Re: [PATCH 0/5 - V2] locks: avoid thundering-herd wake-ups

From: NeilBrown <neilb@suse.com>
To: "J. Bruce Fields" <bfields@fieldses.org>
Cc: Jeff Layton <jlayton@kernel.org>,
	Alexander Viro <viro@zeniv.linux.org.uk>,
	Martin Wilck <mwilck@suse.de>,
	linux-fsdevel@vger.kernel.org,
	Frank Filz <ffilzlnx@mindspring.com>,
	linux-kernel@vger.kernel.org
Subject: Re: [PATCH 0/5 - V2] locks: avoid thundering-herd wake-ups
Date: Fri, 10 Aug 2018 08:12:43 +1000	[thread overview]
Message-ID: <87lg9frxyc.fsf@notabene.neil.brown.name> (raw)
In-Reply-To: <20180809173245.GM23873@fieldses.org>

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On Thu, Aug 09 2018, J. Bruce Fields wrote:

> I think there's also a problem with multiple tasks sharing the same
> lock owner.
>
> So, all locks are exclusive locks for the same range.  We have four
> tasks.  Tasks 1 and 4 share the same owner, the others' owners are
> distinct.
>
> 	- Task 1 gets a lock.
> 	- Task 2 gets a conflicting lock.
> 	- Task 3 gets another conflicting lock.  So now we the tree is
> 		3->2->1.
> 	- Task 1's lock is released.
> 	- Before task 2 is scheduled, task 4 acquires a new lock.
> 	- Task 2 waits on task 4's lock, we now have
> 		3->2->4.
>
> Task 3 shouldn't be waiting--the lock it's requesting has the same owner
> as the lock task 4 holds--but we fail to wake up task 3.

So task 1 and task 4 are threads in the one process - OK.
Tasks 2 and 3 are threads in two other processes.

So 2 and 3 conflict with either 1 or 4 equally - why should task 3 be
woken?

I suspect you got the numbers bit mixed up, but in any case, the
"conflict()" function that is passed around takes ownership into account
when assessing if one lock conflicts with another.

NeilBrown

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