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 timur@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)