On Sat 06-02-16 14:54:24, Tetsuo Handa wrote:

Michal Hocko wrote:

quoted

quoted

But if we consider non system-wide OOM events, it is not very unlikely to hit
this race. This queue is useful for situations where memcg1 and memcg2 hit
memcg OOM at the same time and victim1 in memcg1 cannot terminate immediately.

This can happen of course but the likelihood is _much_ smaller without
the global OOM because the memcg OOM killer is invoked from a lockless
context so the oom context cannot block the victim to proceed.

Suppose mem_cgroup_out_of_memory() is called from a lockless context via
mem_cgroup_oom_synchronize() called from pagefault_out_of_memory(), that
"lockless" is talking about only current thread, doesn't it?

Yes and you need the OOM context to sit on the same lock as the victim
to form a deadlock. So while the victim might be blocked somewhere it is
much less likely it would be deadlocked.

Since oom_kill_process() sets TIF_MEMDIE on first mm!=NULL thread of a
victim process, it is possible that non-first mm!=NULL thread triggers
pagefault_out_of_memory() and first mm!=NULL thread gets TIF_MEMDIE,
isn't it?

I got lost here completely. Maybe it is your usage of thread terminology
again.
 

Then, where is the guarantee that victim1 (first mm!=NULL thread in memcg1
which got TIF_MEMDIE) is not waiting at down_read(&victim2->mm->mmap_sem)
when victim2 (first mm!=NULL thread in memcg2 which got TIF_MEMDIE) is
waiting at down_write(&victim2->mm->mmap_sem)

All threads/processes sharing the same mm are in fact in the same memory
cgroup. That is the reason we have owner in the task_struct

or both victim1 and victim2
are waiting on a lock somewhere in memory reclaim path (e.g.
mutex_lock(&inode->i_mutex))?

Such waiting has to make a forward progress at some point in time
because the lock itself cannot be deadlocked by the memcg OOM context.


-- 
Michal Hocko
SUSE Labs

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