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SubjectRe: [PATCH 1/2] memcg: make oom_lock 0 and 1 based rather than coutner
On Fri 15-07-11 08:47:55, KAMEZAWA Hiroyuki wrote:
> On Thu, 14 Jul 2011 14:55:55 +0200
> Michal Hocko <mhocko@suse.cz> wrote:
>
> > On Thu 14-07-11 20:50:12, KAMEZAWA Hiroyuki wrote:
> > > On Thu, 14 Jul 2011 13:30:09 +0200
> > > Michal Hocko <mhocko@suse.cz> wrote:
> > [...]
> > > > static bool mem_cgroup_oom_lock(struct mem_cgroup *mem)
> > > > {
> > > > - int x, lock_count = 0;
> > > > - struct mem_cgroup *iter;
> > > > + int x, lock_count = -1;
> > > > + struct mem_cgroup *iter, *failed = NULL;
> > > > + bool cond = true;
> > > >
> > > > - for_each_mem_cgroup_tree(iter, mem) {
> > > > - x = atomic_inc_return(&iter->oom_lock);
> > > > - lock_count = max(x, lock_count);
> > > > + for_each_mem_cgroup_tree_cond(iter, mem, cond) {
> > > > + x = !!atomic_add_unless(&iter->oom_lock, 1, 1);
> > > > + if (lock_count == -1)
> > > > + lock_count = x;
> > > > + else if (lock_count != x) {
> > > > + /*
> > > > + * this subtree of our hierarchy is already locked
> > > > + * so we cannot give a lock.
> > > > + */
> > > > + lock_count = 0;
> > > > + failed = iter;
> > > > + cond = false;
> > > > + }
> > > > }
> > >
> > > Hm ? assuming B-C-D is locked and a new thread tries a lock on A-B-C-D-E.
> > > And for_each_mem_cgroup_tree will find groups in order of A->B->C->D->E.
> > > Before lock
> > > A 0
> > > B 1
> > > C 1
> > > D 1
> > > E 0
> > >
> > > After lock
> > > A 1
> > > B 1
> > > C 1
> > > D 1
> > > E 0
> > >
> > > here, failed = B, cond = false. Undo routine will unlock A.
> > > Hmm, seems to work in this case.
> > >
> > > But....A's oom_lock==0 and memcg_oom_wakeup() at el will not able to
> > > know "A" is in OOM. wakeup processes in A which is waiting for oom recover..
> >
> > Hohm, we need to have 2 different states. lock and mark_oom.
> > oom_recovert would check only the under_oom.
> >
>
> yes. I think so, too.
>
> > >
> > > Will this work ?
> >
> > No it won't because the rest of the world has no idea that A is
> > under_oom as well.
> >
> > > ==
> > > # cgcreate -g memory:A
> > > # cgset -r memory.use_hierarchy=1 A
> > > # cgset -r memory.oom_control=1 A
> > > # cgset -r memory.limit_in_bytes= 100M
> > > # cgset -r memory.memsw.limit_in_bytes= 100M
> > > # cgcreate -g memory:A/B
> > > # cgset -r memory.oom_control=1 A/B
> > > # cgset -r memory.limit_in_bytes=20M
> > > # cgset -r memory.memsw.limit_in_bytes=20M
> > >
> > > Assume malloc XXX is a program allocating XXX Megabytes of memory.
> > >
> > > # cgexec -g memory:A/B malloc 30 & #->this will be blocked by OOM of group B
> > > # cgexec -g memory:A malloc 80 & #->this will be blocked by OOM of group A
> > >
> > >
> > > Here, 2 procs are blocked by OOM. Here, relax A's limitation and clear OOM.
> > >
> > > # cgset -r memory.memsw.limit_in_bytes=300M A
> > > # cgset -r memory.limit_in_bytes=300M A
> > >
> > > malloc 80 will end.
> >
> > What about yet another approach? Very similar what you proposed, I
> > guess. Again not tested and needs some cleanup just to illustrate.
> > What do you think?
>
> Hmm, I think this will work. Please go ahead.
> Unfortunately, I'll not be able to make a quick response for a week
> because of other tasks. I'm sorry.
>
> Anyway,
> Reviewed-by: KAMEZAWA Hiroyuki <kamezawa.hiroyu@jp.fujitsu.com>

Thanks

>
> BTW, it's better to add "How-to-test" and the result in description.
> Some test similar to mine will show the result we want.

Agreed. I will hammer it today and repost with cleaned up description
with your example as well as mine that triggered the convoy behavior.

I hope you don't mine if I add you s-o-b to the patch as this was a
cooperative work.

> ==
> Make a hierarchy of memcg, which has 300MB memory+swap limit.
>
> %cgcreate -g memory:A
> %cgset -r memory.limit_in_bytes=300M A
> %cgset -r memory.memsw.limit_in_bytes=300M A
>
> Then, running folloing program under A.
> %cgexec -g memory:A ./fork
> ==
> int main(int argc, char *argv[])
> {
> int i;
> int status;
>
> for (i = 0; i < 5000; i++) {
> if (fork() == 0) {
> char *c;
> c = malloc(1024*1024);
> memset(c, 0, 1024*1024);
> sleep(20);
> fprintf(stderr, "[%d]\n", i);
> exit(0);
> }
> printf("%d\n", i);
> waitpid(-1, &status, WNOHANG);
> }
> while (1) {
> int ret;
> ret = waitpid(-1, &status, WNOHANG);
>
> if (ret == -1)
> break;
> if (!ret)
> sleep(1);
> }
> return 0;
> }
> ==
>
> Thank you for your effort.
> -Kame
>
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--
Michal Hocko
SUSE Labs
SUSE LINUX s.r.o.
Lihovarska 1060/12
190 00 Praha 9
Czech Republic


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