[9fans] libfmt/libutf and long double

[9fans] libfmt/libutf and long double

[suspect]

Hello,

I've run into a behavior with libfmt/libutf on UNIX system
that does not seem intuitive:

	long double ldv = 3.3;
	print("long double var = %llE\n", ldv);

gives:
	long double var = -6.257775E+92

Identical code under Plan 9 gives the expected output of:

	long double var = 3.300000E+00

Is there any particular reason why the behaviors should
differ in this manner ?

cheers,

Re: [9fans] libfmt/libutf and long double

[David Presotto]

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Could it be that we don't know squat about long double?  At least in
our plan9 ape library, we just assume that the 'E' arg is a double
and don't look at the ll flags.  I don't know about the Unix libfmt.

Look at __efgfmt in fltfmt.c and see what the expected arg is.

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From: suspect@suspicious.org
To: 9fans@cse.psu.edu
Subject: [9fans] libfmt/libutf and long double
Date: Mon, 29 Dec 2003 22:10:41 -0500
Message-ID: <C01AD478-3A75-11D8-BD48-000A95C53D8A@suspicious.org>

Hello,

I've run into a behavior with libfmt/libutf on UNIX system
that does not seem intuitive:

	long double ldv = 3.3;
	print("long double var = %llE\n", ldv);

gives:
	long double var = -6.257775E+92

Identical code under Plan 9 gives the expected output of:

	long double var = 3.300000E+00

Is there any particular reason why the behaviors should
differ in this manner ?

cheers,

Re: [9fans] libfmt/libutf and long double

[Russ Cox]

> Hello,
>
> I've run into a behavior with libfmt/libutf on UNIX system
> that does not seem intuitive:
>
> 	long double ldv = 3.3;
> 	print("long double var = %llE\n", ldv);
>
> gives:
> 	long double var = -6.257775E+92
>
> Identical code under Plan 9 gives the expected output of:
>
> 	long double var = 3.300000E+00
>
> Is there any particular reason why the behaviors should
> differ in this manner ?

yes, plan 9 doesn't implement long double and your unix does.

Re: [9fans] libfmt/libutf and long double

[suspect]

Hello,

Actually, my curiosity was about why it worked on my
Plan 9 machine but gave the strange behavior when
using libfmt on a UNIX system.

 From http://www.cs.bell-labs.com/magic/man2html/2/fprintf:
	"... or an optional L specifying that a following e, E, f, g,
	or G conversion specifier applies to a long double argument."

The above made it make sense to me that it worked on
my Plan 9 machine. I'm therefore a bit confused by your
response. Thanks.


cheers,


On Dec 31, 2003, at 11:17 AM, Russ Cox wrote:
>
> yes, plan 9 doesn't implement long double and your unix does.
>




>> Hello,
>>
>> I've run into a behavior with libfmt/libutf on UNIX system
>> that does not seem intuitive:
>>
>> 	long double ldv = 3.3;
>> 	print("long double var = %llE\n", ldv);
>>
>> gives:
>> 	long double var = -6.257775E+92
>>
>> Identical code under Plan 9 gives the expected output of:
>>
>> 	long double var = 3.300000E+00
>>
>> Is there any particular reason why the behaviors should
>> differ in this manner ?

Re: [9fans] libfmt/libutf and long double

[David Presotto]

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On plan 9 double and long double are the same.  Therefore, the
fact that the libfmt library ignores the ll part of the %llE
doesn't break anything on plan 9 but does on Unix.

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From: suspect@suspicious.org
To: 9fans@cse.psu.edu
Subject: Re: [9fans] libfmt/libutf and long double
Date: Wed, 31 Dec 2003 13:43:31 -0500
Message-ID: <3B0EB7C6-3BC1-11D8-906F-000A95C53D8A@suspicious.org>

Hello,

Actually, my curiosity was about why it worked on my
Plan 9 machine but gave the strange behavior when
using libfmt on a UNIX system.

 From http://www.cs.bell-labs.com/magic/man2html/2/fprintf:
	"... or an optional L specifying that a following e, E, f, g,
	or G conversion specifier applies to a long double argument."

The above made it make sense to me that it worked on
my Plan 9 machine. I'm therefore a bit confused by your
response. Thanks.


cheers,


On Dec 31, 2003, at 11:17 AM, Russ Cox wrote:
>
> yes, plan 9 doesn't implement long double and your unix does.
>




>> Hello,
>>
>> I've run into a behavior with libfmt/libutf on UNIX system
>> that does not seem intuitive:
>>
>> 	long double ldv = 3.3;
>> 	print("long double var = %llE\n", ldv);
>>
>> gives:
>> 	long double var = -6.257775E+92
>>
>> Identical code under Plan 9 gives the expected output of:
>>
>> 	long double var = 3.300000E+00
>>
>> Is there any particular reason why the behaviors should
>> differ in this manner  ?