On Thu, Jul 2, 2009 at 10:54 PM, KAMEZAWA
Hiroyuki<kamezawa.hiroyu@xxxxxxxxxxxxxx> 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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