On Tue, Nov 8, 2022 at 2:50 PM Jakub Jelinek <jakub@redhat.com> wrote:
>
> On Tue, Nov 08, 2022 at 02:47:35PM +0100, Aldy Hernandez wrote:
> > 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 :).
>
> Yes, but you had 2 callers, so something needs to be done also if
> in foperator_plus::fold_range.

We can also remove the update_nan_sign() in the other call because the
r.set() before it sets a default NAN (with a varying sign).

Attached patch in testing.

Aldy