Re: [ltt-dev] [RFC git tree] Userspace RCU (urcu) for Linux(repost)
From: Mathieu Desnoyers
Date: Wed Feb 11 2009 - 13:52:21 EST
* Paul E. McKenney (paulmck@xxxxxxxxxxxxxxxxxx) wrote:
> On Wed, Feb 11, 2009 at 01:35:20AM -0500, Mathieu Desnoyers wrote:
> > * Paul E. McKenney (paulmck@xxxxxxxxxxxxxxxxxx) wrote:
> > > On Tue, Feb 10, 2009 at 07:57:01PM -0500, Mathieu Desnoyers wrote:
> > > > * Paul E. McKenney (paulmck@xxxxxxxxxxxxxxxxxx) wrote:
> > > > > On Tue, Feb 10, 2009 at 04:28:33PM -0500, Mathieu Desnoyers wrote:
> > > > > > * Paul E. McKenney (paulmck@xxxxxxxxxxxxxxxxxx) wrote:
> > > > > > > On Tue, Feb 10, 2009 at 02:17:31PM -0500, Mathieu Desnoyers wrote:
> > > > > > > > * Paul E. McKenney (paulmck@xxxxxxxxxxxxxxxxxx) wrote:
> > > > > > > > > On Mon, Feb 09, 2009 at 02:03:17AM -0500, Mathieu Desnoyers wrote:
> > > > > > > > >
> > > > > > > > > [ . . . ]
> > > > > > > > >
> > > > > > > > > > I just added modified rcutorture.h and api.h from your git tree
> > > > > > > > > > specifically for an urcutorture program to the repository. Some results :
> > > > > > > > > >
> > > > > > > > > > 8-way x86_64
> > > > > > > > > > E5405 @2 GHZ
> > > > > > > > > >
> > > > > > > > > > ./urcutorture 8 perf
> > > > > > > > > > n_reads: 1937650000 n_updates: 3 nreaders: 8 nupdaters: 1 duration: 1
> > > > > > > > > > ns/read: 4.12871 ns/update: 3.33333e+08
> > > > > > > > > >
> > > > > > > > > > ./urcutorture 8 uperf
> > > > > > > > > > n_reads: 0 n_updates: 4413892 nreaders: 0 nupdaters: 8 duration: 1
> > > > > > > > > > ns/read: nan ns/update: 1812.46
> > > > > > > > > >
> > > > > > > > > > n_reads: 98844204 n_updates: 10 n_mberror: 0
> > > > > > > > > > rcu_stress_count: 98844171 33 0 0 0 0 0 0 0 0 0
> > > > > > > > > >
> > > > > > > > > > However, I've tried removing the second switch_qparity() call, and the
> > > > > > > > > > rcutorture test did not detect anything wrong. I also did a variation
> > > > > > > > > > which calls the "sched_yield" version of the urcu, "urcutorture-yield".
> > > > > > > > >
> > > > > > > > > My confusion -- I was testing my old approach where the memory barriers
> > > > > > > > > are in rcu_read_lock() and rcu_read_unlock(). To force the failures in
> > > > > > > > > your signal-handler-memory-barrier approach, I suspect that you are
> > > > > > > > > going to need a bigger hammer. In this case, one such bigger hammer
> > > > > > > > > would be:
> > > > > > > > >
> > > > > > > > > o Just before exit from the signal handler, do a
> > > > > > > > > pthread_cond_wait() under a pthread_mutex().
> > > > > > > > >
> > > > > > > > > o In force_mb_all_threads(), refrain from sending a signal to self.
> > > > > > > > >
> > > > > > > > > Then it should be safe in force_mb_all_threads() to do a
> > > > > > > > > pthread_cond_broadcast() under the same pthread_mutex().
> > > > > > > > >
> > > > > > > > > This should raise the probability of seeing the failure in the case
> > > > > > > > > where there is a single switch_qparity().
> > > > > > > > >
> > > > > > > >
> > > > > > > > I just did a mb() version of the urcu :
> > > > > > > >
> > > > > > > > (uncomment CFLAGS=+-DDEBUG_FULL_MB in the Makefile)
> > > > > > > >
> > > > > > > > Time per read : 48.4086 cycles
> > > > > > > > (about 6-7 times slower, as expected)
> > > > > > > >
> > > > > > > > This will be useful especially to increase the chance to trigger races.
> > > > > > > >
> > > > > > > > I tried removing the second parity switch from the writer. The rcu
> > > > > > > > torture test did not find the problem yet (maybe I am not using the
> > > > > > > > correct parameters ? It does not run for more than 5 seconds).
