Re: [RFC v2 3/3] tools/memory-model: litmus: Add two tests for unlock(A)+lock(B) ordering
From: "Paul E. McKenney" <paulmck@kernel.org>
Date: 2021-10-29 15:01:58
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On Fri, Oct 29, 2021 at 10:34:42AM -0400, Alan Stern wrote:
On Fri, Oct 29, 2021 at 07:51:39AM +0800, Boqun Feng wrote:quoted
On Thu, Oct 28, 2021 at 12:11:29PM -0700, Paul E. McKenney wrote:quoted
On Tue, Oct 26, 2021 at 09:01:00AM +0200, Peter Zijlstra wrote:quoted
On Mon, Oct 25, 2021 at 10:54:16PM +0800, Boqun Feng wrote:quoted
diff --git a/tools/memory-model/litmus-tests/LB+unlocklockonceonce+poacquireonce.litmus b/tools/memory-model/litmus-tests/LB+unlocklockonceonce+poacquireonce.litmus new file mode 100644 index 000000000000..955b9c7cdc7f --- /dev/null +++ b/tools/memory-model/litmus-tests/LB+unlocklockonceonce+poacquireonce.litmus@@ -0,0 +1,33 @@ +C LB+unlocklockonceonce+poacquireonce + +(* + * Result: Never + * + * If two locked critical sections execute on the same CPU, all accesses + * in the first must execute before any accesses in the second, even if + * the critical sections are protected by different locks.One small nit; the above "all accesses" reads as if: spin_lock(s); WRITE_ONCE(*x, 1); spin_unlock(s); spin_lock(t); r1 = READ_ONCE(*y); spin_unlock(t); would also work, except of course that's the one reorder allowed by TSO.I applied this series with Peter's Acked-by, and with the above commentThanks!quoted
reading as follows: +(* + * Result: Never + * + * If two locked critical sections execute on the same CPU, all accesses + * in the first must execute before any accesses in the second, even if the + * critical sections are protected by different locks. The one exception + * to this rule is that (consistent with TSO) a prior write can be reordered + * with a later read from the viewpoint of a process not holding both locks.Just want to be accurate, in our memory model "execute" means a CPU commit an memory access instruction to the Memory Subsystem, so if we have a store W and a load R, where W executes before R, it doesn't mean the memory effect of W is observed before the memory effect of R by other CPUs, consider the following case CPU0 Memory Subsystem CPU1 ==== ==== WRITE_ONCE(*x,1); // W ---------->| spin_unlock(s); | spin_lock(t); | r1 = READ_ONCE(*y); // R -------->| // R reads 0 | |<----------------WRITR_ONCE(*y, 1); // W' W' propagates to CPU0 | <-------------------------| | smp_mb(); |<----------------r1 = READ_ONCE(*x); // R' reads 0 | | W progrates to CPU 1 |-----------------> The "->" from CPU0 to the Memory Subsystem shows that W executes before R, however the memory effect of a store can be observed only after the Memory Subsystem propagates it to another CPU, as a result CPU1 doesn't observe W before R is executed. So the original version of the comments is correct in our memory model terminology, at least that's how I understand it, Alan can correct me if I'm wrong.Indeed, that is correct. It is an unfortunate inconsistency with the terminology in Documentation/memory-barriers.txt. I suspect most people think of a write as executing when it is observed by another CPU, even though that really isn't a coherent concept. (For example, it could easily lead somebody to think that a write observed at different times by different CPUs has executed more than once!)
Agreed, the terminology is odd. But the fact that different CPUs can see writes in different orders is probably always going to be a bit counter-intuitive, so it is good to avoid giving that intuition any support.
quoted
Maybe it's better to replace the sentence starting with "The one exception..." into: One thing to notice is that even though a write executes by a read, the memory effects can still be reordered from the viewpoint of a process not holding both locks, similar to TSO ordering. Thoughts?Or more briefly: Note: Even when a write executes before a read, their memory effects can be reordered from the viewpoint of another CPU (the kind of reordering allowed by TSO).
Very good! I took this verbatim in a fixup patch to be combined with the original on my next rebase. Thanx, Paul
Alanquoted
Apologies for responsing late... ("Memory Subsystem" is an abstraction in our memory model, which doesn't mean hardware implements things in the same way.). Regards, Boqunquoted
+ *) Thank you all! Thanx, Paul