LayerMask > GameObject.layer

Unity Engine
1

Embarrassing problem. Anyway. I have a LayerMask that always only has 1 layer activated. How do I get the corresponding integer GameObject layer for this mask value?

Example
00000000000000000000000010000000

This value is 128, but I need the offset which is 8. How do I get this?

Thank you.

NOTE! This is not about getting the layermask for gameobject layer. It’s the other way around.

2

Doing retarded solution until someone replies with something better

    public static int GetGameObjectLayer(this LayerMask mask)
    {
        for (int i = 0; i < 32; ++i)
        {
            if (((1 << i) & mask.value) > 0)
            {
                return i;
            }
        }

        return -1;
    }
1 Like
3

When you have a layermask with a single layer only, the value (eg 128) is 2 raised to the power of that layers number (in this case 2^7 )
What you want is to ask “What power must I raise 2 by to get x”
Luckily, this is what Log does
layer = Mathf.Log(value, 2)
Nb. It returns a float, you might want/need to cast to an int

This will fail when the mask has more layers toggled

ps you said it was 8, but the right-most bit is bit 0, layer zero

2 Likes
4

Edit: I didn’t check out your code before I wrote this because you said it was retarded, causing me to skip over it before trying to help. That is “retarded”, but at least now I know more about myself. Thanks. I will try to avoid making the same mistake in the future.

At least maybe the nullable and the cross-platform calculation of int’s bits will help you.

public static class IntExtensions {
    public static int? GetIndexOfLowestTrueBit(this int flag) {
        for (var index = 0; index < sizeof(int) * 8; index++)
            if ((flag & 1) == 1) return index;
            else flag >>= 1;

        return null;
    }
}
2 Likes
5

Its all based 2 so its simple math really, you can either bit shift or use Algebra…

2^x = LayerMask.value

x = Log(LayerMask.value) / Log(2). Where log represents Log base e or Natural Log. I solved this problem today.

1 Like