> > > > > > > >
> > > > > > > > So I added a "-n" option to test_urcu, so it can make the usleep(1)
> > > > > > > > between the writes optional. I also changed the yield for a usleep with
> > > > > > > > random delay. I also now use a circular buffer rather than malloc so we
> > > > > > > > are sure the memory is not quickly reused by the writer and stays longer
> > > > > > > > in an invalid state.
> > > > > > > >
> > > > > > > > So what really make the problem appear quickly is to add a delay between
> > > > > > > > the rcu_dereference and the assertion on the data validity in thr_reader.
> > > > > > > >
> > > > > > > > It now appears after just a few seconds when running
> > > > > > > > ./test_urcu_yield 20 -r -n
> > > > > > > > Compiled with CFLAGS=+-DDEBUG_FULL_MB
> > > > > > > >
> > > > > > > > It seem to be much harder to trigger with the signal-based version. It's
> > > > > > > > expected, because the writer takes about 50 times longer to execute than
> > > > > > > > with the -DDEBUG_FULL_MB version.
> > > > > > > >
> > > > > > > > So I'll let the ./test_urcu_yield NN -r -n run for a while on the
> > > > > > > > correct version (with DEBUG_FULL_MB) and see what it gives.
> > > > > > >
> > > > > > > Hmmm... I had worse luck this time, took three 10-second tries to
> > > > > > > see a failure:
> > > > > > >
> > > > > > > paulmck@paulmck-laptop:~/paper/perfbook/CodeSamples/defer$ ./rcu_nest32 1 stress
> > > > > > > n_reads: 44682055 n_updates: 9609503 n_mberror: 0
> > > > > > > rcu_stress_count: 44679377 2678 0 0 0 0 0 0 0 0 0
> > > > > > > paulmck@paulmck-laptop:~/paper/perfbook/CodeSamples/defer$ !!
> > > > > > > ./rcu_nest32 1 stress
> > > > > > > n_reads: 42281884 n_updates: 9870129 n_mberror: 0
> > > > > > > rcu_stress_count: 42277756 4128 0 0 0 0 0 0 0 0 0
> > > > > > > paulmck@paulmck-laptop:~/paper/perfbook/CodeSamples/defer$ !!
> > > > > > > ./rcu_nest32 1 stress
> > > > > > > n_reads: 41384304 n_updates: 10040805 n_mberror: 0
> > > > > > > rcu_stress_count: 41380075 4228 1 0 0 0 0 0 0 0 0
> > > > > > > paulmck@paulmck-laptop:~/paper/perfbook/CodeSamples/defer$
> > > > > > >
> > > > > > > This is my prototype version, with read-side memory barriers, no
> > > > > > > signals, and without your initialization-value speedup.
> > > > > > >
> > > > > >
> > > > > > It would be interesting to re-sync our trees, or if you can point me to
> > > > > > a current version of your prototype, I could review it.
> > > > >
> > > > > Look at:
> > > > >
> > > > > CodeSamples/defer/rcu_nest32.[hc]
> > > > >
> > > > > In the git archive:
> > > > >
> > > > > git://git.kernel.org/pub/scm/linux/kernel/git/paulmck/perfbook.git
> > > >
> > > > flip_counter_and_wait : yours do rcu_gp_ctr += RCU_GP_CTR_BOTTOM_BIT
> > > > mine : rcu_gp_ctr ^= RCU_GP_CTR_BOTTOM_BIT.
> > >
> > > Yep, this is before your optimization.
> > >
> >
> > Hrm, and given the RCU_GP_CTR_BOTTOM_BIT is in the MSBs, there is no
> > possible effect on the LSBs. That should work even if it overflows. OK.
> > That should even work with my optimization. But I somehow prefer the xor
> > (if it's not slower), because we really only need 1 bit to flip on and
> > off.
> >
> > > > Another major difference between our tree is the lack of smp_mb() at the
> > > > end of flip_counter_and_wait() (in your tree).
> > > >
> > > > Your code does :
> > > >
> > > > smp_mb()
> > > > switch parity
> > > > smp_mb()
> > > > wait for each thread ongoing old gp
> > > > <<<<<<< ---- missing smp_mb.
> > > > switch parity
> > > > smp_mb()
> > > > wait for each thread ongoing old gp
> > > > smp_mb()
> > >
> > > This should be OK -- or am I missing a failure scenario?
