linux-kernel - Re: [PATCH 25/48] rcu: Mark writes to rcu_sync ->gp

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Date: Fri, 10 May 2024 13:31:49 +0200
From: Oleg Nesterov <oleg@...hat.com>
To: "Paul E. McKenney" <paulmck@...nel.org>
Cc: "Uladzislau Rezki (Sony)" <urezki@...il.com>, RCU <rcu@...r.kernel.org>,
	Neeraj upadhyay <Neeraj.Upadhyay@....com>,
	Boqun Feng <boqun.feng@...il.com>, Hillf Danton <hdanton@...a.com>,
	Joel Fernandes <joel@...lfernandes.org>,
	LKML <linux-kernel@...r.kernel.org>,
	Oleksiy Avramchenko <oleksiy.avramchenko@...y.com>,
	Frederic Weisbecker <frederic@...nel.org>,
	Peter Zijlstra <peterz@...radead.org>
Subject: Re: [PATCH 25/48] rcu: Mark writes to rcu_sync ->gp_count field

On 05/09, Paul E. McKenney wrote:
>
> On Thu, May 09, 2024 at 05:13:12PM +0200, Oleg Nesterov wrote:
> >
> > We can move these WARN_ON()'s into the ->rss_lock protected section.
> >
> > Or perhaps we can use data_race(rsp->gp_count) ? To be honest I thought
> > that READ_ONCE() should be enough...
> >
> > Or we can simply kill these WARN_ON_ONCE()'s.
>
> Or we could move those WARN_ON_ONCE() under the lock.

Sure, see above.

But could you help me to understand this magic? I naively thought
that READ_ONCE() is always "safe"...

So, unless I am totally confused it turns out that, say,

	CPU 0			CPU 1
	-----			-----

	spin_lock(LOCK);
	++X;			READ_ONCE(X); // data race
	spin_unlock(LOCK);

is data-racy accoring to KCSAN, while

	CPU 0			CPU 1
	-----			-----

	spin_lock(LOCK);
	WRITE_ONCE(X, X+1);	READ_ONCE(X); // no data race
	spin_unlock(LOCK);

is not.

Why is that?

Trying to read Documentation/dev-tools/kcsan.rst... it says

	KCSAN is aware of *marked atomic operations* (``READ_ONCE``, WRITE_ONCE``,

	...

	if all accesses to a variable that is accessed concurrently are properly
	marked, KCSAN will never trigger a watchpoint

but how can KCSAN detect that all accesses to X are properly marked? I see nothing
KCSAN-related in the definition of WRITE_ONCE() or READ_ONCE().

And what does the "all accesses" above actually mean? The 2nd version does

	WRITE_ONCE(X, X+1);

but "X + 1" is the plain/unmarked access?

Thanks,

Oleg.