Is this a part of C 99 or just an addon in some windows compilers like
pellesC ?
pellesC ?
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Re: long double
broli wrote:It has been a part of all C language standards: ISO C90, C95, C99.Is this a part of C 99 or just an addon in some windows compilers like
pellesC ?
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Re: long double
"broli" <Broli00@gmail. comwrote in message news:Basically it's a hardware question. Whilst floating point can be emulated inIs this a part of C 99 or just an addon in some windows compilers like
pellesC ?
>
software, if you have to do this it's unlikely that you need much precision.
If the hardware supports 80 bit floats then the C compiler has to support
them, in practise, or people would complain. I think it is actually part of
C99, for what it's worth, but that standard hasn't received wide acceptance,
so it doesn't really matter whether long double is in it or not.
--
Free games and programming goodies.
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Re: long double
Malcolm McLean wrote:You seem to be confused about what 'long double' means. There is no"broli" <Broli00@gmail. comwrote in message news:Basically it's a hardware question. Whilst floating point can be emulated in>Is this a part of C 99 or just an addon in some windows compilers like
>pellesC ?
>>
software, if you have to do this it's unlikely that you need much precision.
If the hardware supports 80 bit floats then the C compiler has to support
them, in practise, or people would complain. I think it is actually part of
C99, for what it's worth, but that standard hasn't received wide acceptance,
so it doesn't really matter whether long double is in it or not.
requirement that long double be at least 80 bits. Although long double
is part of C99, it is not guaranteed to be any larger or store any more
values than double.
Philip
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Re: long double
broli wrote:long double is part of C99Is this a part of C 99 or just an addon in some windows compilers like
pellesC ?
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
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Re: long double
Malcolm McLean wrote:No, the question was about the standard. The questions is>
"broli" <Broli00@gmail. comwrote in message news:Basically it's a hardware question.>Is this a part of C 99 or just an addon in some windows compilers like
>pellesC ?
>>
Are "long double"s part of C99?
There is no hint as to any hardware questions.
No, this is completely wrong. The compiler can IGNORE them and if peopleWhilst floating point can be
emulated in software, if you have to do this it's unlikely that you need
much precision. If the hardware supports 80 bit floats then the C
compiler has to support them, in practise, or people would complain.
complain the compiler builders will not give a damm.
Example at hand: In Microsoft compilers, long double is the same as
double. The users are complaining since 10 years and they go on.
And your opinion about C99 matters even less.I
think it is actually part of C99, for what it's worth, but that standard
hasn't received wide acceptance, so it doesn't really matter whether
long double is in it or not.
>
C99 is the current C standard, even if you persist on trying to
disqualify it.
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
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Re: long double
jacob navia <jacob@nospam.c omwrites:Yes. long double is (was?) also part of C89/C90, and I'm reasonablybroli wrote:>>Is this a part of C 99 or just an addon in some windows compilers like
>pellesC ?
long double is part of C99
sure that the requirements for C90 did not change between C90 and C99.
C99 did make a minor change in the printf formats for long double, and
added math functions for long double, but made no change in the type
itself.
I don't know where this idea that long double is a C99-specific
feature came from, but it's mistaken. (I'm not suggesting that jacob
made this mistake; I'm just providing additional information.)
--
Keith Thompson (The_Other_Keit h) <kst-u@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
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Re: long double
"broli" <Broli00@gmail. comwrote in message
news:eb8adef5-9db8-4440-97e1-c516d35d596e@q2 7g2000prf.googl egroups.com...Long double seems to be part of both C90 and C99 standards. And in both itIs this a part of C 99 or just an addon in some windows compilers like
pellesC ?
seems to be at least 32-bits.
Whether you get any more precision depends on your compiler, but it may not
necessarily be more than double.
--
Bart
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Re: long double
Bartc wrote:Yes."broli" <Broli00@gmail. comwrote in message
news:eb8adef5-9db8-4440-97e1-c516d35d596e@q2 7g2000prf.googl egroups.com...>>Is this a part of C 99 or just an addon in some windows compilers like
>pellesC ?
