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SubjectRe: [PATCH 1/2] Adds a read-only "procs" file similar to "tasks" that shows only unique tgids
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On Thu, Jul 2, 2009 at 10:54 PM, KAMEZAWA
Hiroyuki<kamezawa.hiroyu@jp.fujitsu.com> wrote:
>
> Why we can't do what readdir(/proc) does ? I'm sorry I misunderstand.
> Following is an easy example.
>
>
> 0. at open, inilialize f_pos to 0. f_pos is used as "pid"
>   remember "css_set with hole" as template in f_private?(or somewhere) at open
>   ...like this.
> --
>   struct cgroupfs_root *root = cgrp->root;
>   struct cgroup *template = kzalloc(sizeof(void*) * CGROUP_SUBSYS_COUNT);
>
>   for (i = 0; i < CGROUP_SUBSYS_COUNT; i++)
>        if (root->subsys_bits & (1UL << i))
>                template[i] =  cgrp->subsys[i];
> --
>
>
> 1. at read(), find task_struct of "pid" in f_pos.
> 2. look up task_struct of "pid" and compare with f_private
> --
>   struct cgroup *template = f_private;
>
>   for (i = 0; i < CGROUP_SUBSYS_COUNT; i++) {
>        if (!template[i])
>                contiue;
>        if (template[i] != task_subsys_state(task, i))
>                break;
>   }
>   if (i == CGROUP_SUBSYS_COUNT)
>        print task;

The problem with this is that the time taken to scan a single cgroup
is linear in the total number of threads in the system, so if you have
a lot of threads and a lot of cgroups (even if most of the threads are
concentrated in a single cgroup) the time taken to scan all the tasks
files in O(N^2) in the number of threads in the system. The current
scheme is linear in the number of threads in a cgroup, so looking at
all cgroups is linear in the number of threads in the system. (This
O(N^2) problem is something that we've actually observed as an
overhead on some busy systems at Google).

Paul
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