On Tue, May 17, 2011 at 4:52 PM, KAMEZAWA Hiroyuki <kamezawa.hiroyu@xxxxxxxxxxxxxx> wrote:
On Tue, 17 May 2011 15:25:51 -0700Hmm, If I'm a user, I want to know file-cache is well balanced or where Anon is
Ying Han <yinghan@xxxxxxxxxx> wrote:
> The new API exports numa_maps per-memcg basis. This is a piece of useful
> information where it exports per-memcg page distribution across real numa
> nodes.
>
> One of the usecase is evaluating application performance by combining this
> information w/ the cpu allocation to the application.
>
> The output of the memory.numastat tries to follow w/ simiar format of numa_maps
> like:
>
> <total pages> N0=<node 0 pages> N1=<node 1 pages> ...
>
> $ cat /dev/cgroup/memory/memory.numa_stat
> 292115 N0=36364 N1=166876 N2=39741 N3=49115
>
> Note: I noticed <total pages> is not equal to the sum of the rest of counters.
> I might need to change the way get that counter, comments are welcomed.
>
> Signed-off-by: Ying Han <yinghan@xxxxxxxxxx>
allocated from....Can't we have more precice one rather than total(anon+file) ?
So, I don't like this patch. Could you show total,anon,file at least ?
Ok, then this is really becoming per-memcg numa_maps. Before I go ahead posting the next version, this is something we are looking for:
total=<total pages> N0=<node 0 pages> N1=<node 1 pages> ...
anon=<total anon pages> N0=<node 0 pages> N1=<node 1 pages> ...
file=<total file pages> N0=<node 0 pages> N1=<node 1 pages> ...
please confirm?
thanks
--Ying
Thanks,
-Kame