On Tue, 21 Nov 2006, Oleg Nesterov wrote:
> On 11/20, Alan Stern wrote:
> >
> > Both CPUs execute their "mb" instructions. The mb forces each
> > cache to wait until it receives an Acknowdgement for the
> > Invalidate it sent.
> >
> > Both caches send an Acknowledgement message to the other. The
> > mb instructions complete.
> >
> > "b = B" and "a = A" execute. The caches return A==0 and B==0
> > because they haven't yet invalidated their cache lines.
> >
> > The reason the code failed is because the mb instructions didn't force
> > the caches to wait until the Invalidate messages in their queues had been
> > fully carried out (i.e., the lines had actually been invalidated).
>
> However, from
> http://marc.theaimsgroup.com/?l=linux-kernel&m=113435711112941
>
> Paul E. McKenney wrote:
> >
> > 2. rmb() guarantees that any changes seen by the interconnect
> > preceding the rmb() will be seen by any reads following the
> > rmb().
> >
> > 3. mb() combines the guarantees made by rmb() and wmb().
>
> Confused :(
I'm not certain the odd behavior can occur on systems that use an
interconnect like Paul described. In the context I was describing, rmb()
guarantees only that any changes seen _and acknowledged_ by the cache
preceding the rmb() will be seen by any reads following the rmb(). It's a
weaker guarantee, but it still suffices to show that
A = 1 b = B
wmb rmb
B = 2 a = A
will work as expected.
Alan
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