Messages in this thread | | | Date | Fri, 24 May 2019 18:19:39 +0100 | From | Will Deacon <> | Subject | Re: [PATCH v2] locking/lock_events: Use this_cpu_add() when necessary |
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On Fri, May 24, 2019 at 12:53:46PM -0400, Waiman Long wrote: > The kernel test robot has reported that the use of __this_cpu_add() > causes bug messages like: > > BUG: using __this_cpu_add() in preemptible [00000000] code: ... > > This is only an issue on preempt kernel where preemption can happen in > the middle of a percpu operation. We are still using __this_cpu_*() for > !preempt kernel to avoid additional overhead in case CONFIG_PREEMPT_COUNT > is set. > > v2: Simplify the condition to just preempt or !preempt. > > Fixes: a8654596f0371 ("locking/rwsem: Enable lock event counting") > Signed-off-by: Waiman Long <longman@redhat.com> > --- > kernel/locking/lock_events.h | 23 +++++++++++++++++++++-- > 1 file changed, 21 insertions(+), 2 deletions(-) > > diff --git a/kernel/locking/lock_events.h b/kernel/locking/lock_events.h > index feb1acc54611..05f34068ec06 100644 > --- a/kernel/locking/lock_events.h > +++ b/kernel/locking/lock_events.h > @@ -30,13 +30,32 @@ enum lock_events { > */ > DECLARE_PER_CPU(unsigned long, lockevents[lockevent_num]); > > +/* > + * The purpose of the lock event counting subsystem is to provide a low > + * overhead way to record the number of specific locking events by using > + * percpu counters. It is the percpu sum that matters, not specifically > + * how many of them happens in each cpu. > + * > + * In !preempt kernel, we can just use __this_cpu_*() as preemption > + * won't happen in the middle of the percpu operation. In preempt kernel, > + * preemption happens in the middle of the percpu operation may produce > + * incorrect result. > + */ > +#ifdef CONFIG_PREEMPT > +#define lockevent_percpu_inc(x) this_cpu_inc(x) > +#define lockevent_percpu_add(x, v) this_cpu_add(x, v) > +#else > +#define lockevent_percpu_inc(x) __this_cpu_inc(x) > +#define lockevent_percpu_add(x, v) __this_cpu_add(x, v)
Are you sure this works wrt IRQs? For example, if I take an interrupt when trying to update the counter, and then the irq handler takes a qspinlock which in turn tries to update the counter. Would I lose an update in that scenario?
Will
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