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Message-ID: <20090711210902.GB6641@nowhere>
Date:	Sat, 11 Jul 2009 23:09:04 +0200
From:	Frederic Weisbecker <fweisbec@...il.com>
To:	Ming Lei <tom.leiming@...il.com>
Cc:	a.p.zijlstra@...llo.nl, linux-kernel@...r.kernel.org,
	akpm@...ux-foundation.org, mingo@...e.hu
Subject: Re: [RESEND PATCH 0/11] kernel:lockdep:replace DFS with BFS

On Sat, Jul 11, 2009 at 11:25:29AM +0800, Ming Lei wrote:
> 2009/7/11 Frederic Weisbecker <fweisbec@...il.com>:
> > Hi,
> >
> > On Sun, Jun 28, 2009 at 11:04:35PM +0800, tom.leiming@...il.com wrote:
> >> Hi,Peter
> >>
> >> Currently lockdep uses recursion DFS(depth-first search) algorithm to
> >> search target in checking lock circle(check_noncircular()),irq-safe
> >> -> irq-unsafe(check_irq_usage()) and irq inversion when adding a new
> >> lock dependency. This patches replace the current DFS with BFS, based on
> >> the following consideration:
> >>
> >>     1,no loss of efficiency, no matter DFS or BFS, the running time
> >>     are O(V+E) (V is vertex count, and E is edge count of one
> >>     graph);
> >>
> >>     2,BFS may be easily implemented by circular queue and consumes
> >>     much less kernel stack space than DFS for DFS is implemented by
> >>     recursion.
> >
> >
> >
> > Looks like a valuable argument. check_noncircular() can be called
> > in very random places in the kernel where the stack may be
> > already deep, and this recursive DFS doesn't help there.
> 
> Yes,  BFS uses the preallocated queue buffer as "stack" and removes
> the recursive implementation of DFS, so does decrease kernel stack
> consume
> largely.
> 
> From this point, BFS patch is valuable.


Right!

 
> >
> >
> >
> >>     3,The shortest path can be obtained by BFS if the target is
> >>     found, but can't be got by DFS. By the shortest path, we can
> >>     shorten the lock dependency chain and help to troubleshoot lock
> >>     problem easier than before.
> >
> >
> > But there I don't understand your argument.
> > The shortest path finding doesn't seem to me a need.
> > Example:
> >
> > Task 1 acquires: A B C
> > And Later:
> > Task 2 acquires: C B A
> >
> > DFS will probably report a circular lock dependency
> > with A and C.
> > BFS will probably report a circular lock dependency
> > with B and C.
> >
> > Which one is the most important? Both dependencies must be fixed
> > anyway. Once the developer will fix one of those, the remaining one
> > will be reported and so on...
> >
> > Or am I missing something else?
> 
> Yes, you are right.  By BFS, we can always find the shortest circle, but we
> find a random circle by DFS.   No one can say which circle is the most
> important from the point of deadlock.
> 
> But it is easier to start troubleshooting from the shortest circle
> than a random circle , then from the next shortest circle if other
> circle still exists .
> 
> Right?


I don't have a strong opinion on this. I just don't think the shortest path is
the most important if there are many many paths.
Whatever AB-BA is encountered, all of them must be fixed.
What might give a degree of importance for such bad circle is the window
in which it triggers.

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