On Fri 20-05-16 22:41:27, Tetsuo Handa wrote:
> Michal Hocko wrote:
> > On Fri 20-05-16 20:51:56, Tetsuo Handa wrote:
> > [...]
> > > +static bool has_pending_victim(struct task_struct *p)
> > > +{
> > > + struct task_struct *t;
> > > + bool ret = false;
> > > +
> > > + rcu_read_lock();
> > > + for_each_thread(p, t) {
> > > + if (test_tsk_thread_flag(t, TIF_MEMDIE)) {
> > > + ret = true;
> > > + break;
> > > + }
> > > + }
> > > + rcu_read_unlock();
> > > + return ret;
> > > +}
> >
> > And so you do not speed up anything in the end because you have to
> > iterate all threads anyway yet you add quite some code on top. No I do
> > not like it. This is no longer a cleanup...
>
> I changed for_each_process_thread() to for_each_process(). This means
> O(num_threads^2) task_in_mem_cgroup()
oom_unkillable_task is called with NULL memcg so we do not call
task_in_mem_cgroup.
> and O(num_threads^2)
> has_intersects_mems_allowed() are replaced with O(num_threads)
I am really confused why has_intersects_mems_allowed has to iterate all
threads. Do we really allow different mempolicies for threads in the
same thread group?
> task_in_mem_cgroup() and O(num_threads) has_intersects_mems_allowed()
> at the cost of adding O(num_threads) has_pending_victim().
>
> I expect that O(num_threads) (task_in_mem_cgroup() + has_intersects_mems_allowed() +
> has_pending_victim()) is faster than O(num_threads^2) (task_in_mem_cgroup() +
> has_intersects_mems_allowed()) + O(num_threads) test_tsk_thread_flag().
I thought the whole point of the cleanup was to get rid of O(num_thread)
because num_threads >> num_processes in most workloads. It seems that
we are still not there because of has_intersects_mems_allowed but we
should rather address that than add another O(num_threads) sources.
> > [...]
> > > Note that "[PATCH v3] mm,oom: speed up select_bad_process() loop." temporarily
> > > broke oom_task_origin(task) case, for oom_select_bad_process() might select
> > > a task without mm because oom_badness() which checks for mm != NULL will not be
> > > called.
> >
> > How can we have oom_task_origin without mm? The flag is set explicitly
> > while doing swapoff resp. writing to ksm. We clear the flag before
> > exiting.
>
> What if oom_task_origin(task) received SIGKILL, but task was unable to run for
> very long period (e.g. 30 seconds) due to scheduling priority, and the OOM-reaper
> reaped task's mm within a second. Next round of OOM-killer selects the same task
> due to oom_task_origin(task) without doing MMF_OOM_REAPED test.
Which is actuall the intended behavior. The whole point of
oom_task_origin is to prevent from killing somebody because of
potentially memory hungry operation (e.g. swapoff) and rather kill the
initiator.
--
Michal Hocko
SUSE Labs
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