On Sun, Aug 2, 2009 at 8:00 PM, Li Zefan<lizf@xxxxxxxxxxxxxx> 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.
>> +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.
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