Messages in this thread | | | Subject | Re: [PATCH] x86: Default to ARCH=x86 to avoid overriding CONFIG_64BIT | From | David Woodhouse <> | Date | Fri, 12 Apr 2013 17:38:28 +0100 |
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On Fri, 2013-04-12 at 18:01 +0200, richard -rw- weinberger wrote: > David, > > On Thu, Dec 20, 2012 at 10:51 PM, David Woodhouse <dwmw2@infradead.org> wrote: > > -SUBARCH := $(shell uname -m | sed -e s/i.86/i386/ -e s/sun4u/sparc64/ \ > > +SUBARCH := $(shell uname -m | sed -e s/i.86/x86/ -e s/x86_64/x86/ \ > > + -e s/sun4u/sparc64/ \ > > -e s/arm.*/arm/ -e s/sa110/arm/ \ > > -e s/s390x/s390/ -e s/parisc64/parisc/ \ > > -e s/ppc.*/powerpc/ -e s/mips.*/mips/ \ > > Why is this change needed? Especially the -e s/x86_64/x86. > It causes UML to get build for x86 which is a nasty change > of the default behavior.
Which one? The "new" change of {i.86,x86_64} -> x86, or the pre-existing ones for s390, parisc, powerpc and mips that it's falling into line with?
> Before your change you got a UML kernel for the underlying arch just > by running "make linux ARCH=um". > Now it will always build a x86 UML kernel, also if you run it on > x86_64 and one is forced to override > SUBARCH by hand.
I know this was true a number of years ago when I first submitted this fix. But I thought the ARCH=x86 UML build had been fixed since then. And I was fairly sure I'd *tested* that belief when recently resubmitting...
Isn't it merely a case of setting CONFIG_64BIT to the desired value?
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