On Tue, Apr 10, 2018 at 5:50 PM, Rich Felker wrote: > On Tue, Apr 10, 2018 at 05:41:46PM -0300, dgutson . wrote: > > On Tue, Apr 10, 2018 at 5:32 PM, Rich Felker wrote: > > > > > On Tue, Apr 10, 2018 at 05:23:12PM -0300, dgutson . wrote: > > > > On Tue, Apr 10, 2018 at 4:50 PM, Rich Felker > 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 > I was worried by the case when x is 0 and y is the next (or previous) representable value nearest to 1.0; the y == 1.0 check will fail, but the division may get big; so I did a small program and verified that the result of the division is about 3.24519e+32 when going towards negative and 8.11296e+31 when going towards positive, so everything is OK (I didn't dig in the atan2 arguments though). > 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 > -- Who’s got the sweetest disposition? One guess, that’s who? Who’d never, ever start an argument? Who never shows a bit of temperament? Who's never wrong but always right? Who'd never dream of starting a fight? Who get stuck with all the bad luck?