On Mon, 18 Sep 2023, Florian Weimer via Gcc wrote:
> But the contraction would still be valid after an isfinite check
> (something that ranger might catch these days), or with with
> -ffinite-math-only in general. Right?
Nope, still not valid for negative zero ('x + x - x' would yield
positive zero in the default rounding mode).
> > Contracting 'x + x - x' to fma(x, 2, -x) would be fine.
>
> It still changes the result, doesn't it?
I don't follow. I doesn't change the result for infinities (produces
a NaN). It changes the result when x is so large that 'x + x' is
not representable (exponent would overflow), but that's exactly what
contraction is about?
Alexander
