[PATCH] Fix CFLAGS overwrite by Makefile

[John Baldwin]

On Wednesday, May 25, 2011 11:34:29 am Arnaud Lacombe wrote:
> Hi,
> 
> On Wed, May 25, 2011 at 9:43 AM, John Baldwin <jhb at freebsd.org> wrote:
> >> The original trouble I met, is that building for an i586 target in a
> >> 32bits jail, on top of an amd64 system[0] (I do not have control over
> >> that setup) produces incorrect binaries. The current fix I've got is
> >> to define MACHINE_ARCH=i386 and CPUTYPE=i586. This enforces
> >> `-march=i586' to be passed to the compiler, for all except the
> >> bootloader (because it overwrites CFLAGS). With this, binaries
> >> produced works fine (ie. /bin/sh no longer SIGILL when bringing up the
> >> system). So I suspect that gcc default to i686 in this setup and
> >> corrupt all the binaries, thus the attached patch.
> >
> > Wait.  You must have something wrong in your jail if you can't do a 
buildworld
> > with CPUTYPE set to none and have it do the right thing.  You need to find
> > your root problem.  Forcing CPUCFLAGS for the boot code is a band-aid, 
it's
> > not the right solution to your problem.
> >
> Unless error of my part, I never mentioned it was using `buildworld',
> which it is not. The system uses bare calls to make(1) in the
> sys/boot/ directory. As the jail is 32bits, it was expected not to be
> an issue, but the jail compiler uses /lib/libstand.a to link the
> loader, and it obviously contains i686-only instructions, which
> trigger a reset of an i586-only CPU.
> 
> The more broad issue with the setup is that gcc within that
> environment, without being told -march=i586, produces i686
> instructions which are incompatible with the target CPU.

Huh?  GCC does not generate i686 instructions by default on FreeBSD/i386.  It 
generates i486 instructions but that is all.  Are you sure you aren't running 
the 64-bit gcc (which will generate i686 instructions by default)?

-- 
John Baldwin