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Re: with-headers should be 'build' != 'host'
- From: Ian Lance Taylor <ian at wasabisystems dot com>
- To: gcc at gcc dot gnu dot org
- Cc: dan clark <dlc at ncube dot com>, Felix Lee <felix dot 1 at canids dot net>, gdb-patches at sources dot redhat dot com
- Date: 29 Jan 2004 16:41:48 -0500
- Subject: Re: with-headers should be 'build' != 'host'
- References: <20040129034507.D5314180D@grayscale.canids><Pine.LNX.firstname.lastname@example.org><20040129181152.GA15394@nevyn.them.org>
Daniel Jacobowitz <email@example.com> writes:
> Secondly, you're still confused about --build, --host, and --target.
> If $build != $host == $target, you are building a native toolchain for
> another system. This has never been especially well supported, but
> --with-headers is definitely not the right way to do it. The copied
> headers are used by the built gcc/xgcc, not by the compiler used to
> build GCC.
Hmmm, I don't follow here. I seem to recall that I wrote
--with-header specifically to support the case of $build != $host ==
$target. So I'm surprised that you say that --with-headers is not the
right approach. You're suppose to use --with-headers for the header
files for $target. It's not wrong to use it when $host == $target.
But I'm probably misunderstanding something.