The default implementation of mapping writeX() to __raw_writeX() is wrong. writeX() has stronger ordering semantics. Compiler is allowed to reorder __raw_writeX().
In the abscence of a write barrier or when using a strongly ordered architecture, writeX() should at least have a compiler barrier in it to prevent commpiler from clobbering the execution order. Signed-off-by: Sinan Kaya <[email protected]> --- include/asm-generic/io.h | 3 +++ 1 file changed, 3 insertions(+) diff --git a/include/asm-generic/io.h b/include/asm-generic/io.h index b4531e3..fbbf2bb 100644 --- a/include/asm-generic/io.h +++ b/include/asm-generic/io.h @@ -153,6 +153,7 @@ static inline void writeb(u8 value, volatile void __iomem *addr) static inline void writew(u16 value, volatile void __iomem *addr) { __raw_writew(cpu_to_le16(value), addr); + barrier(); } #endif @@ -161,6 +162,7 @@ static inline void writew(u16 value, volatile void __iomem *addr) static inline void writel(u32 value, volatile void __iomem *addr) { __raw_writel(__cpu_to_le32(value), addr); + barrier(); } #endif @@ -170,6 +172,7 @@ static inline void writel(u32 value, volatile void __iomem *addr) static inline void writeq(u64 value, volatile void __iomem *addr) { __raw_writeq(__cpu_to_le64(value), addr); + barrier(); } #endif #endif /* CONFIG_64BIT */ -- 2.7.4
