x86: explicitly use edx in const delay function.

For x86_64, we can't just use %0, as it would
generate a mul against rdx, which is not really what we
want (note the ">> 32" in x86_64 version).

Using a u64 variable with a shift in i386 generates bad code,
so the solution is to explicitly use %%edx in inline assembly
for both.
Signed-off-by: Glauber Costa <gcosta@redhat.com>
Signed-off-by: H. Peter Anvin <hpa@zytor.com>
Signed-off-by: Ingo Molnar <mingo@elte.hu>

arch/x86/lib/delay_32.c

@@ -114,7 +114,7 @@ inline void __const_udelay(unsigned long xloops)
 	int d0;
 
 	xloops *= 4;
-	__asm__("mull %0"
+	__asm__("mull %%edx"
 		:"=d" (xloops), "=&a" (d0)
 		:"1" (xloops), "0"
 		(cpu_data(raw_smp_processor_id()).loops_per_jiffy * (HZ/4)));

arch/x86/lib/delay_64.c

@@ -103,9 +103,16 @@ EXPORT_SYMBOL(__delay);
 
 inline void __const_udelay(unsigned long xloops)
 {
-	__delay(((xloops * HZ *
-		cpu_data(raw_smp_processor_id()).loops_per_jiffy) >> 32) + 1);
+	int d0;
+	xloops *= 4;
+	__asm__("mull %%edx"
+		:"=d" (xloops), "=&a" (d0)
+		:"1" (xloops), "0"
+		(cpu_data(raw_smp_processor_id()).loops_per_jiffy * (HZ/4)));
+
+	__delay(++xloops);
 }
+
 EXPORT_SYMBOL(__const_udelay);
 
 void __udelay(unsigned long usecs)