Re: : [PATCH v8 1/2] sched/cache: Reduce the overhead of task_cache_work by only scan the visisted cpus

From: Tim Chen

Date: Wed Jul 29 2026 - 14:31:25 EST


On Wed, 2026-07-29 at 17:19 +0800, Luo Gengkun wrote:
>
> On 2026/7/28 20:08, Chen Yu wrote:
> > On Tue, Jul 28, 2026 at 04:53:59PM +0800, Luo Gengkun wrote:
> > >
> > > On 2026/7/27 9:09, Chen, Yu C wrote:
> > > > On 7/23/2026 12:04 PM, Luo Gengkun wrote:
> > > >
> > > > [ ... ]
> > > >
> > > > >       guard(raw_spinlock_irqsave)(&rq->cpu_epoch_lock);
> > > > >       __update_mm_sched(rq, pcpu_sched);
> > > > > +    /* Skip the rq that has not been hit for a long time */
> > > > > +    if ((rq->cpu_epoch - pcpu_sched->epoch_last_visit) > llc_epoch_affinity_timeout) {
> > > >
> > > > In v2 there is a check if the cpu has been set before writing:
> > > > cpumask_test_cpu(cpu_of(rq), &mm->sc_stat.visited_cpus)
> > > > https://lore.kernel.org/all/20260414150745.225416-1-luogengkun2@xxxxxxxxxx/
> > > > do we need to bring that back?
> > > >
> > > I don't think we need it back. Here is why:
> > >
> > > In v2, for_each_cpu was used instead of for_each_cpu_and in the inner loop,
> > > meaning some CPUs being checked might not have been set. Therefore,
> > > cpumask_test_cpu was necessary to filter out those cases.
> > >
> > > Now, with for_each_cpu_and(i, sched_domain_span(sd), &mm->sc_stat.visited_cpus),
> > > we can ensure each scanned CPU is set, so the issue no longer exists.
> > > Furthermore, the only place where the visited_cpus bits are cleared is
> > > task_cache_work(), which is only called once per scan period, there is no
> > > risk of the bit being cleared concurrently mid-loop.
> > >
> >
> > Make sense.
> >
> > > However, is there a possibility that the current task_cache_work() execution
> > > hasn't finished yet when the next scan window arrives? For instance, if the
> > > current task work is heavily delayed or preempted by unexpected interrupt,
> > > jiffies could advance past next_scan before the loop completes.
> > > If we move the `work->next = work;` to the very end of task_cache_work(),
> > > would that resolve this issue? By doing so, the existing `work->next == work`
> > > check in task_tick_cache() should fail and no new task work will be submitted.
> > >
> > > Please let me know if I'm missing something.
> > >
> >
> > There are two layers of protection: first a cheap timeout gate (time_before)
> > that skips scanning until the next period, and then a try_cmpxchg that atomically
> > picks a single winner among the threads that pass the timeout — this actually
> > guarantees only one scanner per mm at a time, no?
>
> What I am worried about is the following scenario:
>
> Thread A (CPU 0) Thread B (CPU 1)
> ================ ================
> task_cache_work()
> |
> +-> try_cmpxchg() == true
> | (Sets next_scan = now + 10)
> |
> +-> Enters Scanning Loop (jiffies = 100)
> | [ Delayed / Preempted ]
> | jiffies advances 100 -> 115.
> | Thread A STILL in the loop!
> | task_cache_work() (jiffies = 115)
>
> | |
> | +-> next_scan == 110 (pass timeout check)
>
> | |
> | +-> try_cmpxchg() == true
>
> | | (Sets next_scan = 115 + 10)
> | |
>
> In other words, concurrent execution of task_cache_work() from two adjacent periods can
> occur under extreme conditions; do we need to take this scenario into consideration?
>
> Perhaps we need an explicit state flag (e.g., using test_and_set_bit) to ensure
> that the previous task_cache_work() execution has fully completed before allowing
> a new thread to proceed, regardless of whether the next scan window has arrived.
> What do you think?

First, the EPOCH_PERIOD is fairly long (10 msec) so it is unlikely that Thread A
hasn't completed its work.

Second, suppose the above scenario happened, the two thread above
are serialized in the work on updating occupancy
and visited cpus under the cpus_read_lock when looping through the cpus
in task_cache_work().

So I think we should be okay without an explicit state flag.

That said, I think we should use mm->sc_stat.lock instead of cpus_read_lock
in task_cache_work()'s cpu loop. That will improve scalability.

Tim

>
> thanks,
> Gengkun
>
> >
> > thanks,
> > Chenyu