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

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Date:	Sun, 30 Sep 2007 16:02:09 -0700 (PDT)
From:	Davide Libenzi <davidel@...ilserver.org>
To:	Oleg Nesterov <oleg@...sign.ru>
cc:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	linux-rt-users@...r.kernel.org, Ingo Molnar <mingo@...e.hu>,
	Andrew Morton <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 Sun, 30 Sep 2007, Oleg Nesterov wrote:

> 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 ?

That can't be. CPU_2 sees X=1, that happened after (or same time at most - 
from a cache inv. POV) to *P=1, that must have happened after P=&B (in 
order for *P to assign B). So P=&B happened, from a pure time POV, before 
the rmb(), and the rmb() should guarantee that CPU_2 sees P=&B too.



- Davide


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