On Wed, May 23, 2018 at 01:04:58PM -0700, Paul E. McKenney wrote:
> On Wed, May 23, 2018 at 03:13:37PM -0400, Steven Rostedt wrote:
> > 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
>
> Looks plausible to me!
>
> Joel, do you see any gotchas in Steve's patch?
I see one but I hope I'm not day dreaming.. :D
> > > > > 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.
I agree with your suggestions and Steven's patch is better.
> > 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;
Shouldn't this assignment be done outside the loop? I believe the variable
will be initialized on each iteration.
A program like this doesn't terminate:
#include<stdio.h>
int main() {
for (;;) {
int i = 10;
if (!(i--))
break;
}
return 0;
}
Otherwise looks good to me, I would initialize fract to 10 so its consistent
with "HZ/10" in other parts of the code but I'm ok with either number.
thanks!
- Joel
