std.math.isPowerOf2

[Andrei Alexandrescu via Digitalmars-d]

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