On Wed, 23 May 2018 10:03:03 -0700
"Paul E. McKenney" <paul...@linux.vnet.ibm.com> wrote:
> > > > diff --git a/kernel/rcu/update.c b/kernel/rcu/update.c
> > > > index 5783bdf86e5a..a28698e44b08 100644
> > > > --- a/kernel/rcu/update.c
> > > > +++ b/kernel/rcu/update.c
> > > > @@ -743,6 +743,12 @@ static int __noreturn rcu_tasks_kthread(void *arg)
> > > > */
> > > > synchronize_srcu(&tasks_rcu_exit_srcu);
> > > >
> > > > + /*
> > > > + * Wait a little bit incase held tasks are released
> > >
> > > in case
> > >
> > > > + * during their next timer ticks.
> > > > + */
> > > > + schedule_timeout_interruptible(HZ/10);
> > > > +
> > > > /*
> > > > * Each pass through the following loop scans the list
> > > > * of holdout tasks, removing any that are no longer
> > > > @@ -755,7 +761,6 @@ static int __noreturn rcu_tasks_kthread(void *arg)
> > > > int rtst;
> > > > struct task_struct *t1;
> > > >
> > > > - schedule_timeout_interruptible(HZ);
> > > > rtst = READ_ONCE(rcu_task_stall_timeout);
> > > > needreport = rtst > 0 &&
> > > > time_after(jiffies, lastreport +
> > > > rtst);
> > > > @@ -768,6 +773,11 @@ static int __noreturn rcu_tasks_kthread(void *arg)
> > > > check_holdout_task(t, needreport,
> > > > &firstreport);
> > > > cond_resched();
> > > > }
> > > > +
> > > > + if (list_empty(&rcu_tasks_holdouts))
> > > > + break;
> > > > +
> > > > + schedule_timeout_interruptible(HZ);
> >
> > Why is this a full second wait and not the HZ/10 like the others?
>
> The idea is to respond quickly on small idle systems and to reduce the
> number of possibly quite lengthy traversals of the task list otherwise.
> I actually considered exponential backoff, but decided to keep it simple,
> at least to start with.
Ah, now it makes sense. Reading what you wrote, we can still do a
backoff and keep it simple. What about the patch below. It appears to
have the same performance improvement as Joel's
-- Steve
> > >
> > > Is there a better way to do this? Can this be converted into a for-loop?
> > > Alternatively, would it make sense to have a firsttime local variable
> > > initialized to true, to keep the schedule_timeout_interruptible() at
> > > the beginning of the loop, but skip it on the first pass through the loop?
> > >
> > > Don't get me wrong, what you have looks functionally correct, but
> > > duplicating the condition might cause problems later on, for example,
> > > should a bug fix be needed in the condition.
> > >
diff --git a/kernel/rcu/update.c b/kernel/rcu/update.c
index 68fa19a5e7bd..c6df9fa916cf 100644
--- a/kernel/rcu/update.c
+++ b/kernel/rcu/update.c
@@ -796,13 +796,22 @@ static int __noreturn rcu_tasks_kthread(void *arg)
* holdouts. When the list is empty, we are done.
*/
lastreport = jiffies;
- while (!list_empty(&rcu_tasks_holdouts)) {
+ for (;;) {
bool firstreport;
bool needreport;
int rtst;
struct task_struct *t1;
+ int fract = 15;
+
+ /* Slowly back off waiting for holdouts */
+ schedule_timeout_interruptible(HZ/fract);
+
+ if (list_empty(&rcu_tasks_holdouts))
+ break;
+
+ if (fract > 1)
+ fract--;
- schedule_timeout_interruptible(HZ);
rtst = READ_ONCE(rcu_task_stall_timeout);
needreport = rtst > 0 &&
time_after(jiffies, lastreport + rtst);
