[arch-general] problem compiling for i586 with new makepkg

Jan de Groot jan at jgc.homeip.net
Wed Dec 19 13:07:25 EST 2007

On Sat, 2007-12-15 at 11:54 +0100, Attila wrote:
> On Samstag, 15. Dezember 2007 09:43 Karolina Lindqvist wrote:
> > That is allright, but why abort "makepkg" on non-authorized architectures?
> It seems that makepkg test only right or wrong because a "arch=(aai686
> aax86_64" brings the same break. I'm not a dev but from my view this is too
> strict. On the other side only a warning at the beginning is too simple
> because it can get overseen very easy but i think a warning at the end will
> be read in the most cases. Perhaps everybody can lives with this suggestion
> better than with the actual situation.
> See you, Attila

Why is checking strict wrong? If the official name of an architecture is
i686, and we make all packages have the i686 extension, why would aai686
be allowed then?

We could decide to use x86 as architecture, when an x86_64-only package
doesn't work on 32bit, does that mean that the check should pass because
x86 is in the architecture name?

More information about the arch-general mailing list