Long double seems to be part of both C90 and C99 standards.
This needs some explanation. There must be at leastAnd in both it
seems to be at least 32-bits.
LDBL_DIG * lg(10) bit-equivalents in the fractional part
of a long double, and since LDBL_DIG must be at least 10
this implies a fraction of at least 33.2+ bit-equivalents.
There must also be enough distinct exponent values to cover
the range LDBL_MIN_10_EXP through LDBL_MAX_10_EXP , which
must be at least the range -37 through +37; 75 or more
distinct values implies lg(75) = 6.2+ bit-equivalents.
Finally there's the sign, for one more bit-equivalent. All
in all, long double needs at least 40.4+ bit-equivalents.
Since 40.4+ is "at least 32," you're right.
Right: the minimum requirements for long double's rangeWhether you get any more precision depends on your compiler, but it may not
necessarily be more than double.
and precision are the same as those for plain double, so the
implementation is not obliged to do any better. It must not
do any worse, though.
--
Eric.Sosman@sun .com
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Re: long double
Eric Sosman wrote:No, you only need (LDBL_DIG * lg(10) - 1) bit-equivalents in binary. InBartc wrote:>>"broli" <Broli00@gmail. comwrote in message
>news:eb8adef 5-9db8-4440-97e1-c516d35d596e@q2 7g2000prf.googl egroups.com...>Long double seems to be part of both C90 and C99 standards.>>Is this a part of C 99 or just an addon in some windows compilers like
>>pellesC ?
Yes.
>>>And in both it
>seems to be at least 32-bits.
This needs some explanation. There must be at least
LDBL_DIG * lg(10) bit-equivalents in the fractional part
of a long double, and since LDBL_DIG must be at least 10
this implies a fraction of at least 33.2+ bit-equivalents.
binary floating point, the most significant bit is always 1 except for
unnormalised numbers, infinities, NaNs, and zero. The first three aren't
required to exist, and the minimum possible value can be defined to be zero.
And hence I think you only need 39.4 bits in binary (and even in otherThere must also be enough distinct exponent values to cover
the range LDBL_MIN_10_EXP through LDBL_MAX_10_EXP , which
must be at least the range -37 through +37; 75 or more
distinct values implies lg(75) = 6.2+ bit-equivalents.
Finally there's the sign, for one more bit-equivalent. All
in all, long double needs at least 40.4+ bit-equivalents.
>
Since 40.4+ is "at least 32," you're right.
radices, you could probably exploit the fact that the most significant
digit is nonzero to reduce the requirement below 40.4 bits).
Indeed. The set of double values are a subset of long double values.>>Whether you get any more precision depends on your compiler, but it may not
>necessarily be more than double.
Right: the minimum requirements for long double's range
and precision are the same as those for plain double, so the
implementation is not obliged to do any better. It must not
do any worse, though.
Philip
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Re: long double
On Mon, 07 Apr 2008 14:40:55 -0400, Eric Sosman <Eric.Sosman@su n.com>
wrote:
Possibly minus 1 for hidden-bit, as already discussed.Bartc wrote:><snip[long double] seems to be at least 32-bits.
This needs some explanation. There must be at least
LDBL_DIG * lg(10) bit-equivalents in the fractional part
of a long double, and since LDBL_DIG must be at least 10
this implies a fraction of at least 33.2+ bit-equivalents.
There is no need for the exponent's base to be 10. Commonly it is 2There must also be enough distinct exponent values to cover
the range LDBL_MIN_10_EXP through LDBL_MAX_10_EXP , which
must be at least the range -37 through +37; 75 or more
distinct values implies lg(75) = 6.2+ bit-equivalents.
