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Date:      Tue, 18 Sep 2007 09:40:54 -0700
From:      Maxim Sobolev <sobomax@FreeBSD.org>
To:        Dmitry Morozovsky <marck@rinet.ru>
Cc:        current@FreeBSD.org
Subject:   Re: i386 package building on an amd64 system
Message-ID:  <46EFFF96.1040101@FreeBSD.org>
In-Reply-To: <20070918011258.L96165@woozle.rinet.ru>
References:  <20070914124630.D14000@woozle.rinet.ru>	<46EEC481.5010503@FreeBSD.org> <20070918011258.L96165@woozle.rinet.ru>

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Dmitry Morozovsky wrote:
> On Mon, 17 Sep 2007, Maxim Sobolev wrote:
> 
> MS> > possibly stupid question: it there a way to fool jail (real, not
> MS> > "tinderbox" one) that it's working under i386 kernel? This would be
> MS> > extremely useful for local package building.
> MS> > 
> MS> > Quick googling does not reveal anything, or did I miss something obvious?
> MS> > 
> MS> > Thanks in advance.
> MS> 
> MS> Works here like a charm. The only thing is that you need to set UNAME_p=i386
> MS> UNAME_m=i386 and CPUTYPE=i686 variables in the chroot/jail.
> 
> I know aboot uname tricks ;-)
> 
> What about sysctls revealing amd64 system internals? Did you compare packages 
> built by this jailed system with native?

Any package that uses some sysctls from build machine to somehow alter 
build process should be considered b0rken.

-Maxim



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