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

Frederic Weisbecker
2014-08-15T02:52:39Z

2014-08-14 16:39 GMT+02:00 Oleg Nesterov <oleg@redhat.com>:
> On 08/14, Frederic Weisbecker wrote:
>>
>> 2014-08-14 3:57 GMT+02:00 Rik van Riel <riel@redhat.com>:
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>> > On 08/13/2014 08:43 PM, Frederic Weisbecker wrote:
>> >> On Wed, Aug 13, 2014 at 05:03:24PM -0400, Rik van Riel wrote:
>> >>
>> >> I'm worried about such lockless solution based on RCU or read
>> >> seqcount because we lose the guarantee that an update is
>> >> immediately visible by all subsequent readers.
>> >>
>> >> Say CPU 0 updates the thread time and both CPU 1 and CPU 2 right
>> >> after that call clock_gettime(), with the spinlock we were
>> >> guaranteed to see the new update. Now with a pure seqlock read
>> >> approach, we guarantee a read sequence coherency but we don't
>> >> guarantee the freshest update result.
>> >>
>> >> So that looks like a source of non monotonic results.
>> >
>> > Which update are you worried about, specifically?
>> >
>> > The seq_write_lock to update the usage stat in p->signal will lock out
>> > the seqlock read side used to check those results.
>> >
>> > Is there another kind of thing read by cpu_clock_sample_group that you
>> > believe is not excluded by the seq_lock?
>>
>> 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.