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Date:	Wed, 20 Jan 2010 10:23:50 +0900
From:	Tejun Heo <tj@...nel.org>
To:	Jeff Layton <jlayton@...hat.com>
CC:	torvalds@...ux-foundation.org, mingo@...e.hu, peterz@...radead.org,
	awalls@...ix.net, linux-kernel@...r.kernel.org, jeff@...zik.org,
	akpm@...ux-foundation.org, jens.axboe@...cle.com,
	rusty@...tcorp.com.au, cl@...ux-foundation.org,
	dhowells@...hat.com, arjan@...ux.intel.com, avi@...hat.com,
	johannes@...solutions.net, andi@...stfloor.org,
	Steve French <sfrench@...ba.org>
Subject: Re: [PATCH 38/40] cifs: use workqueue instead of slow-work

On 01/20/2010 09:56 AM, Jeff Layton wrote:
>> Hmmm... Why is locking GlobalSMBSeslock necessary?
>> cifs_oplock_break_get() can never fail and it seems that
>> is_valid_oplock_break() should be holding valid reference by the time
>> it enqueues the work, so wouldn't the following be sufficient?
>>
>> 	if (queue_work(system_single_wq, &netfile->oplock_break))
>> 		cifs_oplock_break_get(netfile);
>>
> 
> I guess I wasn't sure I could count on that. It seems unlikely that the
> work would run before you did the "get", but unlikely races are even
> harder to troubleshoot when they do get hit.
> 
> FWIW, it's not a terribly hot codepath, so taking and dropping the
> spinlock shouldn't be too bad for performance. If you're certain that
> we don't need to worry about it though, then maybe we can just skip
> doing that.

Oooh, I don't know the code very well and can't guarantee any of that.
Ah, okay, I was confused.  Work can run and finish before the
reference is incremented.  So, yeap, I'll add the spinlocking in the
work function.

>>> Also, this change seems to have changed the logic a bit. The
>>> oplock_break_cancelled flag is being set to false unconditionally, and
>>> the printk was dropped. Not a big deal on the last part, but we can't
>>> really do much with errors in this codepath so it might be helpful to
>>> have some indication that there are problems here.
>>
>> The thing is that slow_work_enqueue() can only fail if getting a
>> reference fails.  In cifs' case, it always succeeds so there's no
>> failure case to handle there.
>>
> 
> Ok, but here we're changing this to queue_work. Is that also guaranteed
> to succeed here? If so, then dropping the printk is fine. If not, then
> I think we should keep it in.

Yeap, queue_work() is guaranteed to succeed.  The only possible
outcomes are 1. queued or 2. is already queued.

> It's been a while since I overhauled this code, so I'll need to look
> again at the semantics for the oplock_break_cancelled flag. It may be
> ok to just set it unconditionally like this, but I'll need to check
> and be sure.

Thanks.

-- 
tejun
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