Re: [PATCH v2] locking/lock_events: Use this_cpu_add() when necessary

[Waiman Long]

On 5/24/19 1:19 PM, Will Deacon wrote:
> 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@xxxxxxxxxx>
>> ---
>>  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

Good point! But this will be an issue even if we use the non-underscore
version as I don't think it will disable interrupt. Also it is only a
problem if the percpu operation is more than 1 instruction. It is a
single instruction for x86. Other architectures may require more than 1
instruction. In those cases, we may lose count, but it is still better
than getting the count from one CPU and put it into another CPU.

Cheers,
Longman