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

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Date:	Wed, 4 Jun 2014 10:16:12 -0400
From:	Steven Rostedt <rostedt@...dmis.org>
To:	"Brad Mouring" <bmouring@...com>
Cc:	linux-rt-users@...r.kernel.org,
	Thomas Gleixner <tglx@...utronix.de>,
	LKML <linux-kernel@...r.kernel.org>,
	Peter Zijlstra <peterz@...radead.org>,
	Ingo Molnar <mingo@...nel.org>,
	Clark Williams <williams@...hat.com>
Subject: Re: [PATCH 1/1] rtmutex: Handle when top lock owner changes

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