Is there a good reason not to detect single expression multiply adds and
just emit a new FMA bytecode?
Is our goal for floats to strictly match the result of the same operations
coded in unoptimized C using doubles?
Or can we be more precise on occasion?
I guess a similar question may be asked of all C compilers as they too
could emit an FMA instruction on such expressions... If they don't do it by
default, that suggests we match them and not do it either.
Regardless +1 on adding math.fma() either way as it is an expression of
precise intent.
-gps
On Mon, Jan 16, 2017, 10:44 AM David Mertz <me...@gnosis.cx> wrote:
> My understanding is that NumPy does NOT currently support a direct FMA
> operation "natively." However, higher-level routines like
> `numpy.linalg.solve` that are linked to MKL or BLAS DO take advantage of
> FMA within the underlying libraries.
>
> On Mon, Jan 16, 2017 at 10:06 AM, Guido van Rossum <gvanros...@gmail.com>
> wrote:
>
> Does numpy support this?
>
> --Guido (mobile)
>
> On Jan 16, 2017 7:27 AM, "Stephan Houben" <stephan...@gmail.com> wrote:
>
> Hi Steve,
>
> Very good!
> Here is a version which also handles the nan's, infinities,
> negative zeros properly.
>
> ===============
> import math
> from fractions import Fraction
>
> def fma2(x, y, z):
> if math.isfinite(x) and math.isfinite(y) and math.isfinite(z):
> result = float(Fraction(x)*Fraction(y) + Fraction(z))
> if not result and not z:
> result = math.copysign(result, x*y+z)
> else:
> result = x * y + z
> assert not math.isfinite(result)
> return result
> ===========================
>
> Stephan
>
>
> 2017-01-16 12:04 GMT+01:00 Steven D'Aprano <st...@pearwood.info>:
>
> On Mon, Jan 16, 2017 at 11:01:23AM +0100, Stephan Houben wrote:
>
> [...]
> > So the following would not be a valid FMA fallback
> >
> > double bad_fma(double x, double y, double z) {
> > return x*y + z;
> > }
> [...]
> > Upshot: if we want to provide a software fallback in the Python code, we
> > need to do something slow and complicated like musl does.
>
> I don't know about complicated. I think this is pretty simple:
>
> from fractions import Fraction
>
> def fma(x, y, z):
> # Return x*y + z with only a single rounding.
> return float(Fraction(x)*Fraction(y) + Fraction(z))
>
>
> When speed is not the number one priority and accuracy is important,
> its hard to beat the fractions module.
>
>
> --
> Steve
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