linux-kernel - Re: [PATCH 17/17] RCU'd vfsmounts

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Message-ID: <CA+55aFxvZV0brpjRERYj6Bb-i2kF+nhfCVvmXV5Bdn2QhEt5sw@mail.gmail.com>
Date:	Thu, 3 Oct 2013 16:51:00 -0700
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	Josh Triplett <josh@...htriplett.org>
Cc:	Al Viro <viro@...iv.linux.org.uk>,
	linux-fsdevel <linux-fsdevel@...r.kernel.org>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	Paul McKenney <paulmck@...ux.vnet.ibm.com>
Subject: Re: [PATCH 17/17] RCU'd vfsmounts

On Thu, Oct 3, 2013 at 4:28 PM, Josh Triplett <josh@...htriplett.org> wrote:
> On Thu, Oct 03, 2013 at 01:52:45PM -0700, Linus Torvalds wrote:
>> On Thu, Oct 3, 2013 at 1:41 PM, Al Viro <viro@...iv.linux.org.uk> wrote:
>> >
>> > The problem is this:
>> > A = 1, B = 1
>> > CPU1:
>> > A = 0
>> > <full barrier>
>> > synchronize_rcu()
>> > read B
>> >
>> > CPU2:
>> > rcu_read_lock()
>> > B = 0
>> > read A
>> >
>> > Are we guaranteed that we won't get both of them seeing ones, in situation
>> > when that rcu_read_lock() comes too late to be noticed by synchronize_rcu()?
>>
>> Yeah, I think we should be guaranteed that, because the
>> synchronize_rcu() will guarantee that all other CPU's go through an
>> idle period. So the "read A" on CPU2 cannot possibly see a 1 _unless_
>> it happens so early that synchronize_rcu() definitely sees it (ie it's
>> a "preexisting reader" by definition), in which case synchronize_rcu()
>> will be waiting for a subsequent idle period, in which case the B=0 on
>> CPU2 is not only guaranteed to happen but also be visible out, so the
>> "read B" on CPU1 will see 0. And that's true even if CPU2 doesn't have
>> an explicit memory barrier, because the "RCU idle" state implies that
>> it has gone through a barrier.
>
> I think the reasoning in one direction is actually quite a bit less
> obvious than that.
>
> rcu_read_unlock() does *not* necessarily imply a memory barrier

Don't think of it in those terms.

The only thing that matters is semantics. The semantics of
synchronize_rcu() is that it needs to wait for all RCU users. It's
that simple. By definition, anything inside a "rcu_read_lock()" is a
RCU user, so if we have a read of memory (memory barrier or not), then
synchronize_rcu() needs to wait for it. Otherwise serialize_rcu() is
clearly totally broken.

Now, the fact that the normal rcu_read_lock() is just a compiler
barrier may make you think "oh, it cannot work", but the thing is, the
way things happens is that synchronize_rcu() ends up relying on the
_scheduler_ data structures, rather than anything else. It requires
seeing an idle scheduler state for each CPU after being called.

                  Linus
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