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);