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From "Lucian Adrian Grijincu" <lucian.griji...@gmail.com>
Subject Re: freezing 1.3 tonight
Date Fri, 02 May 2008 02:31:10 GMT
On Fri, May 2, 2008 at 2:52 AM, William A. Rowe, Jr.
<wrowe@rowe-clan.net> wrote:
> Lucian Adrian Grijincu wrote:
> >
> > On Fri, May 2, 2008 at 2:18 AM, Roy T. Fielding <fielding@gbiv.com> wrote:
> >
> >
> > >  Why?  The type char is defined by the C standard to be an 8bit signed
> integer.
> > >  The type unsigned char is defined to be an 8bit unsigned integer.  Why
> would
> > >  we want to add a bunch of unnecessary casting?
> > >
> >
> > Not quite: http://home.att.net/~jackklein/c/inttypes.html
> >
>  That doesn't resolve Roy's question of "why overload signed char and
>  unsigned char"?
>  Can anyone point to a platform where int8_t/uint8_t != signed/unsigned
> char?

I've searched through the Linux kernel sources. This
   typedef __signed__ char __s8;
   typedef unsigned char __u8;
  typedef         __u8            uint8_t;
  typedef         __s8            int8_t;
seems to be the general way to define int8_t and uint8_t (if I didn't
skip any by mistake) regardless of cpu architecture.

This doesn't prove anything, but it puts things in perspective:)
There was a talk (flame) on comp.lang.c a few months ago. I've skimmed
through quite a few, but found no mention of a C conforming
implementation that had sizeof(char) != 8. Wherever the machine's
smallest addressable unit was greater than 8 bits, the compiler would
emulate smaller types by inserting shifts and bitwise operations (a
supposed machine would be the Cray 90
http://groups.google.com/group/comp.lang.c/msg/6cd4a4c8b2b0806c )

Based on the above I think that
   typedef   signed   char int8_t;
   typedef unsigned char uint8_t;
Inserted directly in apr.h.in would do just fine, without needing any
support from the autotools.


PS: At least maybe this way we'd break an implementation somewhere and
someone would file a bug report on this :P
I'm really curious to find a machine with chars!=8bits

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