Convert bitmask back to 9 booleans
bitmask, boolean, c#
Solution
You want to use logical AND (`&`) to mask out your answers, and reverse the way you made `r`:
a = (r & (1 << 0)) != 0;
b = (r & (1 << 1)) != 0;
c = (r & (1 << 2)) != 0;
d = (r & (1 << 3)) != 0;
e = (r & (1 << 4)) != 0;
f = (r & (1 << 5)) != 0;
g = (r & (1 << 6)) != 0;
h = (r & (1 << 7)) != 0;
i = (r & (1 << 8)) != 0;
What this does:
say r was = 14 (i.e. b, c, and d were true). Then the bits of `r` are
00000000 00000000 00000000 00001110
For `a`, 1 << 0 = 1, so take those bits and AND it with
00000000 00000000 00000000 00000001
The result for a is 0 so the bool should be false.
For `b`, 1 << 1 = 2, so take the bits of `r` and AND it with
00000000 00000000 00000000 00000010
You see there's 1 bit in common, so the result of ANDing it is 2, so it's true.
Problem
I have the following code which converts a set of 9 booleans into an int for me, which I'm using for a tilemap sprite array to quickly find the tiles. Just curious, is there a simple way of reversing this? i.e. if all the bool's are true, the result below would be 511. Is there an easy way to convert that back to assign the booleans all true? Or another example, a is false, the rest are true, result would be 510. How can I convert that back to a = true and the rest false? ``` int GetBitmask(bool a, bool b, bool c, bool d, bool e, bool f, bool g, bool h, bool i) { int r = (a ? 1 << 0 : 0) | (b ? 1 << 1 : 0) | (c ? 1 << 2 : 0) | (d ? 1 << 3 : 0) | (e ? 1 << 4 : 0) | (f ? 1 << 5 : 0) | (g ? 1 << 6 : 0) | (h ? 1 << 7 : 0) | (i ? 1 << 8 : 0); return r; } ```