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

[Luo Gengkun]

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?

thanks,
Gengkun

> thanks,
> Chenyu