linux-kernel - Re: [PATCH RFC 3/9] RCU: Preemptible RCU

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Message-ID: <20070930163102.GA374@tv-sign.ru>
Date:	Sun, 30 Sep 2007 20:31:02 +0400
From:	Oleg Nesterov <oleg@...sign.ru>
To:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Cc:	linux-kernel@...r.kernel.org, linux-rt-users@...r.kernel.org,
	mingo@...e.hu, akpm@...ux-foundation.org, dipankar@...ibm.com,
	josht@...ux.vnet.ibm.com, tytso@...ibm.com, dvhltc@...ibm.com,
	tglx@...utronix.de, a.p.zijlstra@...llo.nl, bunk@...nel.org,
	ego@...ibm.com, srostedt@...hat.com
Subject: Re: [PATCH RFC 3/9] RCU: Preemptible RCU

On 09/28, Paul E. McKenney wrote:
>
> On Fri, Sep 28, 2007 at 06:47:14PM +0400, Oleg Nesterov wrote:
> > Ah, I was confused by the comment,
> > 
> > 	smp_mb();  /* Don't call for memory barriers before we see zero. */
> > 	                                             ^^^^^^^^^^^^^^^^^^
> > So, in fact, we need this barrier to make sure that _other_ CPUs see these
> > changes in order, thanks. Of course, _we_ already saw zero.
> 
> Fair point!
> 
> Perhaps: "Ensure that all CPUs see their rcu_mb_flag -after- the
> rcu_flipctrs sum to zero" or some such?
> 
> > But in that particular case this doesn't matter, rcu_try_flip_waitzero()
> > is the only function which reads the "non-local" per_cpu(rcu_flipctr), so
> > it doesn't really need the barrier? (besides, it is always called under
> > fliplock).
> 
> The final rcu_read_unlock() that zeroed the sum was -not- under fliplock,
> so we cannot necessarily rely on locking to trivialize all of this.

Yes, but still I think this mb() is not necessary. Becasue we don't need
the "if we saw rcu_mb_flag we must see sum(lastidx)==0" property. When another
CPU calls rcu_try_flip_waitzero(), it will use another lastidx. OK, minor issue,
please forget.

> > OK, the last (I promise :) off-topic question. When CPU 0 and 1 share a
> > store buffer, the situation is simple, we can replace "CPU 0 stores" with
> > "CPU 1 stores". But what if CPU 0 is equally "far" from CPUs 1 and 2?
> > 
> > Suppose that CPU 1 does
> > 
> > 	wmb();
> > 	B = 0
> > 
> > Can we assume that CPU 2 doing
> > 
> > 	if (B == 0) {
> > 		rmb();
> > 
> > must see all invalidations from CPU 0 which were seen by CPU 1 before wmb() ?
> 
> Yes.  CPU 2 saw something following CPU 1's wmb(), so any of CPU 2's
> reads following its rmb() must therefore see all of CPU 1's stores
> preceding the wmb().

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 ?

> The other approach would be to simply have a separate thread for this
> purpose.  Batching would amortize the overhead (a single trip around the
> CPUs could satisfy an arbitrarily large number of synchronize_sched()
> requests).

Yes, this way we don't need to uglify migration_thread(). OTOH, we need
another kthread ;)

Oleg.

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