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Date: Fri, 18 Sep 2015 15:28:44 +0200
From: Oleg Nesterov <oleg@redhat.com>
To: Peter Zijlstra <peterz@infradead.org>
Cc: Dmitry Vyukov <dvyukov@google.com>, ebiederm@xmission.com,
        Al Viro <viro@zeniv.linux.org.uk>,
        Andrew Morton <akpm@linux-foundation.org>,
        Ingo Molnar <mingo@kernel.org>,
        Paul McKenney <paulmck@linux.vnet.ibm.com>, mhocko@suse.cz,
        LKML <linux-kernel@vger.kernel.org>, ktsan@googlegroups.com,
        Kostya Serebryany <kcc@google.com>,
        Andrey Konovalov <andreyknvl@google.com>,
        Alexander Potapenko <glider@google.com>,
        Hans Boehm <hboehm@google.com>
Subject: Re: [PATCH] kernel: fix data race in put_pid
Message-ID: <20150918132844.GA10241@redhat.com>
References: <1442496268-47803-1-git-send-email-dvyukov@google.com> <20150917160837.GA26050@redhat.com> <CACT4Y+YKmhz-+sugKt86CF2715YtoVz4ee+sRfDp2Wu==A4zPQ@mail.gmail.com> <20150917174456.GA30178@redhat.com> <CACT4Y+bnqMGjMV6FHHJWsx0YaxWmRqAM+YY7b5wz-7JmFBnwjA@mail.gmail.com> <20150917180919.GA32116@redhat.com> <20150918085156.GS3816@twins.programming.kicks-ass.net>
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On 09/18, Peter Zijlstra wrote:
>
> On Thu, Sep 17, 2015 at 08:09:19PM +0200, Oleg Nesterov wrote:
>
> > I need to recheck, but afaics this is not possible. This optimization
> > is fine, but probably needs a comment.
>
> For sure, this code doesn't make any sense to me.

So yes, after a sleep I am starting to agree that in theory this fast-path
check is wrong. I'll write another email..

> As an alternative patch, could we not do:
>
>   void put_pid(struct pid *pid)
>   {
> 	struct pid_namespace *ns;
>
> 	if (!pid)
> 		return;
>
> 	ns = pid->numbers[pid->level].ns;
> 	if ((atomic_read(&pid->count) == 1) ||
> 	     atomic_dec_and_test(&pid->count)) {
>
> +		smp_read_barrier_depends(); /* ctrl-dep */

Not sure... Firstly it is not clear what this barrier pairs with. And I
have to admit that I can not understand if _CTRL() logic applies here.
The same for atomic_read_ctrl().

OK, please forget about put_pid() for the moment. Suppose we have

	X = 1;
	synchronize_sched();
	Y = 1;

Or
	X = 1;
	call_rcu_sched( func => { Y = 1; } );



Now. In theory this this code is wrong:

	if (Y) {
		BUG_ON(X == 0);
	}

But this is correct:

	if (Y) {
		rcu_read_lock_sched();
		rcu_read_unlock_sched();
		BUG_ON(X == 0);
	}

So perhaps something like this

	/*
	 * Comment to explain it is eq to read_lock + read_unlock,
	 * in a sense that this guarantees a full barrier wrt to
	 * the previous synchronize_sched().
	 */
	#define rcu_read_barrier_sched()	barrier()

make sense?


And again, I simply can't understand if this code

	if (READ_ONCE_CTRL(Y))
		BUG_ON(X == 0);

to me it does _not_ look correct in theory.
		
Oleg.