For some reason, the implementation of out_8() does not match the other
output accessors, nor does it match equivalent Linux accessor.  This can
cause a problem in a situation like this:

        __raw_readb(p1);
        out_8(p2, x);

In this case, there is no barrier between the read and the write.
If the out_8() is changed to out_be16(), then there is a barrier.

Signed-off-by: Timur Tabi <ti...@freescale.com>
---
 arch/powerpc/include/asm/io.h |    5 ++++-
 1 files changed, 4 insertions(+), 1 deletions(-)

diff --git a/arch/powerpc/include/asm/io.h b/arch/powerpc/include/asm/io.h
index 4ddad26..56ac9fe 100644
--- a/arch/powerpc/include/asm/io.h
+++ b/arch/powerpc/include/asm/io.h
@@ -175,7 +175,10 @@ extern inline int in_8(const volatile unsigned char 
__iomem *addr)
 
 extern inline void out_8(volatile unsigned char __iomem *addr, int val)
 {
-       __asm__ __volatile__("stb%U0%X0 %1,%0; eieio" : "=m" (*addr) : "r" 
(val));
+       __asm__ __volatile__("sync;\n"
+                            "stb%U0%X0 %1,%0;\n"
+                            : "=m" (*addr)
+                            : "r" (val));
 }
 
 extern inline int in_le16(const volatile unsigned short __iomem *addr)
-- 
1.7.3.4


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