On Tue, Apr 10, 2018 at 05:41:46PM -0300, dgutson . wrote:
> On Tue, Apr 10, 2018 at 5:32 PM, Rich Felker <dalias@libc.org> wrote:
>
> > On Tue, Apr 10, 2018 at 05:23:12PM -0300, dgutson . wrote:
> > > On Tue, Apr 10, 2018 at 4:50 PM, Rich Felker <dalias@libc.org> wrote:
> > >
> > > > The OpenBSD catan implementation we're using has a number of
> > > > nonsensical "overflow" (goto ovrf) conditions that aren't errors,
> > > > reported by mepholic on irc. I think the attached patch fixes them
> > > > without introducing new problems, but I'm not sure if any other
> > > > problems remain.
> > > >
> > > > Note that, of the three cases removed:
> > > >
> > > > 1. Is not an exceptional case at all, and made no sense to begin with..
> > > >
> > > > 2. Is only exceptional if x and a are both zero; atan(2x,0) is
> > > > perfectly well-defined.
> > > >
> > > > 3. Is only possible if y==1.0 and x==0.0, which is the only real
> > > > exceptional case for atan: z==I.
> > > >
> > >
> > >
> > > Besides the trigonometric case, are you considering de-normalized
> > numbers,
> > > such as 4.94066e-324 as divisor?
> > > For example:
> > > double x = 1.0;
> > > double y = 5E-324;
> > > x / y is inf, and y != 0.0.
> > > Shouldn't 'a' be checked against that number or its absolute value >=
> > > minimum?
> >
> > Can you clarify where you think something goes wrong?
> >
>
> - if (a == 0.0)
> - goto ovrf;
>
> t = y + 1.0;
> a = (x2 + t * t)/a;
>
>
> The check you removed does not look correct for me because what I mentioned..
> However, shouldn't you check, before the division, that a is not the
> nearest to zero (+ or -) denormalized representable double,
> in order to avoid ending in inf?
Here a=x²+(y-1)², so unless both x==0 and y==1, the smallest a can be
is DBL_EPSILON². When a is small, the numerator in the last line is
also small (x²+(1+y)² < 2) so dividing by a does not overflow.
Rich