On Sunday 01 October 2006 02:45, Nikola Ciprich wrote:
> Jeff,
> I'm terribly sorry, please ignore previous two emails. Problem was not
> in UML itself, but in my scripts, as it happened that I was running it
> with setarch i386, which caused guest uml report i386 architecture, and
> it made quite a mess when I was testing it...
> once again I'm sorry, I guess I really should rather have some sleep :(
> n.
> PS: but anyways, I wonder, is this behavior bug or feature? that running
> x86_64 uml using setarch i386 then causes guest UML report i686 arch?
> (even it's still x86_64)
It is a bug. Under setarch i386, uname returns i686; however, UML should
correct this. A 32bit UML says i686 on a 64bit host. Patch attached
(compile-tested).
Please report whether it works or not.
--
Inform me of my mistakes, so I can keep imitating Homer Simpson's "Doh!".
Paolo Giarrusso, aka Blaisorblade
http://www.user-mode-linux.org/~blaisorblade
Index: linux-2.6.git/arch/um/os-Linux/util.c
===================================================================
--- linux-2.6.git.orig/arch/um/os-Linux/util.c
+++ linux-2.6.git/arch/um/os-Linux/util.c
@@ -81,11 +81,18 @@ void setup_machinename(char *machine_out
struct utsname host;
uname(&host);
-#if defined(UML_CONFIG_UML_X86) && !defined(UML_CONFIG_64BIT)
+#ifdef UML_CONFIG_UML_X86
+# ifndef UML_CONFIG_64BIT
if (!strcmp(host.machine, "x86_64")) {
strcpy(machine_out, "i686");
return;
}
+# else
+ if (!strcmp(host.machine, "i686")) {
+ strcpy(machine_out, "x86_64");
+ return;
+ }
+# endif
#endif
strcpy(machine_out, host.machine);
}
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