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Message-ID:	<513900D2.6040106@gmx.de>
Date:	Thu, 07 Mar 2013 22:04:18 +0100
From:	Juan Manuel Guerrero <juan DOT guerrero AT gmx DOT de>
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To:	Eli Zaretskii <eliz AT gnu DOT org>
CC:	djgpp-workers AT delorie DOT com
Subject:	Re: Printing sign of NaN.
References:	<51364BCD DOT 1030807 AT gmx DOT de> <83y5e11pn7 DOT fsf AT gnu DOT org> <5137961A DOT 4000505 AT gmx DOT de> <83mwug1bnn DOT fsf AT gnu DOT org>
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Reply-To:	djgpp-workers AT delorie DOT com

Am 06.03.2013 22:06, schrieb Eli Zaretskii:
>> Date: Wed, 06 Mar 2013 20:16:42 +0100
>> From: Juan Manuel Guerrero <juan DOT guerrero AT gmx DOT de>
>> CC: Eli Zaretskii <eliz AT gnu DOT org>
>>
>>   > Isn't NaN always negative?  Do you succeed in printing both negative
>>   > and positive NaN with this patch?
>>
>> Yes, I can print both signs with this change.
> And what about reading a signed NaN?  Does it produce a value of that
> same sign when printed afterwards?
>
Sorry but I do not fully understand what exactly I shall test.
I have run the code snippet below on linux and using djgpp.
I think when the additions are passed as arguments to printf
the "signed " NaN is read.  Both give the same result like below:

NaN + NaN = -nan   NaN + NaN = -nan   NaN + v = -nan
NaN + NaN = +nan   NaN + NaN = nan   NaN + v = nan

Regards,
Juan M. Guerrero



#include <stdio.h>


typedef struct {
   unsigned mantissal:32;
   unsigned mantissah:32;
   unsigned exponent:15;
   unsigned sign:1;
} long_double_t;

typedef union
{
   long double ld;
   long_double_t ldt;
} _longdouble_union_t;


int main(void)
{
   long double v = 1.2345;
   _longdouble_union_t nan;


   nan.ldt.mantissal = 0x00000001U;
   nan.ldt.mantissah = 0xC0000000U;
   nan.ldt.exponent  = 0x7FFFU;


   nan.ldt.sign      = 1;
   printf("NaN + NaN = %+Lg   NaN + NaN = %-Lg   NaN + v = %Lg\n", 
nan.ld + nan.ld, nan.ld + nan.ld, nan.ld + v);

   nan.ldt.sign      = 0;
   printf("NaN + NaN = %+Lg   NaN + NaN = %-Lg   NaN + v = %Lg\n", 
nan.ld + nan.ld, nan.ld + nan.ld, nan.ld + v);

   return 0;
}

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