On 5/19/15 7:41 PM, deadalnix wrote:
On Tuesday, 19 May 2015 at 22:43:46 UTC, Steven Schveighoffer wrote:
On 5/19/15 5:32 PM, deadalnix wrote:
On Tuesday, 19 May 2015 at 20:09:23 UTC, Steven Schveighoffer wrote:
On 5/19/15 4:01 PM, deadalnix wrote:
Have you tried things like :
(x >> bsr(x)) == 1 ?
I have no idea if this is faster or not, but worth trying.
Hm.. I think this would always succeed. Perhaps you mean:
1 << bsr(x) == x;
Both work as long as you use a fully defined instruction, like tzcnt.
Hm... I messed up, (x >> bsr(x)) is always zero. I think you meant to
write:
x >> (bsr(x) - 1)
which always is 1.
Either way, it doesn't work.
No.
bsr(1) is 0.
1 >> bsr(1) is 1.
Gah, I messed up, used output from old code that wasn't doing what I
thought it was. You are right about that.
But my whole point there was that x >> bsr(x) is ALWAYS 1.
2 >> bsr(2) == 1
3 >> bsr(3) == 1
4 >> bsr(4) == 1
17 >> bsr(17) == 1
So really, your test checks to see if a value is zero or not, not
whether it's a power of 2.
BUT, the opposite mechanism would work:
1 << bsr(x) == x;
0 >> anything is 0.
Hah, good point :) Even if bsr(0) is undefined it doesn't matter. Didn't
think of that.
But that means the opposite solution I mentioned above, doesn't work,
still back to square one.
-Steve
