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:
"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."
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
