Re: [PATCH 1/1] rtmutex: Handle when top lock owner changes

From: Steven Rostedt <rostedt@goodmis.org>
To: "Brad Mouring" <bmouring@ni.com>
Cc: linux-rt-users@vger.kernel.org,
	Thomas Gleixner <tglx@linutronix.de>,
	LKML <linux-kernel@vger.kernel.org>,
	Peter Zijlstra <peterz@infradead.org>,
	Ingo Molnar <mingo@kernel.org>,
	Clark Williams <williams@redhat.com>
Subject: Re: [PATCH 1/1] rtmutex: Handle when top lock owner changes
Date: Wed, 4 Jun 2014 10:16:12 -0400	[thread overview]
Message-ID: <20140604101612.0d47b399@gandalf.local.home> (raw)
In-Reply-To: <20140604130525.GA1621@linuxgetsreal>

On Wed, 4 Jun 2014 08:05:25 -0500
"Brad Mouring" <bmouring@ni.com> wrote:

 >          A->L2
> 
> This is a slight variation on what I was seeing. To use the nomenclature
> that you proposed at the start, rewinding to the point
> 
>    A->L2->B->L3->C->L4->D
> 
> Let's assume things continue to unfold as you explain. Task is D,
> top_waiter is C. A is scheduled out and the chain shuffles.
> 
>        A->L2->B
> C->L4->D->'

But isn't that a lock ordering problem there?

If B can block on L3 owned by C, I see the following:

  B->L3->C->L4->D->L2->B

Deadlock!

In my scenario I was very careful to point out that the lock ordering
was: L1->L2->L3->L4

But you show that we can have both:

   L2-> ... ->L4

    and

   L4-> ... ->L2

Which is a reverse of lock ordering and a possible deadlock can occur.

-- Steve

> 
> So, we now have D blocking on L2 and L4 has waiters, C again. Also,
> since the codepath to have C block on L4 again is the same as the
> codepath from when it blocked on it in the first place, the location
> is the same since the stack (for what we care about) is the same.
> 