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 <okaya@xxxxxxxxxxxxxx>
---
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
--
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