i386: cpu/transmeta.c: fix definition of USER686
authorH. Peter Anvin <hpa@zytor.com>
Wed, 9 May 2007 07:02:00 +0000 (00:02 -0700)
committerLinus Torvalds <torvalds@woody.linux-foundation.org>
Wed, 9 May 2007 19:49:33 +0000 (12:49 -0700)
The definition of USER686 is supposed to be a mask of feature bits,
not an OR of feature numbers!  It happened to work anyway on the only
processor affected, simply by pure coincidence.  Fix.

Signed-off-by: H. Peter Anvin <hpa@zytor.com>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
arch/i386/kernel/cpu/transmeta.c

index 6471a5a13202db49b8f0fbd79d355047ac008a89..200fb3f9ebfbda2ec232deec234ded94128f4cfb 100644 (file)
@@ -77,8 +77,10 @@ static void __cpuinit init_transmeta(struct cpuinfo_x86 *c)
        set_bit(X86_FEATURE_CONSTANT_TSC, c->x86_capability);
        
        /* If we can run i686 user-space code, call us an i686 */
-#define USER686 (X86_FEATURE_TSC|X86_FEATURE_CX8|X86_FEATURE_CMOV)
-        if ( c->x86 == 5 && (c->x86_capability[0] & USER686) == USER686 )
+#define USER686 ((1 << X86_FEATURE_TSC)|\
+                (1 << X86_FEATURE_CX8)|\
+                (1 << X86_FEATURE_CMOV))
+        if (c->x86 == 5 && (c->x86_capability[0] & USER686) == USER686)
                c->x86 = 6;
 
 #ifdef CONFIG_SYSCTL