linux-kernel - Re: [PATCH RFC] time,signal: protect resource use statistics with seqlock

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Date:	Fri, 15 Aug 2014 16:26:01 +0200
From:	Oleg Nesterov <oleg@...hat.com>
To:	Frederic Weisbecker <fweisbec@...il.com>
Cc:	Rik van Riel <riel@...hat.com>,
	LKML <linux-kernel@...r.kernel.org>,
	Peter Zijlstra <peterz@...radead.org>,
	Hidetoshi Seto <seto.hidetoshi@...fujitsu.com>,
	Frank Mayhar <fmayhar@...gle.com>,
	Frederic Weisbecker <fweisbec@...hat.com>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Sanjay Rao <srao@...hat.com>,
	Larry Woodman <lwoodman@...hat.com>
Subject: Re: [PATCH RFC] time,signal: protect resource use statistics with
	seqlock

On 08/15, Frederic Weisbecker wrote:
>
> 2014-08-14 16:39 GMT+02:00 Oleg Nesterov <oleg@...hat.com>:
> > On 08/14, Frederic Weisbecker wrote:
> >>
> >> I mean the read side doesn't use a lock with seqlocks. It's only made
> >> of barriers and sequence numbers to ensure the reader doesn't read
> >> some half-complete update. But other than that it can as well see the
> >> update n - 1 since barriers don't enforce latest results.
> >
> > Yes, sure, read_seqcount_begin/read_seqcount_retry "right after"
> > write_seqcount_begin-update-write_seqcount_begin can miss "update" part
> > along with ->sequence modifications.
> >
> > But I still can't understand how this can lead to non-monotonic results,
> > could you spell?
>
> Well lets say clock = T.
> CPU 0 updates at T + 1.
> Then I call clock_gettime() from CPU 1 and CPU 2. CPU 1 reads T + 1
> while CPU 1 still reads T.
> If I do yet another round of clock_gettime() on CPU 1 and CPU 2, it's
> possible that CPU 2 still sees T. With the spinlocked version that
> thing can't happen, the second round would read at least T + 1 for
> both CPUs.

But this is fine? And CPU 2 doesn't see a non-monotonic result?

OK, this could be wrong if, say,

	void print_clock(void)
	{
		lock(SOME_LOCK);
		printk(..., clock_gettime());
		unlock(SOME_LOCK);
	}
	
printed the non-monotonic numbers if print_clock() is called on CPU_1 and
then on CPU_2. But in this case CPU_2 can't miss the changes on CPU_0 if
they were already visible to CPU_1 under the same lock. IOW,

	int T = 0;	/* can be incremented at any time */

	void check_monotony(void)
	{
		static int t = 0;

		lock(SOME_LOCK);
		BUG(t > T);
		T = t;
		unlock(SOME_LOCK);
	}

must work corrrectly (ignoring overflow) even if T is changed without
SOME_LOCK.

Otherwise, without some sort of synchronization the different results on
CPU_1/2 should be fine.

Or I am still missing your point?

Oleg.

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