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

[Michael Matz]

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.

While the interpretation of sign bits in NaNs isn't specified in 
IEEE754/854/P754, its existence is a given (in particular it talks about 
"the sign of a NaN", in order to say that their interpretation isn't 
determined :) )

Further IEEE754 recommends implementations to provide a negate(x) 
operation that copies x with reversed sign, that is to work on NaNs (and 
due to copysign needs to have observable behaviour).  C99 and up specify 
that the unary '-' operator maps to that operation.

So, I think it's pretty clear, that whatever the sign bit of NaNs is 
supposed to mean, it must be switchable by unary '-' when IEEE754 
conformance is claimed.  We currently don't claim so, but we aim for it if 
possible :)

So the current "-0.0-x" expansion of unary '-' needs a change.  It turned 
out to be a bit uglier than I wished for, but alas, fixed in mob.

Thanks for the report, Arnold.


Ciao,
Michael.