...10110110 & ...00000001
Of course, this is 0, so you can say that the original input was even.
Alternatively, consider an odd number. For example, add 1 to what we had above. Then
...10110111 & ...00000001
Is equal to 1, and is therefore, not equal to zero. Voila.