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Date:   Mon, 1 Jul 2019 08:55:25 +0900
From:   Byungchul Park <byungchul.park@lge.com>
To:     Steven Rostedt <rostedt@goodmis.org>
Cc:     "Paul E. McKenney" <paulmck@linux.ibm.com>,
        Scott Wood <swood@redhat.com>,
        Joel Fernandes <joel@joelfernandes.org>,
        Sebastian Andrzej Siewior <bigeasy@linutronix.de>,
        rcu <rcu@vger.kernel.org>, LKML <linux-kernel@vger.kernel.org>,
        Thomas Gleixner <tglx@linutronix.de>,
        Ingo Molnar <mingo@redhat.com>,
        Peter Zijlstra <peterz@infradead.org>,
        Josh Triplett <josh@joshtriplett.org>,
        Mathieu Desnoyers <mathieu.desnoyers@efficios.com>,
        Lai Jiangshan <jiangshanlai@gmail.com>
Subject: Re: [RFC] Deadlock via recursive wakeup via RCU with threadirqs
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On Fri, Jun 28, 2019 at 11:44:11AM -0400, Steven Rostedt wrote:
> On Fri, 28 Jun 2019 19:40:45 +0900
> Byungchul Park <byungchul.park@lge.com> wrote:
> 
> > Wait.. I got a little bit confused on recordering.
> > 
> > This 'STORE rcu_read_lock_nesting = 0' can happen before
> > 'STORE rcu_read_unlock_special.b.exp_hint = false' regardless of the
> > order a compiler generated to by the barrier(), because anyway they
> > are independent so it's within an arch's right.
> > 
> > Then.. is this scenario possible? Or all archs properly deal with
> > interrupts across this kind of reordering?
> 
> As Paul stated, interrupts are synchronization points. Archs can only
> play games with ordering when dealing with entities outside the CPU
> (devices and other CPUs). But if you have assembly that has two stores,
> and an interrupt comes in, the arch must guarantee that the stores are
> done in that order as the interrupt sees it.
> 
> If this is not the case, there's a hell of a lot more broken in the
> kernel than just this, and "barrier()" would also be meaningless, as
> that is used mostly to deal with interrupts.

Clear. Dear Paul and Steve, Thank you.

> -- Steve