Message-ID: <20151006195727.GI11639@twins.programming.kicks-ass.net>
Date:	Tue, 6 Oct 2015 21:57:27 +0200
From:	Peter Zijlstra <peterz@...radead.org>
To:	Oleg Nesterov <oleg@...hat.com>
Cc:	Boqun Feng <boqun.feng@...il.com>,
	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>,
	linux-kernel@...r.kernel.org, linux-doc@...r.kernel.org,
	Jonathan Corbet <corbet@....net>,
	Michal Hocko <mhocko@...nel.org>,
	David Howells <dhowells@...hat.com>,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	Will Deacon <will.deacon@....com>
Subject: Re: [PATCH] Documentation: Remove misleading examples of the
 barriers in wake_*()

> On Mon, Sep 21, 2015 at 07:46:11PM +0200, Oleg Nesterov wrote:

> > In short, I got lost ;) Now I don't even understand why we do not need
> > another rmb() between p->on_rq and p->on_cpu. Suppose a thread T does
> > 
> > 	set_current_state(...);
> > 	schedule();
> > 
> > it can be preempted in between, after that we have "on_rq && !on_cpu".
> > Then it gets CPU again and calls schedule() which clears on_rq.
> > 
> > What guarantees that if ttwu() sees on_rq == 0 cleared by schedule()
> > then it can _not_ still see the old value of on_cpu == 0?

I think you're right. Does the below adequately explain things?

I'll have another look tomorrow to see if I still agree with myself, but
for now I think I've convinced myself you're right.

---
Subject: sched: Fix race in try_to_wake_up() vs schedule()

Oleg noticed that its possible to falsely observe p->on_cpu == 0 such
that we'll prematurely continue with the wakeup and effectively run p on
two CPUs at the same time.

Even though the overlap is very limited; the task is in the middle of
being scheduled out; it could still result in corruption of the
scheduler data structures.


	CPU0				CPU1

	set_current_state(...)

	<preempt_schedule>
	  context_switch(X, Y)
	    prepare_lock_switch(Y)
	      Y->on_cpu = 1;
	    finish_lock_switch(X)
	      store_release(X->on_cpu, 0);

					try_to_wake_up(X)
					  LOCK(p->pi_lock);

					  t = X->on_cpu; // 0

	  context_switch(Y, X)
	    prepare_lock_switch(X)
	      X->on_cpu = 1;
	    finish_lock_switch(Y)
	      store_release(Y->on_cpu, 0);
	</preempt_schedule>

	schedule();
	  deactivate_task(X);
	  X->on_rq = 0;

					  if (X->on_rq) // false

					  if (t) while (X->on_cpu)
					    cpu_relax();

	  context_switch(X, ..)
	    finish_lock_switch(X)
	      store_release(X->on_cpu, 0);


Avoid the load of X->on_cpu being hoisted over the X->on_rq load.

Reported-by: Oleg Nesterov <oleg@...hat.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@...radead.org>
---
 kernel/sched/core.c |   19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

--- a/kernel/sched/core.c
+++ b/kernel/sched/core.c
@@ -2084,6 +2084,25 @@ try_to_wake_up(struct task_struct *p, un
 
 #ifdef CONFIG_SMP
 	/*
+	 * Ensure we load p->on_cpu _after_ p->on_rq, otherwise it would be
+	 * possible to, falsely, observe p->on_cpu == 0.
+	 *
+	 * One must be running (->on_cpu == 1) in order to remove oneself
+	 * from the runqueue.
+	 *
+	 *  [S] ->on_cpu = 1;	[L] ->on_rq
+	 *      UNLOCK rq->lock
+	 *			RMB
+	 *      LOCK   rq->lock
+	 *  [S] ->on_rq = 0;    [L] ->on_cpu
+	 *
+	 * Pairs with the full barrier implied in the UNLOCK+LOCK on rq->lock
+	 * from the consecutive calls to schedule(); the first switching to our
+	 * task, the second putting it to sleep.
+	 */
+	smp_rmb();
+
+	/*
 	 * If the owning (remote) cpu is still in the middle of schedule() with
 	 * this task as prev, wait until its done referencing the task.
 	 */
--
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