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SubjectRe: Re: [RFC][PATCH 2/5] mips/atomic: Fix loongson_llsc_mb() wreckage
On Thu, Apr 25, 2019 at 09:33:48AM +0200, Peter Zijlstra wrote:

> > Let me explain the bug more specific:
> >
> > the bug ONLY matters in following situation:
> >
> > #. more than one cpu (assume cpu A and B) doing ll/sc on same shared
> > var V
> >
> > #. speculative memory access from A cause A erroneously succeed sc
> > operation, since the erroneously successful sc operation violate the
> > coherence protocol. (here coherence protocol means the rules that CPU
> > follow to implement ll/sc right)
> >
> > #. B succeed sc operation too, but this sc operation is right both
> > logically and follow the coherence protocol, and makes A's sc wrong
> > logically since only ONE sc operation can succeed.
> (I know your coherence protocol is probably more complicated than MESI,
> but bear with me)
> So A speculatively gets V's line in Exclusive mode, speculates the Lock
> flag is still there and completes the Store. This speculative store then
> leaks out and violates MESI because there _should_ only be one Exclusive
> owner of a line (B).
> Something like that?

So B gets E (from LL), does I on A, then SC succeeds and get M. A got
I, speculates E, speculates M and lets the M escape.

That gets us with 2 competing Ms (which is of course completely
insane), one wins one looses (at random I presume).

And this violates atomic guarantees because one operation got lost.

> > If it is not LL/SC but other memory access from B on V, A's ll/sc can
> > follow the atomic semantics even if A violate the coherence protocol
> > in the same situation.
> *shudder*...
> C atomic-set
> {
> atomic_set(v, 1);
> }
> P1(atomic_t *v)
> {
> atomic_add_unless(v, 1, 0);
> }
> P2(atomic_t *v)
> {
> atomic_set(v, 0);
> }
> exists
> (v=2)
> So that one will still work? (that is, v=2 is forbidden)

But then in this case, P1 has E from LL, P2 does M from the STORE, which
should cause I on P1. P1 speculates E, speculates M and lets M escape.

We again have two competing Ms, one wins at random, and v==2 if P1
wins. This again violates the atomic guarantees and would invalidate
your claim of it only mattering for competing LL/SC.

Or am I missing something? (quite likely, I always get confused with
these things)

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