Re: nfsd oops on Linus' current tree.

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On Thu, Jan 03, 2013 at 04:28:37PM +0000, Adamson, Dros wrote:
> Hey, sorry for the late response, I've been on vacation.
> 
> On Dec 21, 2012, at 6:45 PM, J. Bruce Fields <bfields@xxxxxxxxxxxx>
> wrote:
> 
> > On Fri, Dec 21, 2012 at 11:36:51PM +0000, Myklebust, Trond wrote:
> >> Please reread what I said. There was no obvious circular
> >> dependency, because nfsiod and rpciod are separate workqueues, both
> >> created with WQ_MEM_RECLAIM.
> > 
> > Oh, sorry, I read "rpciod" as "nfsiod"!
> > 
> >> Dros' experience shows, however that a call to rpc_shutdown_client
> >> in an nfsiod work item will deadlock with rpciod if the RPC task's
> >> work item has been assigned to the same CPU as the one running the
> >> rpc_shutdown_client work item.
> >> 
> >> I can't tell right now if that is intentional (in which case the
> >> WARN_ON in the rpc code is correct), or if it is a bug in the
> >> workqueue code. For now, we're assuming the former.
> > 
> > Well, Documentation/workqueue.txt says:
> > 
> > 	"Each wq with WQ_MEM_RECLAIM set has an execution context
> > 	reserved for it.  If there is dependency among multiple work
> > 	items used during memory reclaim, they should be queued to
> > 	separate wq each with WQ_MEM_RECLAIM."
> 
> The deadlock we were seeing was:
> 
> - task A gets queued on rpciod workqueue and assigned kworker-0:0 -
> task B gets queued on rpciod workqueue and assigned the same kworker
> (kworker-0:0) - task A gets run, calls rpc_shutdown_client(), which
> will loop forever waiting for task B to run rpc_async_release() - task
> B will never run rpc_async_release() - it can't run until kworker-0:0
> is free, which won't happen until task A (rpc_shutdown_client) is done
> 
> The same deadlock happened when we tried queuing the tasks on a
> different workqueues -- queue_work() assigns the task to a kworker
> thread and it's luck of the draw if it's the same kworker as task A.
> We tried the different workqueue options, but nothing changed this
> behavior.
> 
> Once a work struct is queued, there is no way to back out of the
> deadlock.  From kernel/workqueue.c:insert_wq_barrier comment:
> 
>  * Currently, a queued barrier can't be canceled.  This is because *
>  try_to_grab_pending() can't determine whether the work to be *
>  grabbed is at the head of the queue and thus can't clear LINKED *
>  flag of the previous work while there must be a valid next work *
>  after a work with LINKED flag set.
> 
> So once a work struct is queued and there is an ordering dependency
> (i.e. task A is before task B), there is no way to back task B out -
> so we can't just call cancel_work() or something on task B in
> rpc_shutdown_client.
> 
> The root of our issue is that rpc_shutdown_client is never safe to
> call from a workqueue context - it loops until there are no more
> tasks, marking tasks as killed and waiting for them to be cleaned up
> in each task's own workqueue context. Any tasks that have already been
> assigned to the same kworker thread will never have a chance to run
> this cleanup stage.
> 
> When fixing this deadlock, Trond and I discussed changing how
> rpc_shutdown_client works (making it workqueue safe), but Trond felt
> that it'd be better to just not call it from a workqueue context and
> print a warning if it is.
> 
> IIRC we tried using different workqueues with WQ_MEM_RECLAIM (with no
> success), but I'd argue that even if that did work it would still be
> very easy to call rpc_shutdown_client from the wrong context and MUCH
> harder to detect it.  It's also unclear to me if setting rpciod
> workqueue to WQ_MEM_RECLAIM would limit it to one kworker, etc...

Both rpciod and nfsiod already set WQ_MEM_RECLAIM.

But, right, looking at kernel/workqueue.c, it seems that the dedicated
"rescuer" threads are invoked only in the case when work is stalled
because a new worker thread isn't allocated quickly enough.

So, what to do that's simplest enough that it would work for
post-rc2/stable?  I was happy having just a simple dedicated
thread--these are only started when nfsd is, so there's no real thread
proliferation problem.

I'll go quietly weep for a little while and then think about it some
more....

--b.
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