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Date:   Thu, 15 Dec 2022 12:13:56 -0800
From:   "Paul E. McKenney" <paulmck@...nel.org>
To:     Joel Fernandes <joel@...lfernandes.org>
Cc:     Frederic Weisbecker <frederic@...nel.org>, boqun.feng@...il.com,
        neeraj.iitr10@...il.com, urezki@...il.com, rcu@...r.kernel.org,
        linux-kernel@...r.kernel.org
Subject: Re: [PATCH RFC] srcu: Yet more detail for
 srcu_readers_active_idx_check() comments

On Thu, Dec 15, 2022 at 05:58:14PM +0000, Joel Fernandes wrote:
> On Thu, Dec 15, 2022 at 5:48 PM Joel Fernandes <joel@...lfernandes.org> wrote:
> >
> > On Thu, Dec 15, 2022 at 5:08 PM Paul E. McKenney <paulmck@...nel.org> wrote:
> >
> > > > Scenario for the reader to increment the old idx once:
> > > >
> > > > _ Assume ssp->srcu_idx is initially 0.
> > > > _ The READER reads idx that is 0
> > > > _ The updater runs and flips the idx that is now 1
> > > > _ The reader resumes with 0 as an index but on the next srcu_read_lock()
> > > >   it will see the new idx which is 1
> > > >
> > > > What could be the scenario for it to increment the old idx twice?
> > >
> > > Unless I am missing something, the reader must reference the
> > > srcu_unlock_count[old_idx] and then do smp_mb() before it will be
> > > absolutely guaranteed of seeing the new value of ->srcu_idx.
> >
> > I think both of you are right depending on how the flip raced with the
> > first reader's unlock in that specific task.
> >
> > If the first read section's srcu_read_unlock() and its corresponding
> > smp_mb()  happened before the flip, then the increment of old idx
> > would happen only once. The next srcu_read_lock() will read the new
> > index. If the srcu_read_unlock() and it's corresponding smp_mb()
> > happened after the flip, the old_idx will be sampled again and can be
> > incremented twice. So it depends on how the flip races with
> > srcu_read_unlock().
> 
> I am sorry this is inverted, but my statement's gist stands I believe:
> 
> 1. Flip+smp_mb() happened before unlock's smp_mb() -- reader will not
> increment old_idx the second time.

By "increment old_idx" you mean "increment ->srcu_lock_count[old_idx]",
correct?

Again, the important ordering isn't the smp_mb(), but the accesses,
in this case, the accesses to ->srcu_unlock_count[idx].

> 2. unlock()'s smp_mb() happened before Flip+smp_mb() , now the reader
> has no new smp_mb() that happens AFTER the flip happened. So it can
> totally sample the old idx again -- that particular reader will
> increment twice, but the next time, it will see the flipped one.

I will let you transliterate both.  ;-)

> Did I get that right? Thanks.

So why am I unhappy with orderings of smp_mb()?

To see this, let's take the usual store-buffering litmus test:

	CPU 0			CPU 1
	WRITE_ONCE(x, 1);	WRITE_ONCE(y, 1);
	smp_mb();		smp_mb();
	r0 = READ_ONCE(y);	r1 = READ_ONCE(x);

Suppose CPU 0's smp_mb() happens before that of CPU 1:

	CPU 0			CPU 1
	WRITE_ONCE(x, 1);	WRITE_ONCE(y, 1);
	smp_mb();
				smp_mb();
	r0 = READ_ONCE(y);	r1 = READ_ONCE(x);

We get r0 == r1 == 1.

Compare this to CPU 1's smp_mb() happening before that of CPU 0:

	CPU 0			CPU 1
	WRITE_ONCE(x, 1);	WRITE_ONCE(y, 1);
				smp_mb();
	smp_mb();
	r0 = READ_ONCE(y);	r1 = READ_ONCE(x);

We still get r0 == r1 == 1.  Reversing the order of the two smp_mb()
calls changed nothing.

But, if we order CPU 1's write to follow CPU 0's read, then we have
this:

	CPU 0			CPU 1
	WRITE_ONCE(x, 1);
	smp_mb();
	r0 = READ_ONCE(y);
				WRITE_ONCE(y, 1);
				smp_mb();
				r1 = READ_ONCE(x);

Here, given that r0 had the final value of zero, we know that
r1 must have a final value of 1.

And suppose we reverse this:

	CPU 0			CPU 1
				WRITE_ONCE(y, 1);
				smp_mb();
				r1 = READ_ONCE(x);
	WRITE_ONCE(x, 1);
	smp_mb();
	r0 = READ_ONCE(y);

Now there is a software-visible difference in behavior.  The value of
r0 is now 1 instead of zero and the value of r1 is now 0 instead of 1.

Does this make sense?

							Thanx, Paul

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