«

Here's an excerpt from the Dinkumware library reference:
__________
| uint8_t, uint16_t, uint32_t, uint64_t
|
| The types each specify an unsigned integer type
| whose representation has exactly eight, 16, 32,
| or 64 bits, respectively.
__________

How could that be possible if we had a 36-Bit system such as the following?

char: 9-Bit
short: 18-Bit
int: 36-Bit
long: 36-Bit

--

Frederick Gotham

Frederick Gotham said:

> Here's an excerpt from the Dinkumware library reference:
> __________
> | uint8_t, uint16_t, uint32_t, uint64_t
> |
> | The types each specify an unsigned integer type
> | whose representation has exactly eight, 16, 32,
> | or 64 bits, respectively.
> __________
>
> How could that be possible if we had a 36-Bit system such as the
> following?
>
> char: 9-Bit
> short: 18-Bit
> int: 36-Bit
> long: 36-Bit

Well, it couldn't, obviously. And even if it could, would it matter?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Richard Heathfield posted:

> Frederick Gotham said:
>
>> Here's an excerpt from the Dinkumware library reference:
>> __________
>> | uint8_t, uint16_t, uint32_t, uint64_t
>> |
>> | The types each specify an unsigned integer type
>> | whose representation has exactly eight, 16, 32,
>> | or 64 bits, respectively.
>> __________
>>
>> How could that be possible if we had a 36-Bit system such as the
>> following?
>>
>> char: 9-Bit
>> short: 18-Bit
>> int: 36-Bit
>> long: 36-Bit
>
> Well, it couldn't, obviously. And even if it could, would it matter?


I would understand if it read:

Where possible, the types "uint8_t, uint16_t, uint32_t, uint64_t"
denote an unsigned integer type whose representation has exactly...

But it doesn't say "where possible" -- it plainly states that these types
are available... ?

--

Frederick Gotham

Frederick Gotham wrote:
> | uint8_t, uint16_t, uint32_t, uint64_t
[...]
> How could that be possible if we had a 36-Bit system such as the
> following?
>
> char: 9-Bit
> short: 18-Bit
> int: 36-Bit
> long: 36-Bit

IIRC, the terms state that those typedefs are present _if_ there are
suitable types on the system. In the case of a 9 bit byte, they just
wouldn't be there.

What you would get would be int_least8_t etc though, I think.

Uli

Frederick Gotham wrote:
> Here's an excerpt from the Dinkumware library reference:
> __________
> | uint8_t, uint16_t, uint32_t, uint64_t
> |
> | The types each specify an unsigned integer type
> | whose representation has exactly eight, 16, 32,
> | or 64 bits, respectively.
> __________
>
> How could that be possible if we had a 36-Bit system such as the following?
>
> char: 9-Bit
> short: 18-Bit
> int: 36-Bit
> long: 36-Bit
>
I suppose in order to warrant C99 standard compliance the compiler for
such a system would have to cater for the necessary translation and
storage methods. How it does that may not be the programmer's concern,
as long as he/she sticks to standard C.

A similar 'problem' exists for 64 bit variables on 32-bit machines. To
the programmer the long longs are available as if they were a system
default, yet the compiler translates all relevant code to multiple
32-bit manipulations.

Is there a particular (practical) reason for this question??
Sh.

Frederick Gotham wrote:
> Richard Heathfield posted:
>
>> Frederick Gotham said:
>>
>>> Here's an excerpt from the Dinkumware library reference:
>>> __________
>>> | uint8_t, uint16_t, uint32_t, uint64_t
>>> |
>>> | The types each specify an unsigned integer type
>>> | whose representation has exactly eight, 16, 32,
>>> | or 64 bits, respectively.
>>> __________
>>>
>>> How could that be possible if we had a 36-Bit system such as the
>>> following?
>>>
>>> char: 9-Bit
>>> short: 18-Bit
>>> int: 36-Bit
>>> long: 36-Bit
>> Well, it couldn't, obviously. And even if it could, would it matter?
>
>
> I would understand if it read:
>
> Where possible, the types "uint8_t, uint16_t, uint32_t, uint64_t"
> denote an unsigned integer type whose representation has exactly...
>
> But it doesn't say "where possible" -- it plainly states that these types
> are available... ?
>
Perhaps Dinkumware don't support any 36-bit machines.

"Tim Prince" <> wrote in message
news:...

> Frederick Gotham wrote:
>> Richard Heathfield posted:
>>
>>> Frederick Gotham said:
>>>
>>>> Here's an excerpt from the Dinkumware library reference:
>>>> __________
>>>> | uint8_t, uint16_t, uint32_t, uint64_t
>>>> |
>>>> | The types each specify an unsigned integer type
>>>> | whose representation has exactly eight, 16, 32,
>>>> | or 64 bits, respectively.
>>>> __________
>>>>
>>>> How could that be possible if we had a 36-Bit system such as the
>>>> following?
>>>>
>>>> char: 9-Bit
>>>> short: 18-Bit
>>>> int: 36-Bit
>>>> long: 36-Bit
>>> Well, it couldn't, obviously. And even if it could, would it matter?
>>
>>
>> I would understand if it read:
>>
>> Where possible, the types "uint8_t, uint16_t, uint32_t, uint64_t"
>> denote an unsigned integer type whose representation has exactly...
>>
>> But it doesn't say "where possible" -- it plainly states that these types
>> are available... ?
>>
> Perhaps Dinkumware don't support any 36-bit machines.

We don't in our prepackaged libraries, to be sure. Our OEM customers
are free to modify our headers, and documentation, to meet more
specific needs.

P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com

Frederick Gotham <> writes:
> Here's an excerpt from the Dinkumware library reference:
> __________
> | uint8_t, uint16_t, uint32_t, uint64_t
> |
> | The types each specify an unsigned integer type
> | whose representation has exactly eight, 16, 32,
> | or 64 bits, respectively.
> __________
>
> How could that be possible if we had a 36-Bit system such as the following?
>
> char: 9-Bit
> short: 18-Bit
> int: 36-Bit
> long: 36-Bit

It isn't. The uintN_t types are optional.

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

On Sat, 05 Aug 2006 21:45:52 GMT, Keith Thompson <kst-> wrote
in comp.lang.c:

> Frederick Gotham <> writes:
> > Here's an excerpt from the Dinkumware library reference:
> > __________
> > | uint8_t, uint16_t, uint32_t, uint64_t
> > |
> > | The types each specify an unsigned integer type
> > | whose representation has exactly eight, 16, 32,
> > | or 64 bits, respectively.
> > __________
> >
> > How could that be possible if we had a 36-Bit system such as the following?
> >
> > char: 9-Bit
> > short: 18-Bit
> > int: 36-Bit
> > long: 36-Bit
>
> It isn't. The uintN_t types are optional.

Actually, "semi optional" would be more accurate. If an
implementation provides any or all types which meet the definition,
namely exactly that many bits, no padding, and 2's complement
representation for negative values in the signed types, it is required
to provide the appropriate typedefs.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

On Sat, 05 Aug 2006 13:56:08 GMT, Frederick Gotham
<> wrote in comp.lang.c:

> Here's an excerpt from the Dinkumware library reference:
> __________
> | uint8_t, uint16_t, uint32_t, uint64_t
> |
> | The types each specify an unsigned integer type
> | whose representation has exactly eight, 16, 32,
> | or 64 bits, respectively.
> __________
>
> How could that be possible if we had a 36-Bit system such as the following?
>
> char: 9-Bit
> short: 18-Bit
> int: 36-Bit
> long: 36-Bit

Purchase and read a copy of the C standard, 1999 or later, and all
will be revealed to you.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html