From: Paolo 'Blaisorblade' Giarrusso <[EMAIL PROTECTED]>
On a 64bit Uml, if run under "setarch i386" (which a user did), uname()
currently returns the obtained i686 as machine - fix that. Btw, I'm quite
surprised that under setarch i386 a 64-bit binary can run.
Cc: Andi Kleen <[EMAIL PROTECTED]>
Signed-off-by: Paolo 'Blaisorblade' Giarrusso <[EMAIL PROTECTED]>
---
arch/um/os-Linux/util.c | 9 ++++++++-
1 files changed, 8 insertions(+), 1 deletions(-)
diff --git a/arch/um/os-Linux/util.c b/arch/um/os-Linux/util.c
index 3f5b151..56b8a50 100644
--- a/arch/um/os-Linux/util.c
+++ b/arch/um/os-Linux/util.c
@@ -80,11 +80,18 @@ 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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