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Message-ID: <4A778A49.6040302@cn.fujitsu.com>
Date:	Tue, 04 Aug 2009 09:09:29 +0800
From:	Li Zefan <lizf@...fujitsu.com>
To:	Benjamin Blum <bblum@...gle.com>
CC:	linux-kernel@...r.kernel.org,
	containers@...ts.linux-foundation.org, akpm@...ux-foundation.org,
	serue@...ibm.com, menage@...gle.com
Subject: Re: [PATCH 6/6] Makes procs file writable to move all threads by
 tgid 	at once

Benjamin Blum wrote:
> On Sun, Aug 2, 2009 at 8:00 PM, Li Zefan<lizf@...fujitsu.com> wrote:
>> Ben Blum wrote:
>>> +     }
>>> +     write_unlock(&css_set_lock);
>>> +
>>> +     /*
>>> +      * We just gained a reference on oldcg by taking it from the task. As
>> This comment is incorrect, the ref we just got has been dropped by
>> the above put_css_set(oldcg).
> 
> No, the idea is that even though we had a reference that we already
> dropped, we in effect "traded" the newcg to the task for its oldcg,
> giving it our reference on newcg and gaining its reference on oldcg. I
> believe the cgroup_mutex guarantees that it'll still be there when we
> do the trade - perhaps a BUG_ON(tsk->cgroups != oldcg) is wanted
> inside the second task_lock section there? At the very least, a
> clearer comment.
> 

Maybe my English sucks..

By "gained a reference", doesn't it mean get_css_set()? But this
put_css_set() is not against the get() just called.

And in fact the ref can be 0 before this put(), because task_exit
can drop the last ref, but put_css_set() will check this case,
so it's Ok.

>>> +static int css_set_check_fetched(struct cgroup *cgrp, struct task_struct *tsk,
>>> +                              struct css_set *cg,
>>> +                              struct list_head *newcg_list)
>>> +{
>>> +     struct css_set *newcg;
>>> +     struct cg_list_entry *cg_entry;
>>> +     struct cgroup_subsys_state *template[CGROUP_SUBSYS_COUNT];
>>> +     read_lock(&css_set_lock);
>>> +     newcg = find_existing_css_set(cg, cgrp, template);
>>> +     if (newcg)
>>> +             get_css_set(newcg);
>>> +     read_unlock(&css_set_lock);
>>> +     /* doesn't exist at all? */
>>> +     if (!newcg)
>>> +             return 1;
>> I think it's more intuitive to return 1 if found and 0 if not found.
> 
> I was sticking with the convention of nonzero return values indicating
> failure, as is used everywhere else in this context.
> 

Quoted from Documentation/CodingStyle:

...Such a value can be represented as an error-code integer
(-Exxx = failure, 0 = success) or a "succeeded" boolean (0 = failure,
non-zero = success).
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