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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