linux-kernel - Re: [PATCH 1/1] percpu_rw_semaphore: reimplement to not block the readers unnecessarily

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Date:	Thu, 1 Nov 2012 19:33:24 +0100
From:	Oleg Nesterov <oleg@...hat.com>
To:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Cc:	Mikulas Patocka <mpatocka@...hat.com>,
	Peter Zijlstra <peterz@...radead.org>,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	Ingo Molnar <mingo@...e.hu>,
	Srikar Dronamraju <srikar@...ux.vnet.ibm.com>,
	Ananth N Mavinakayanahalli <ananth@...ibm.com>,
	Anton Arapov <anton@...hat.com>, linux-kernel@...r.kernel.org
Subject: Re: [PATCH 1/1] percpu_rw_semaphore: reimplement to not block the
	readers unnecessarily

Paul, thanks.

Sorry, I can't reply today, just one note...

On 11/01, Paul E. McKenney wrote:
>
> OK, so it looks to me that this code relies on synchronize_sched()
> forcing a memory barrier on each CPU executing in the kernel.

No, the patch tries to avoid this assumption, but probably I missed
something.

> 1.	A task running on CPU 0 currently write-holds the lock.
>
> 2.	CPU 1 is running in the kernel, executing a longer-than-average
> 	loop of normal instructions (no atomic instructions or memory
> 	barriers).
>
> 3.	CPU 0 invokes percpu_up_write(), calling up_write(),
> 	synchronize_sched(), and finally mutex_unlock().

And my expectation was, this should be enough because ...

> 4.	CPU 1 executes percpu_down_read(), which calls update_fast_ctr(),

since update_fast_ctr does preempt_disable/enable it should see all
modifications done by CPU 0.

IOW. Suppose that the writer (CPU 0) does

	percpu_done_write();
	STORE;
	percpu_up_write();

This means

	STORE;
	synchronize_sched();
	mutex_unlock();

Now. Do you mean that the next preempt_disable/enable can see the
result of mutex_unlock() but not STORE?

Oleg.

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