On Tue, 11 Dec 2018, Paul E. McKenney wrote: > > Rewriting the litmus test in these terms gives: > > > > P0 P1 P2 P3 P4 P5 > > Wa=2 Wb=2 Wc=2 [mb23] [mb14] [mb05] > > mb0s mb1s mb2s Wd=2 We=2 Wf=2 > > mb0e mb1e mb2e Re=0 Rf=0 Ra=0 > > Rb=0 Rc=0 Rd=0 > > > > Here the brackets in "[mb23]", "[mb14]", and "[mb05]" mean that the > > positions of these barriers in their respective threads' program > > orderings is undetermined; they need not come at the top as shown. > > > > (Also, in case David is unfamiliar with it, the "Wa=2" notation is > > shorthand for "Write 2 to a" and "Rb=0" is short for "Read 0 from b".) > > > > Finally, here are a few facts which may be well known and obvious, but > > I'll state them anyway: > > > > A CPU cannot reorder instructions across a memory barrier. > > If x is po-after a barrier then x executes after the barrier > > is finished. > > > > If a store is po-before a barrier then the store propagates > > to every CPU before the barrier finishes. > > > > If a store propagates to some CPU before a load on that CPU > > reads from the same location, then the load will obtain the > > value from that store or a co-later store. This implies that > > if a load obtains a value co-earlier than some store then the > > load must have executed before the store propagated to the > > load's CPU. > > > > The proof consists of three main stages, each requiring three steps. > > Using the facts that b - f are all read as 0, I'll show that P1 > > executes Rc before P3 executes Re, then that P0 executes Rb before P4 > > executes Rf, and lastly that P5's Ra must obtain 2, not 0. This will > > demonstrate that the litmus test is not allowed. > > > > 1. Suppose that mb23 ends up coming po-later than Wd in P3. > > Then we would have: > > > > Wd propagates to P2 < mb23 < mb2e < Rd, > > > > and so Rd would obtain 2, not 0. Hence mb23 must come > > po-before Wd (as shown in the listing): mb23 < Wd. > > > > 2. Since mb23 therefore occurs po-before Re and instructions > > cannot be reordered across barriers, mb23 < Re. > > > > 3. Since Rc obtains 0, we must have: > > > > Rc < Wc propagates to P1 < mb2s < mb23 < Re. > > > > Thus Rc < Re. > > > > 4. Suppose that mb14 ends up coming po-later than We in P4. > > Then we would have: > > > > We propagates to P3 < mb14 < mb1e < Rc < Re, > > > > and so Re would obtain 2, not 0. Hence mb14 must come > > po-before We (as shown in the listing): mb14 < We. > > > > 5. Since mb14 therefore occurs po-before Rf and instructions > > cannot be reordered across barriers, mb14 < Rf. > > > > 6. Since Rb obtains 0, we must have: > > > > Rb < Wb propagates to P0 < mb1s < mb14 < Rf. > > > > Thus Rb < Rf. > > > > 7. Suppose that mb05 ends up coming po-later than Wf in P5. > > Then we would have: > > > > Wf propagates to P4 < mb05 < mb0e < Rb < Rf, > > > > and so Rf would obtain 2, not 0. Hence mb05 must come > > po-before Wf (as shown in the listing): mb05 < Wf. > > > > 8. Since mb05 therefore occurs po-before Ra and instructions > > cannot be reordered across barriers, mb05 < Ra. > > > > 9. Now we have: > > > > Wa propagates to P5 < mb0s < mb05 < Ra, > > > > and so Ra must obtain 2, not 0. QED. > > Like this, then, with maximal reordering of P3-P5's reads? > > P0 P1 P2 P3 P4 P5 > Wa=2 > mb0s > [mb05] > mb0e Ra=0 > Rb=0 Wb=2 > mb1s > [mb14] > mb1e Rf=0 > Rc=0 Wc=2 Wf=2 > mb2s > [mb23] > mb2e Re=0 > Rd=0 We=2 > Wd=2
Yes, that's right. This shows how P5's Ra must obtain 2 instead of 0. > But don't the sys_membarrier() calls affect everyone, especially given > the shared-variable communication? They do, but the other effects are irrelevant for this proof. > If so, why wouldn't this more strict > variant hold? > > P0 P1 P2 P3 P4 P5 > Wa=2 > mb0s > [mb05] [mb05] [mb05] You have misunderstood the naming scheme. mb05 is the barrier injected by P0's sys_membarrier call into P5. So the three barriers above should be named "mb03", "mb04", and "mb05". And you left out mb01 and mb02. > mb0e > Rb=0 Wb=2 > mb1s > [mb14] [mb14] [mb14] > mb1e > Rc=0 Wc=2 > mb2s > [mb23] [mb23] [mb23] > mb2e Re=0 Rf=0 Ra=0 > Rd=0 We=2 Wf=2 > Wd=2 Yes, this does hold. But since it doesn't affect the end result, there's no point in mentioning all those other barriers. > In which case, wouldn't this cycle be forbidden even if it had only one > sys_membarrier() call? No, it wouldn't. I don't understand why you might think it would. This is just like RCU, if you imagine a tiny critical section between each adjacent pair of instructions. You wouldn't expect RCU to enforce ordering among six CPUs with only one synchronize_rcu call. > Ah, but the IPIs are not necessarily synchronized across the CPUs, > so that the following could happen: > > P0 P1 P2 P3 P4 P5 > Wa=2 > mb0s > [mb05] [mb05] [mb05] > mb0e Ra=0 > Rb=0 Wb=2 > mb1s > [mb14] [mb14] > Rf=0 > Wf=2 > [mb14] > mb1e > Rc=0 Wc=2 > mb2s > [mb23] > Re=0 > We=2 > [mb23] [mb23] > mb2e > Rd=0 > Wd=2 Yes it could. But even in this execution you would end up with Ra=2 instead of Ra=0. > I guess in light of this post in 2001, I really don't have an excuse, > do I? ;-) > > https://lists.gt.net/linux/kernel/223555 > > Or am I still missing something here? You tell me... Alan
