Re: [Tinycc-devel] x86_64 tcc doesn't set sign bit on NaNs

[ian]

Hi to all, and happy new year.

AFAIK neither -NaN nor +NaN have sense (how to put a sign at a 'not a number' ????).

So, d=-d is a non-sense too.

For instance, in my own language NaN have no sign, but on the other hand I use -Inf and +Inf :

? (/ -1 0)
-Inf.
? (/ 1 0)
+Inf.
? (* 0 (/ -1 0))
NaN

It seems to me that using a sign bit on NaN is a design error.....

Regards to all, and best wishes.

ian

Le 05/01/2021 à 10:27, Vincent Lefevre a écrit :

> On 2021-01-04 04:59:28 +0100, Michael Matz wrote:
> > Hello,
> >
> > On Mon, 4 Jan 2021, Vincent Lefevre wrote:
> >
> > > > -----------------------------
> > > > #include <stdio.h>
> > > > #include <math.h>
> > > > #include <stdlib.h>
> > > >
> > > > int main(int argc, char **argv)
> > > > {
> > > > 	double d = strtod("-nan", NULL);
> > > > 	d = -d;
> > > > 	printf("%g, signbit(d) = %d\n", d, signbit(d));
> > > > 	return 0;
> > > > }
> > > > -----------------------------
> > > >
> > > > Results:
> > > >
> > > > 	$ gcc foo.c -o foo && ./foo
> > > > 	-nan, signbit(d) = 1
> > > >
> > > > 	$ tcc foo.c -o foo2 && ./foo2
> > > > 	nan, signbit(d) = 0
> > > >
> > > > I get the same results as gcc with clang and pcc. tcc is the outlier.
> > > >           
> > > AFAIK, the status of the sign bit of a NaN is unspecified, except
> > > for some particular functions, but not strtod. So I don't see a
> > > bug in tcc.
> > >
> > > Note: for GCC, there's an inconsistency between your testcase
> > > and the result.
> > >         
> > Yeah, I think that's merely a typo in Arnolds email.  The inconsistency is
> > there, applying unary '-' to a NaN doesn't change the sign of it in TCC.
> >       
> But my point is that with the above testcase, you cannot know whether
> the difference between gcc and tcc comes from strtod (which would be
> valid, as strtod doesn't specify the sign or NaN) or the "d = -d;"
> (which would be invalid). A printf should have been added between
> the strtod and the "d = -d;" to be sure.
>
>

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