Checking if an Integer is an Exact Binary Power

[Dominikus Dittes Scherkl via Digitalmars-d]

On Monday, 25 April 2016 at 22:42:38 UTC, deadalnix wrote:
> x & -x is the smallest power of 2 that divides x. Basically, if 
> x = xxxx1000 , x & -x = 00001000 . This is easy to proves 
> considering -x = ~x + 1 .
>
> Now, x >> 1 will be of the form 0xxxx100 . If one of these 
> digit is one, then (x >> 1) >= (x & -x) . If they are all 
> zeros, (x >> 1) < (x & -x) which is the case where you have a 
> power of 2.
>
> 0 is a special case, can it can be checked that the function 
> return false for this specific input.
>
> This looks like it is correct.

Yes. Except for the case 0x80000000 (= int.min), because this is 
negative so NOT smaller than 0x40000000 (= int.min>>>1), which is 
considered positive.
So the algorithm doesn't work for signed integers (without extra 
cast).