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.
>
--
-- sibian0...@gmail.com
-- Développeur compulsif
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