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

From: ian
Subject: Re: [Tinycc-devel] x86_64 tcc doesn't set sign bit on NaNs
Date: Wed, 6 Jan 2021 02:13:57 +0100
User-agent: Mozilla/5.0 (X11; Linux i686; rv:68.0) Gecko/20100101 Thunderbird/68.2.2

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)
? (/ 1 0)
? (* 0 (/ -1 0))

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

Regards to all, and best wishes.


Le 05/01/2021 à 10:27, Vincent Lefevre a écrit :
On 2021-01-04 04:59:28 +0100, Michael Matz wrote:

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;


	$ 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.

-- sibian0218@gmail.com
-- Développeur compulsif

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