On Mon, Sep 19, 2022 at 4:04 PM Michael Matz <m...@suse.de> wrote:
>
> Hello,
>
> On Mon, 19 Sep 2022, Richard Biener via Gcc-patches wrote:
>
> > > but I guess it's good we do the right thing for correctness sake, and
> > > if it ever gets used by someone else.
> > >
> > > >
> > > > That said, 'set_nonnegative' could be interpreted to be without
> > > > NaNs?
> > >
> > > Sounds good to me. How's this?
> >
> > Hmm, I merely had lots of questions above so I think to answer them
> > we at least should document what 'set_nonnegative' implies in the
> > abstract vrange class.
> >
> > It's probably safer to keep NaN in. For example unfolded copysign (x, 1.)
> > will return true for x == -NaN but the result will be a NaN.
>
> FWIW, in IEEE, 'abs' (like 'copy, 'copysign' and 'negate') are not
> arithmetic, they are quiet-computational. Hence they don't rise
> anything, not even for sNaNs; they copy the input bits and appropriately
> modify the bit pattern according to the specification (i.e. fiddle the
> sign bit).
>
> That also means that a predicate like negative_p(x) that would be
> implemented ala
>
> copysign(1.0, x) < 0.0
I suppose this means -0.0 is not considered negative, though it has
the signbit set? FWIW, on real_value's real_isneg() returns true for
-0.0 because it only looks at the sign.
>
> deal with NaNs just fine and is required to correctly capture the sign of
> 'x'. If frange::set_nonnegative is supposed to be used in such contexts
> (and I think it's a good idea if that were the case), then set_nonnegative
> does _not_ imply no-NaN.
>
> In particular I would assume that, given an VAYRING frange FR, that
> FR.set_nonnegative() would result in an frange {[+0.0,+inf],+nan} .
That was my understanding as well, and what my original patch did.
But again, I'm just the messenger.
Aldy
