On Saturday, 2 July 2016 at 20:17:59 UTC, Andrei Alexandrescu
wrote:
So what's the fastest way to figure that an integral is
convertible to a floating point value precisely (i.e. no other
integral converts to the same floating point value)? Thanks! --
Andrei
bool isConvertible(T) (long n) if (is(T == float) || is(T ==
double)) {
pragma(inline, true);
static if (is(T == float)) {
return (((n+(n>>63))^(n>>63))&0xffffffffff000000UL) == 0;
} else {
return (((n+(n>>63))^(n>>63))&0xffe0000000000000UL) == 0;
}
}