> > > Keep in mind that I get failures only when omitting a counter
> > > flip, not with the above code.
> > >
> >
> > OK, it's good that you point out that the failure only occurs when
> > omitting the counter flip.
> >
> > So if we leave out the mb() we can end up in a situation where a reader
> > thread is still in an ongoing old gp and we switch the parity. The big
> > question is : should we be concerned about this ?
> >
> > From the writer point of view :
> >
> > Given there is no data dependency between the parity update and the
> > per_thread(rcu_reader_gp, t) read done in the while loop waiting for
> > threads, and given even the compiler barrier() has no effect wrt the
> > last test done after the last iteration of the loop, we could think of
> > compiler optimizations doing the following to our code (let's focus on a
> > single loop of for_each_thread) :
> >
> > transforming
> >
> > while (rcu_old_gp_ongoing(t))
> > barrier();
> > rcu_gp_ctr += RCU_GP_CTR_BOTTOM_BIT;
> >
> > into
> >
> > if (!rcu_old_gp_ongoing(t))
> > goto end;
> > while (rcu_old_gp_ongoing(t))
> > barrier();
> > end:
> > rcu_gp_ctr += RCU_GP_CTR_BOTTOM_BIT;
> >
> > This leaves the choice to the compiler to perform the rcu_gp_ctr
> > increment before the per_thread(rcu_reader_gp, t) read, because there is
> > no barrier.
> >
> > Not only does this apply to the compiler, but also to the memory
> > barriers. We can end up in a situation where the CPU decides to to the
> > rcu_gp_ctr increment before reading the last rcu_old_gp_ongoing value,
> > given there is no data dependency between those two.
> >
> > You could argue that ACCESS_ONCE() around the per_thread(rcu_reader_gp,
> > t) read will order reads, but I don't think we should rely on this on
> > SMP. This is really supposed to be there just to make sure we don't end
> > up doing multiple variable reads on UP wrt to local interrupts.
> >
> > You could also argue that rcu_gp_ctr is read within
> > rcu_old_gp_ongoing(), which should normally order the memory accesses.
> > It actually does only order memory access to the rcu_gp_ctr variable,
> > not the per_thread(rcu_reader_gp, t), because, here again, there if no
> > data dependency whatsoever between per_thread(rcu_reader_gp, t) and
> > rcu_gp_ctr. A possible scenario : rcu_gp_ctr could be read, then we have
> > the rcu_gp_ctr increment, and only then could the
> > per_thread(rcu_reader_gp, t) variable be read to perform the test.
> >
> > But I see that even in rcu_read_lock, there is no strict ordering
> > between __get_thread_var(rcu_reader_gp) and rcu_gp_ctr read. Therefore,
> > I conclude that ordering between those two variables does not matter at
> > all. I also suspect that this is the core reason for doing 2 q.s. period
> > flip at each update.
> >
> > Am I correct ?
>
> I do not believe so -- please see my earlier email calling out the
> sequence of events leading to failure in the single-flip case:
>
> http://lkml.org/lkml/2009/2/7/67
>
Hrm, let me present it in a different, more straightfoward way :
In you Promela model (here : http://lkml.org/lkml/2009/2/10/419)
There is a memory barrier here in the updater :
do
:: 1 ->
if
:: (urcu_active_readers & RCU_GP_CTR_NEST_MASK) != 0 &&
(urcu_active_readers & ~RCU_GP_CTR_NEST_MASK) !=
(urcu_gp_ctr & ~RCU_GP_CTR_NEST_MASK) ->
skip;
:: else -> break;
fi
od;
need_mb = 1;
do
:: need_mb == 1 -> skip;
:: need_mb == 0 -> break;
od;
urcu_gp_ctr = urcu_gp_ctr + RCU_GP_CTR_BIT;
do
:: 1 ->
if
:: (urcu_active_readers & RCU_GP_CTR_NEST_MASK) != 0 &&
(urcu_active_readers & ~RCU_GP_CTR_NEST_MASK) !=
(urcu_gp_ctr & ~RCU_GP_CTR_NEST_MASK) ->
skip;
:: else -> break;
fi;
od;
However, in your C code of nest_32.c, there is none. So it is at the
very least an inconsistency between your code and your model.
> > > > I also wonder why you have a smp_mb() after spin_unlock() in your
> > > > synchronize_rcu() -> if you follow the Linux kernel semantics for
> > > > spinlocks, the smp_mb() should be implied. (but I have not looked at
> > > > your spin_lock/unlock primitives yet).
