Re: [PATCH] [PR24021] Implement PLUS_EXPR range-op entry for floats.

[Aldy Hernandez <aldyh@redhat.com>]

On Tue, Nov 8, 2022 at 2:25 PM Jakub Jelinek <jakub@redhat.com> wrote:
>
> On Tue, Nov 08, 2022 at 02:06:58PM +0100, Aldy Hernandez wrote:
> > +  gcc_checking_assert (!r.nan_signbit_p (sign1));
> > +  if (op1_nan && op2_nan)
> > +    {
> > +      // If boths signs agree, we could use that sign, but IEEE754
> > +      // does not guarantee this for a binary operator.  The x86_64
> > +      // architure does keep the common known sign, but further tests
> > +      // are needed to see if other architectures do the same (i387
> > +      // long double, ARM/aarch64, PowerPC, s390,{,x}, RSICV, etc).
>
> s/RSICV/RISCV/
>
> > +      // In the meantime, keep sign VARYING.
> > +      ;
> > +    }
> > +  else if (op1_nan)
> > +    {
> > +      if (op1.nan_signbit_p (sign1))
> > +     r.update_nan (sign1);
> > +    }
> > +  else if (op2_nan)
> > +    {
> > +      if (op2.nan_signbit_p (sign2))
> > +     r.update_nan (sign2);
> > +    }
>
> Well, these cases also aren't guaranteed for binary operator.
> I think a conforming implementation can say copy the NaN operand
> to result and toggle the sign.  Or, if the operand would be a sNaN,
> it must turn it into a qNaN (don't remember right now if there are
> requirements on what it can do with the mantissa which needs to change
> for the sNaN -> qNaN difference at least, but whether it can just
> generate a canonical qNaN or needs to preserve at least some bits),
> but could e.g. clear or toggle the sign of the NaN as well.
> Whether there are any such implementations or not is a question.
> For the single qNaN operand case, it would surprise me if anybody
> bothered to tweak the sign bit in any way, just copying the input
> seems simpler to me, but for the sNaN -> qNaN conversion it wouldn't
> surprise me that much.

Well, perhaps we should just nuke update_nan_sign() altogether, and
always keep the sign varying?

inline bool
propagate_nans (frange &r, const frange &op1, const frange &op2)
{
  if (op1.known_isnan () || op2.known_isnan ())
    {
      r.set_nan (op1.type ());
      return true;
    }
  return false;
}

I'm fine either way.  The less code the better :).

Aldy