Messages in this thread | | | Date | Wed, 23 May 2018 17:51:19 -0700 | From | Joel Fernandes <> | Subject | Re: [PATCH 1/4] rcu: Speed up calling of RCU tasks callbacks |
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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" <paulmck@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
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