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SubjectRe: [PATCH RT v2 2/3] sched: migrate_enable: Use sleeping_lock to indicate involuntary sleep
On Tue, Aug 27, 2019 at 11:23:33AM +0200, Sebastian Andrzej Siewior wrote:
> On 2019-08-26 09:29:45 [-0700], Paul E. McKenney wrote:
> > > The mechanism that is used here may change in future. I just wanted to
> > > make sure that from RCU's side it is okay to schedule here.
> >
> > Good point.
> >
> > The effect from RCU's viewpoint will be to split any non-rcu_read_lock()
> > RCU read-side critical section at this point. This alrady happens in a
> > few places, for example, rcu_note_context_switch() constitutes an RCU
> > quiescent state despite being invoked with interrupts disabled (as is
> > required!). The __schedule() function just needs to understand (and does
> > understand) that the RCU read-side critical section that would otherwise
> > span that call to rcu_node_context_switch() is split in two by that call.
>
> Okay. So I read this as invoking schedule() at this point is okay.

As long as no one is relying on a non-rcu_read_lock() RCU
read-side critical section (local_bh_disable(), preempt_disable(),
local_irq_disable(), ...) spanning this call. But that depends on the
calling code and on other code it interacts with it, not on any specific
need on the part of RCU itself.

> Looking at this again, this could also happen on a PREEMPT=y kernel if
> the kernel decides to preempt a task within a rcu_read_lock() section
> and put it back later on another CPU.

This is an rcu_read_lock() critical section, so yes, on a PREEMPT=y
kernel, executing schedule() will cause the corresponding RCU read-side
critical section to persist, following the preempted tasks. Give or
take lockdep complaints.

On a PREEMPT=n kernel, schedule() within an RCU read-side critical
section instead results in that critical section being split in two.
And this will also results in lockdep complaints.

> > However, if this was instead an rcu_read_lock() critical section within
> > a PREEMPT=y kernel, then if a schedule() occured within stop_one_task(),
> > RCU would consider that critical section to be preempted. This means
> > that any RCU grace period that is blocked by this RCU read-side critical
> > section would remain blocked until stop_one_cpu() resumed, returned,
> > and so on until the matching rcu_read_unlock() was reached. In other
> > words, RCU would consider that RCU read-side critical section to span
> > the call to stop_one_cpu() even if stop_one_cpu() invoked schedule().
>
> Isn't that my example from above and what we do in RT? My understanding
> is that this is the reason why we need BOOST on RT otherwise the RCU
> critical section could remain blocked for some time.

At this point, I must confess that I have lost track of whose example
it is. It was first reported in 2006, if I remember correctly. ;-)

But yes, you are correct, the point of RCU priority boosting is to
cause tasks that have been preempted while within RCU read-side critical
sections to be scheduled so that they can reach their rcu_read_unlock()
calls, thus allowing the current grace period to end.

> > On the other hand, within a PREEMPT=n kernel, the call to schedule()
> > would split even an rcu_read_lock() critical section. Which is why I
> > asked earlier if sleeping_lock_inc() and sleeping_lock_dec() are no-ops
> > in !PREEMPT_RT_BASE kernels. We would after all want the usual lockdep
> > complaints in that case.
>
> sleeping_lock_inc() +dec() is only RT specific. It is part of RT's
> spin_lock() implementation and used by RCU (rcu_note_context_switch())
> to not complain if invoked within a critical section.

Then this is being called when we have something like this, correct?

DEFINE_SPINLOCK(mylock); // As opposed to DEFINE_RAW_SPINLOCK().

...

rcu_read_lock();
do_something();
spin_lock(&mylock); // Can block in -rt, thus needs sleeping_lock_inc()
...
rcu_read_unlock();

Without sleeping_lock_inc(), lockdep would complain about a voluntary
schedule within an RCU read-side critical section. But in -rt, voluntary
schedules due to sleeping on a "spinlock" are OK.

Am I understanding this correctly?

Thanx, Paul

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