Re: [Qestion] Is preempt_disable/enable needed in non-preemption code path

[Xu, Yanfei]

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