> > >
> > > Perhaps things have changed, but last I knew, spin_lock() and
> > > spin_unlock() were only required to keep the critical section in, not
> > > to keep things out of the critical section.
> > >
> >
> > Hrm, reading Documentation/memory-barriers.txt again tells me things
> > might have changed (if I am reading correctly the section LOCKS VS
> > MEMORY ACCESSES).
>
> In the 2.6.26 version of Documentation/memory-barriers.txt, there is
> the following near line 366:
>
> (5) LOCK operations.
>
> This acts as a one-way permeable barrier. It guarantees that all memory
> operations after the LOCK operation will appear to happen after the LOCK
> operation with respect to the other components of the system.
>
> Memory operations that occur before a LOCK operation may appear to happen
> after it completes.
>
> A LOCK operation should almost always be paired with an UNLOCK operation.
>
>
> (6) UNLOCK operations.
>
> This also acts as a one-way permeable barrier. It guarantees that all
> memory operations before the UNLOCK operation will appear to happen before
> the UNLOCK operation with respect to the other components of the system.
>
> Memory operations that occur after an UNLOCK operation may appear to
> happen before it completes.
>
> LOCK and UNLOCK operations are guaranteed to appear with respect to each
> other strictly in the order specified.
>
> The use of LOCK and UNLOCK operations generally precludes the need for
> other sorts of memory barrier (but note the exceptions mentioned in the
> subsection "MMIO write barrier").
>
> > Correct me if I am wrong, but I don't think it makes sense to insure
> > memory barriers to keep accesses within the critical section and not
> > outside, because such memory access could well be another spinlock.
>
> Almost, but not quite. ;-)
>
> > Therefore, we could end up in a situation where we have two locks, A and
> > B, taken in the following order in the source code :
> >
> > LOCK A
> >
> > UNLOCK A
> >
> > LOCK B
> >
> > UNLOCK B
> >
> > Then, following your assumption, it would be possible for a CPU to do
> > the memory accesses associated to lock A and B in a random order one vs
> > the other. Given there would be no requirement to keep things out of
> > those respective critical sections, LOCK A could be taken within LOCK B,
> > and the opposite would also be valid.
> >
> > Valid memory access orders :
> >
> > 1)
> > LOCK A
> > LOCK B
> > UNLOCK B
> > UNLOCK A
> >
> > 2)
> > LOCK B
> > LOCK A
> > UNLOCK A
> > UNLOCK B
>
> #2 is wrong -- LOCK A is guaranteed to prohibit LOCK B from passing it,
> as that would be equivalent to letting LOCK A's critical section leak out.
>
> > The only constraint that ensures we won't end up in this situation is
> > the fact that memory accesses done outside of the critical section stays
> > outside of the critical section.
>
> Let's take it one transformation at a time:
>
> 1. LOCK A; UNLOCK A; LOCK B; UNLOCK B
>
> 2. LOCK A; LOCK B; UNLOCK A; UNLOCK B
>
> This one is OK, because both the LOCK B and the UNLOCK A
> are permitted to allow more stuff to enter their respective
> critical sections.
>
> 3. LOCK B; LOCK A; UNLOCK A; UNLOCK B
>
> This is -not- legal! LOCK A is forbidden to allow LOCK B
> to escape its critical section.
>
> Does this make sense?
>
Ah, yes. Thanks for the explanation.
Mathieu
> Thanx, Paul
>
> > Mathieu
> >
> >
> >
> > > Thanx, Paul
> > >
> > > > Mathieu
> > > >
> > > > > Thanx, Paul
> > > > >
> > > > > _______________________________________________
> > > > > ltt-dev mailing list
> > > > > ltt-dev@xxxxxxxxxxxxxxxxxxxxx
> > > > > http://lists.casi.polymtl.ca/cgi-bin/mailman/listinfo/ltt-dev
> > > > >
> > > >
> > > > --
> > > > Mathieu Desnoyers
> > > > OpenPGP key fingerprint: 8CD5 52C3 8E3C 4140 715F BA06 3F25 A8FE 3BAE 9A68
> > >
> >
> > --
> > Mathieu Desnoyers
> > OpenPGP key fingerprint: 8CD5 52C3 8E3C 4140 715F BA06 3F25 A8FE 3BAE 9A68
>
--
Mathieu Desnoyers
OpenPGP key fingerprint: 8CD5 52C3 8E3C 4140 715F BA06 3F25 A8FE 3BAE 9A68
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