Re: Why is wmb() a no-op on x86_64?

[Andi Kleen]

On Wednesday 18 January 2006 17:23, Bryan O'Sullivan wrote:
> Hi, Andi -
> 
> I notice that wmb() is a no-op 

Actually it is a compiler optimizer barrier, not a no-op.

> on x86_64 kernels unless 
> CONFIG_UNORDERED_IO is set. 

Because x86 is architecturally defined as having ordered writes (unless you use 
write combining or non temporal stores which normal kernel code doesn't). So it's 
not needed.

> Is there any particular reason for this? 
> It's not similarly conditional on other platforms, and as a consequence,
> in our driver (which requires a write barrier in some situations for
> correctness), I have to add the following piece of ugliness:
> 
> #if defined(CONFIG_X86_64) && !defined(CONFIG_UNORDERED_IO)
> #define ipath_wmb() asm volatile("sfence" ::: "memory")
> #else
> #define ipath_wmb() wmb()
> #endif

Hmm, I suppose one could add a wc_wmb() or somesuch, but WC 
is currently deeply architecture specific so I'm not sure
how you can even use it portably.

Why do you need the barrier?

-Andi
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