(which allows maximum density of a binary significand, and hidden-bit)
and a dynamic range of 1e-/+37 is roughly 2 up -/+123 for 8 bits of
exponent. But S/360 uses 16 (as you noted) and needs only 6 bits of
exponent for the same range. And it would be legal to use a larger
base, but the loss in significand precision grows much faster than the
gain in reduced exponent size, so people don't.
<snip rest>
- formerly david.thompson1 || achar(64) || worldnet.att.ne t
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Re: long double
"David Thompson" <dave.thompson2 @verizon.netwro te in message
news:digl04533h 6k100034q1mj1c9 1ek3qjh5d@4ax.c om...I guess it is not true on a binary machine, but I once saw a proof thatOn Mon, 07 Apr 2008 14:40:55 -0400, Eric Sosman <Eric.Sosman@su n.com>
wrote:
>>>Bartc wrote:>><snip[long double] seems to be at least 32-bits.
> This needs some explanation. There must be at least
>LDBL_DIG * lg(10) bit-equivalents in the fractional part
>of a long double, and since LDBL_DIG must be at least 10
>this implies a fraction of at least 33.2+ bit-equivalents.
Possibly minus 1 for hidden-bit, as already discussed.
>>>There must also be enough distinct exponent values to cover
>the range LDBL_MIN_10_EXP through LDBL_MAX_10_EXP , which
>must be at least the range -37 through +37; 75 or more
>distinct values implies lg(75) = 6.2+ bit-equivalents.
There is no need for the exponent's base to be 10. Commonly it is 2
(which allows maximum density of a binary significand, and hidden-bit)
and a dynamic range of 1e-/+37 is roughly 2 up -/+123 for 8 bits of
exponent. But S/360 uses 16 (as you noted) and needs only 6 bits of
exponent for the same range. And it would be legal to use a larger
base, but the loss in significand precision grows much faster than the
gain in reduced exponent size, so people don't.
maximum information density requires base e (and therefore, for integral
bases, 3 is a bit more compact than binary since it is closer to e than 2
is.) I guess that a ternary machine would be needed to express the data
most compactly using this representation.
** Posted from http://www.teranews.com **
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Re: long double
Dann Corbit wrote:.... snip ...>And then you have to consider the added physical complexity of>
I guess it is not true on a binary machine, but I once saw a proof
that maximum information density requires base e (and therefore,
for integral bases, 3 is a bit more compact than binary since it
is closer to e than 2 is.) I guess that a ternary machine would
be needed to express the data most compactly using this
representation.
building a flip-flap-flop (in place of a flip-flop), and
implementing tri-level gates, etc.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.
** Posted from http://www.teranews.com **
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Re: long double
On Apr 22, 2:21 pm, CBFalconer <cbfalco...@yah oo.comwrote:With +voltage, ground, -voltage the idea is so obvious I would guessDann Corbit wrote:
>
... snip ...
>>I guess it is not true on a binary machine, but I once saw a proof
that maximum information density requires base e (and therefore,
for integral bases, 3 is a bit more compact than binary since it
is closer to e than 2 is.) I guess that a ternary machine would
be needed to express the data most compactly using this
representation.
And then you have to consider the added physical complexity of
building a flip-flap-flop (in place of a flip-flop), and
implementing tri-level gates, etc.
someone has tried it.
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Re: long double
On Apr 22, 3:31 pm, user923005 <dcor...@connx. comwrote:On Apr 22, 2:21 pm, CBFalconer <cbfalco...@yah oo.comwrote:
>>Dann Corbit wrote:>... snip ...>I guess it is not true on a binary machine, but I once saw a proof
that maximum information density requires base e (and therefore,
for integral bases, 3 is a bit more compact than binary since it
is closer to e than 2 is.) I guess that a ternary machine would
be needed to express the data most compactly using this
representation.>And then you have to consider the added physical complexity of
building a flip-flap-flop (in place of a flip-flop), and
implementing tri-level gates, etc.
With +voltage, ground, -voltage the idea is so obvious I would guess
someone has tried it.
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