std.math.isPowerOf2

[Manu via Digitalmars-d]

On 2 October 2016 at 13:46, Andrei Alexandrescu via Digitalmars-d
<digitalmars-d at puremagic.com> wrote:
> On 10/01/2016 11:05 PM, Manu via Digitalmars-d wrote:
>>
>> Unsigned case is:
>>   return (x & -x) > (x - 1);
>>
>> Wouldn't this be better:
>>   return (sz & (sz-1)) == 0;
>>
>> I also don't understand the integer promotion and recursive call in
>> the integer case. Can someone explain how the std.math implementation
>> is ideal?
>
>
> The intent is to return 0 when the input is 0. Looking at
> https://github.com/dlang/phobos/blob/master/std/math.d, the implementation
> for signed integers might be simplified a bit. -- Andrei

Yeah, I feel that's probably sub-optimal, but I haven't tried to solve
with that case in mind.
I have a feeling that if you have to handle x == 0, then it might be
possible to make the signed and unsigned cases identical... it smells
like there's a '>' in there in that case, which should be able to
eliminate negative cases aswell as the 0 case.

I'm not sure this is written in a way where, if you're able to
convince the optimiser that x > 0, that the optimiser is able to
eliminate the unnecessary work.
It's pretty easy to convince the optimiser of valid value ranges, and
in the case you demonstrate x > 0, it should empower the optimiser to
produce the most efficient version.