Re: rcu_read_lock lost its compiler barrier

From: "Paul E. McKenney" <paulmck@linux.ibm.com>
To: Herbert Xu <herbert@gondor.apana.org.au>
Cc: Alan Stern <stern@rowland.harvard.edu>,
	Boqun Feng <boqun.feng@gmail.com>,
	Linus Torvalds <torvalds@linux-foundation.org>,
	Frederic Weisbecker <fweisbec@gmail.com>,
	Fengguang Wu <fengguang.wu@intel.com>, LKP <lkp@01.org>,
	LKML <linux-kernel@vger.kernel.org>,
	Netdev <netdev@vger.kernel.org>,
	"David S. Miller" <davem@davemloft.net>,
	Andrea Parri <andrea.parri@amarulasolutions.com>,
	Luc Maranget <luc.maranget@inria.fr>,
	Jade Alglave <j.alglave@ucl.ac.uk>
Subject: Re: rcu_read_lock lost its compiler barrier
Date: Thu, 6 Jun 2019 06:48:56 -0700	[thread overview]
Message-ID: <20190606134856.GL28207@linux.ibm.com> (raw)
In-Reply-To: <20190606133824.aibysezb5qdo3x27@gondor.apana.org.au>

On Thu, Jun 06, 2019 at 09:38:24PM +0800, Herbert Xu wrote:
> On Thu, Jun 06, 2019 at 03:58:17AM -0700, Paul E. McKenney wrote:
> >
> > I cannot immediately think of a way that the compiler could get this
> > wrong even in theory, but similar code sequences can be messed up.
> > The reason for this is that in theory, the compiler could use the
> > stored-to location as temporary storage, like this:
> > 
> > 	a = whatever;	// Compiler uses "a" as a temporary
> > 	do_something();
> > 	whatever = a;
> > 	a = 1;		// Intended store
> 
> Well if the compiler is going to do this then surely it would
> continue to do this even if you used WRITE_ONCE.  Remember a is
> not volatile, only the access of a through WRITE_ONCE is volatile.

I disagree.  Given a volatile store, the compiler cannot assume that the
stored-to location is normal memory at that point in time, and therefore
cannot assume that it is safe to invent a store to that location (as
shown above).  Thus far, the C++ standards committee seems on-board with
this, though time will tell.

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/p1382r1.pdf

							Thanx, Paul