On Tue, Sep 20, 2022 at 5:10 PM Jakub Jelinek <ja...@redhat.com> wrote:
>
> On Tue, Sep 20, 2022 at 04:58:38PM +0200, Aldy Hernandez wrote:
> > > > > 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.
> > >
> > > Ah, I obviously haven't followed the thread carefully then. If that's
> > > what it was doing then IMO it was the right thing.
> >
> > This brings me back to my original patch :).
> >
> > Richard, do you agree nonnegative should be [0.0, +INF] U +NAN.
>
> I agree with that. And similarly if there is negative that does the
> opposite [-INF, -0.0] U -NAN.
> Though, in most other places when we see that something may be a NaN, I
> think we need to set both +NAN and -NAN, because at least the 2008 version
> of IEEE 754 says:
Yeah, every other place does update_nan() with no arguments which sets
+-NAN. The only use of update_nan(bool signbit) is this patch.
>
> "When either an input or result is NaN, this standard does not interpret the
> sign of a NaN. Note, however,
> that operations on bit strings — copy, negate, abs, copySign — specify the
> sign bit of a NaN result,
> sometimes based upon the sign bit of a NaN operand. The logical predicate
> totalOrder is also affected by
> the sign bit of a NaN operand. For all other operations, this standard does
> not specify the sign bit of a NaN
> result, even when there is only one input NaN, or when the NaN is produced
> from an invalid
> operation."
Ughh, that means that my upcoming PLUS_EXPR implementation will have
to keep better track of NAN signs.
Pushed original patch.
Thanks.
Aldy
>
> So not sure if we should count on what NaN sign bit we get normally and what
> we get for canonical NaN. If we could rely on it, then the rule is
> that if at least one input to binary operation is NaN, then that NaN is
> copied to result, but if both are NaNs, which one is picked isn't specified,
> so we might need just union the +NAN and -NAN bits from the operands.
> But there are still sNaNs and those ought to be turned into some qNaN and
> dunno if that can change the NaN bit (say turn the sNaN into canonical
> qNaN).
> If neither operand is NaN, but result is NaN because of invalid operation
> (0/0, inf-inf, inf+-inf, sqrt (-1) and the like),
> the result is qNaN, but dunno if we can rely that it will be one with
> positive sign.
>
> Jakub
>
