Re: x86_64 and x86 userland

[Richard B. Kreckel]

On Sat, 7 May 2005, Ralf Wildenhues wrote:
> Sorry for the long response delay.  And thank you everyone for providing
> useful information.

Sorry too, for not popping into this discussion earlier.

> * Noah Misch wrote on Tue, May 03, 2005 at 03:57:07AM CEST:
> > On Mon, May 02, 2005 at 10:31:57AM +0200, Ralf Wildenhues wrote:
> > > I have a question regarding systems with more than one ABI, specifically
> > > x86_64.  If you consider for example the Debian distribution which has a
> > > x86_64 kernel, but a completely x86 userland, config.guess still gives
> > > you x86_64-unknown-linux-gnu as output.  (I have been told this, but not
> > > tried it myself).
> > >
> > > Now, if you configure a package and forget to add
> > >   --build=i686-pc-linux-gnu
> > > or maybe use the setarch tool to set personality (I do not even know how
> > > portable/available this is -- it exists in RedHat Fedora), it may break,
> > > e.g. because of the __x86_64__ preprocessor define.
> >
> > The compiler generates x86 binaries but defines __x86_64__?  Weird.
>
> Now that you mention it, I don't think it does.  It might just be a bug
> in the configuration of the package, but I'm not sure yet.  I'll go and
> check this.

No, the compiler does not define __x86_64__.  The application's
config.h.in contains

#ifndef __x86_64__
#undef __x86_64__
#endif

and config.guess (through configure) generates

#ifndef __x86_64__
#define __x86_64__ 1
#endif

from that.  The compiler internally defines __i386__.  The presence of the
two defines causes compile-time problems because certain sections in the
source code try to do special things depending on which architecture is
defined.

One could go ahead and write

#if defined(__i386__) && !defined(__x86_64__)

wherever there was a

#ifdef __i386__

but that looks rather silly.

Regards
  -richy.
-- 
Richard B. Kreckel
<http://www.ginac.de/~kreckel/>