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Message-ID: <Pine.LNX.4.58.0801152238130.19680@gandalf.stny.rr.com>
Date: Wed, 16 Jan 2008 08:17:58 -0500 (EST)
From: Steven Rostedt <rostedt@...dmis.org>
To: Mathieu Desnoyers <mathieu.desnoyers@...ymtl.ca>
cc: LKML <linux-kernel@...r.kernel.org>, Ingo Molnar <mingo@...e.hu>,
Linus Torvalds <torvalds@...ux-foundation.org>,
Andrew Morton <akpm@...ux-foundation.org>,
Peter Zijlstra <a.p.zijlstra@...llo.nl>,
Christoph Hellwig <hch@...radead.org>,
Gregory Haskins <ghaskins@...ell.com>,
Arnaldo Carvalho de Melo <acme@...stprotocols.net>,
Thomas Gleixner <tglx@...utronix.de>,
Tim Bird <tim.bird@...sony.com>,
Sam Ravnborg <sam@...nborg.org>,
"Frank Ch. Eigler" <fche@...hat.com>,
Steven Rostedt <srostedt@...hat.com>,
Paul Mackerras <paulus@...ba.org>,
Daniel Walker <dwalker@...sta.com>
Subject: Re: [RFC PATCH 16/22 -v2] add get_monotonic_cycles
[ CC'd Daniel Walker, since he had problems with this code ]
On Tue, 15 Jan 2008, Mathieu Desnoyers wrote:
>
> I agree with you that I don't see how the compiler could reorder this.
> So we forget about compiler barriers. Also, the clock source used is a
> synchronized clock source (get_cycles_sync on x86_64), so it should make
> sure the TSC is read at the right moment.
>
> However, what happens if the clock source is, say, the jiffies ?
>
> Is this case, we have :
>
> static cycle_t jiffies_read(void)
> {
> return (cycle_t) jiffies;
> }
>
> Which is nothing more than a memory read of
>
> extern unsigned long volatile __jiffy_data jiffies;
Yep, and that's not my concern.
>
> I think it is wrong to assume that reads from clock->cycle_raw and from
> jiffies will be ordered correctly in SMP. I am tempted to think that
> ordering memory writes to clock->cycle_raw vs jiffies is also needed in this
> case (where clock->cycle_raw is updated, or where jiffies is updated).
>
> We can fall in the same kind of issue if we read the HPET, which is
> memory I/O based. It does not seems correct to assume that MMIO vs
> normal memory reads are ordered. (pointing back to this article :
> http://lwn.net/Articles/198988/)
That and the dread memory barrier thread that my head is still spinning
on.
Ok, lets take a close look at the code in question. I may be wrong, and if
so, great, we can fix it.
We have this in get_monotonic_cycles:
{
cycle_t cycle_now, cycle_delta, cycle_monotonic, cycle_last;
do {
cycle_monotonic = clock->cycle_monotonic;
cycle_last = clock->cycle_last;
cycle_now = clocksource_read(clock);
cycle_delta = (cycle_now - cycle_last) & clock->mask;
} while (cycle_monotonic != clock->cycle_monotonic ||
cycle_last != clock->cycle_last);
return cycle_monotonic + cycle_delta;
}
and this in clocksource.h
static inline void clocksource_accumulate(struct clocksource *cs, cycle_t now)
{
cycle_t offset = (now - cs->cycle_last) & cs->mask;
cs->cycle_last = now;
cs->cycle_accumulated += offset;
cs->cycle_monotonic += offset;
}
now is usually just a clocksource_read() passed in.
The goal is to have clock_monotonic always return something that is
greater than what was read the last time.
Let's make a few assumptions now (for others to shoot them down). One
thing is that we don't need to worry too much about MMIO, because we are
doing a read. This means we need the data right now to contiune. So the
read being a function call should keep gcc from moving stuff around, and
since we are doing an IO read, the order of events should be pretty much
synchronized. in
1. load cycle_last and cycle_monotonic (we don't care which order)*
2. read clock source
3. calculate delta and while() compare (order doesn't matter)
* we might care (see below)
If the above is incorrect, then we need to fix get_monotonic_cycles.
in clocksource_accumulate, we have:
offset = ((now = cs->read()) - cycle_last) & cs->mask;
cycle_last = now;
cycle_accumulate += offset;
cycle_monotonic += offset;
The order of events here are. Using the same reasoning as above, the read
must be first and completed because for gcc it's a function, and for IO,
it needs to return data.
1. cs->read
2. update cycle_last, cycle_accumulate, cycle_monotonic.
Can we assume, if the above for get_monotonic_cycles is correct, that
since we read and compare cycle_last and cycle_monotonic, that neither of
them have changed over the read? So we have a snapshot of the
clocksource_accumulate.
So the worst thing that I can think of, is that cycle_monotonic is update
*before* cycle_last:
cycle_monotonic += offest;
<get_monotonic_cycles run on other CPU>
cycle_last = now;
cycle_last = 5
cycle_monotonic = 0
CPU 0 CPU 1
---------- -------------
cs->read() = 10
offset = 10 - 5 = 5
cycle_monotonic = 5
cycle_monotonic = 5
cycle_last = 5
cs->read() = 11
delta = 11 - 5 = 6
cycle_monotonic and cycle_last still same
return 5 + 6 = 11
cycle_last = 10
cycle_monotonic = 5
cycle_last = 10
cs->read() = 12
delta = 12 - 10 = 2
cycle_monotonic and cycle_last still same
return 5 + 2 = 7
**** ERROR *****
So, we *do* need memory barriers. Looks like cycle_last and
cycle_monotonic need to be synchronized.
OK, will this do?
cycle_t notrace get_monotonic_cycles(void)
{
cycle_t cycle_now, cycle_delta, cycle_monotonic, cycle_last;
do {
cycle_monotonic = clock->cycle_monotonic;
smp_rmb();
cycle_last = clock->cycle_last;
cycle_now = clocksource_read(clock);
cycle_delta = (cycle_now - cycle_last) & clock->mask;
} while (cycle_monotonic != clock->cycle_monotonic ||
cycle_last != clock->cycle_last);
return cycle_monotonic + cycle_delta;
}
and this in clocksource.h
static inline void clocksource_accumulate(struct clocksource *cs, cycle_t now)
{
cycle_t offset = (now - cs->cycle_last) & cs->mask;
cs->cycle_last = now;
smp_wmb();
cs->cycle_accumulated += offset;
cs->cycle_monotonic += offset;
}
We may still get to a situation where cycle_monotonic is of the old value
and cycle_last is of the new value. That would give us a smaller delta
than we want.
Lets look at this, with a slightly different situation.
cycle_last = 5
cycle_monotonic = 0
CPU 0 CPU 1
---------- -------------
cs->read() = 10
offset = 10 - 5 = 5
cycle_last = 10
cycle_monotonic = 5
cycle_monotonic = 5
cycle_last = 10
cs->read() = 12
delta = 12 - 10 = 2
cycle_monotonic and cycle_last still same
return 5 + 2 = 7
cs->read() = 13
offset = 13 - 10 = 2
cycle_last = 13
cycle_monotonic = 5
cycle_last = 13
cs->read() = 14
delta = 14 - 13 = 1
cycle_monotonic and cycle_last still same
return 5 + 1 = 6
**** ERROR ****
Crap, looks like we do need a stronger locking here :-(
Hmm, I might as well just use seq_locks, and make sure that tracing
does not hit them.
Thanks!
-- Steve
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