[systemd-devel] Where did DefaultControllers= option go?

Umut Tezduyar umut at tezduyar.com
Tue Oct 29 18:45:40 CET 2013


On Tue, Oct 29, 2013 at 5:52 PM, Kay Sievers <kay at vrfy.org> wrote:
> On Tue, Oct 29, 2013 at 5:39 PM, Thomas Bächler <thomas at archlinux.org> wrote:
>> Am 29.10.2013 17:21, schrieb Colin Guthrie:
>
>>> 'Twas brillig, and Umut Tezduyar at 29/10/13 15:17 did gyre and gimble:
>>>> I have noticed DefaultControllers= option is no longer in system.conf
>>>> file. Has it been moved to somewhere else or are all controllers
>>>> default controllers by default?
>>>
>>> See the various posts on this list and in blogs regarding the cgroups
>>> rework. In particular see the mails with the subject "[HEADSUP] cgroup..."
>>>
>>> See also:
>>> http://www.freedesktop.org/wiki/Software/systemd/ControlGroupInterface/
>>
>> None of this explains why systemd no longer applies certain controllers
>> by default. Previously, systemd would attach cpu controllers to each
>> service by default. Now, it only groups your processes in the systemd
>> tree, but does not touch any cgroup controllers.
>>
>> The new default behaviour (or lack thereof) seems like a step back to me.
>
> It's a property of a unit now which cgroup controllers get attached:
>   http://www.freedesktop.org/software/systemd/man/systemd.resource-control.html
>
> The global mirroring across trees makes no sense in the future. The
> several independent trees will go away in the kernel next year, and
> then systemd would not know what to do with an instruction like
> DefaultControllers.

I don't disagree that mirroring will not make sense in the future.
Though, without global mirroring on cpuacct I don't understand how
systemd-cgtop is expected to give cpu accounting information of
individual services. Also without global mirroring on cpu control
group, how would CPUShares=weight even work on any service. I am just
confused.

Umut

>
> All implementation details of kernel cgroups are no longer exposed in
> systemd, which is a huge step forward, not back.
>
> Kay
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