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SubjectRe: [PATCH RFC 3/9] RCU: Preemptible RCU
On Sun, 30 Sep 2007, Paul E. McKenney wrote:

> On Sun, Sep 30, 2007 at 04:02:09PM -0700, Davide Libenzi wrote:
> > On Sun, 30 Sep 2007, Oleg Nesterov wrote:
> >
> > > Ah, but I asked the different question. We must see CPU 1's stores by
> > > definition, but what about CPU 0's stores (which could be seen by CPU 1)?
> > >
> > > Let's take a "real life" example,
> > >
> > > A = B = X = 0;
> > > P = Q = &A;
> > >
> > > CPU_0 CPU_1 CPU_2
> > >
> > > P = &B; *P = 1; if (X) {
> > > wmb(); rmb();
> > > X = 1; BUG_ON(*P != 1 && *Q != 1);
> > > }
> > >
> > > So, is it possible that CPU_1 sees P == &B, but CPU_2 sees P == &A ?
> >
> > That can't be. CPU_2 sees X=1, that happened after (or same time at most -
> > from a cache inv. POV) to *P=1, that must have happened after P=&B (in
> > order for *P to assign B). So P=&B happened, from a pure time POV, before
> > the rmb(), and the rmb() should guarantee that CPU_2 sees P=&B too.
>
> Actually, CPU designers have to go quite a ways out of their way to
> prevent this BUG_ON from happening. One way that it would happen
> naturally would be if the cache line containing P were owned by CPU 2,
> and if CPUs 0 and 1 shared a store buffer that they both snooped. So,
> here is what could happen given careless or sadistic CPU designers:

Ohh, I misinterpreted that rmb(), sorry. Somehow I gave it for granted
that it was a cross-CPU sync point (ala read_barrier_depends). If that's a
local CPU load ordering only, things are different, clearly. But ...



>
> o CPU 0 stores &B to P, but misses the cache, so puts the
> result in the store buffer. This means that only CPUs 0 and 1
> can see it.
>
> o CPU 1 fetches P, and sees &B, so stores a 1 to B. Again,
> this value for P is visible only to CPUs 0 and 1.
>
> o CPU 1 executes a wmb(), which forces CPU 1's stores to happen
> in order. But it does nothing about CPU 0's stores, nor about CPU
> 1's loads, for that matter (and the only reason that POWER ends
> up working the way you would like is because wmb() turns into
> "sync" rather than the "eieio" instruction that would have been
> used for smp_wmb() -- which is maybe what Oleg was thinking of,
> but happened to abbreviate. If my analysis is buggy, Anton and
> Paulus will no doubt correct me...)

If a store buffer is shared between CPU_0 and CPU_1, it is very likely
that a sync done on CPU_1 is going to sync even CPU_0 stores that are
held in the buffer at the time of CPU_1's sync.



- Davide


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