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SubjectRe: [Qestion] Is preempt_disable/enable needed in non-preemption code path
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On 4/16/21 1:07 AM, Paul E. McKenney wrote:
> [Please note: This e-mail is from an EXTERNAL e-mail address]
>
> On Fri, Apr 16, 2021 at 12:18:42AM +0800, Xu, Yanfei wrote:
>>
>>
>> On 4/15/21 11:43 PM, Paul E. McKenney wrote:
>>> [Please note: This e-mail is from an EXTERNAL e-mail address]
>>>
>>> On Thu, Apr 15, 2021 at 11:04:05PM +0800, Xu, Yanfei wrote:
>>>> Hi experts,
>>>>
>>>> I am learning rcu mechanism and its codes. When looking at the
>>>> rcu_blocking_is_gp(), I found there is a pair preemption disable/enable
>>>> operation in non-preemption code path. And it has been a long time. I can't
>>>> understand why we need it? Is there some thing I missed? If not, can we
>>>> remove the unnecessary operation like blow?
>>>
>>> Good point, you are right that preemption is disabled anyway in that block
>>> of code. However, preempt_disable() and preempt_enable() also prevent the
>>> compiler from moving that READ_ONCE() around. So my question to you is
>>> whether it is safe to remove those statements entirely or whether they
>>> should instead be replaced by barrier() or similar.
>>
>> Thanks for your reply! :)
>>
>> Yes, preempt_disable() and preempt_enable() defined in !preemption are
>> barrier(). barrier can prevent from reordering that READ_ONCE(), but base on
>> my current understanding, volatile in READ_ONCE can also tell the compiler
>> not to reorder it. So, I think it's safe?
>
> Maybe.
>
> Please keep in mind that although the compiler is prohibited from
> reordering volatile accesses with each other, there is nothing stopping
> it from reordering volatile accesses with non-volatile accesses.

Thanks for your patient explanation!

I am trying to absorb what you said. Blow are my understanding:
1. "the compiler is prohibited from reordering volatile accesses with
each other" means these situations:
int a;
foo()
{
for(;;)
READ_ONCE(a);
}

or

int a,b;
foo()
{
int c,d;
c = READ_ONCE(a);
d = READ_ONCE(b);
}

2. "volatile accesses with non-volatile accesses" means d=b may happen
before c=READ_ONCE(a) :
int a;
foo()
{
int b = 2
int c,d;
c = READ_ONCE(a);
d = b;
}
if we want to keep the ordering of volatile access "c=READ_ONCE(a)" and
non-volatile access "d=b", we should use stronger barrier like barrier().

Hope I didn't misunderstand.

Back to rcu_blocking_is_gp(), I find this link today
https://www.spinics.net/lists/rcu/msg03985.html
With the content in this link, I still haven't got the meaning of these
two barrier(). I think I should learn knowledge about cpu-hotplug and
things which talked in the link first to make sure if I am missing
something, and then consult you. :)

Best regards,
Yanfei

>
> Thanx, Paul
>
>> Best regards,
>> Yanfei
>>
>>>
>>> Thanx, Paul
>>>
>>>> diff --git a/kernel/rcu/tree.c b/kernel/rcu/tree.c
>>>> index da6f5213fb74..c6d95a00715e 100644
>>>> --- a/kernel/rcu/tree.c
>>>> +++ b/kernel/rcu/tree.c
>>>> @@ -3703,7 +3703,6 @@ static int rcu_blocking_is_gp(void)
>>>> if (IS_ENABLED(CONFIG_PREEMPTION))
>>>> return rcu_scheduler_active == RCU_SCHEDULER_INACTIVE;
>>>> might_sleep(); /* Check for RCU read-side critical section. */
>>>> - preempt_disable();
>>>> /*
>>>> * If the rcu_state.n_online_cpus counter is equal to one,
>>>> * there is only one CPU, and that CPU sees all prior accesses
>>>> @@ -3718,7 +3717,6 @@ static int rcu_blocking_is_gp(void)
>>>> * Those memory barriers are provided by CPU-hotplug code.
>>>> */
>>>> ret = READ_ONCE(rcu_state.n_online_cpus) <= 1;
>>>> - preempt_enable();
>>>> return ret;
>>>> }
>>>>
>>>>
>>>>
>>>> Best regards,
>>>> Yanfei

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