On 10/1/2016 8:46 PM, Andrei Alexandrescu 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
Interestingly, this is one of the few algorithms that can be tested with an
exhaustive test of all possibilities!
