Re: [PATCH 1/3] sched: don't repeat the initialization in sched_init()
From: Michael Wang
Date: Tue Jun 04 2013 - 03:23:42 EST
Hi, Paul
On 06/04/2013 02:52 PM, Paul Turner wrote:
> On Mon, Jun 3, 2013 at 11:23 PM, Michael Wang
[snip]
>
> This comment has become unglued from what it's supposed to be attached
> to (it's tied to root_task_group.shares & init_tg_cfs_entry, not
> init_cfs_bandwidth).
Thanks for your review and notify :)
What about put the comment with init_tg_cfs_entry()?
'root_task_group.shares' may not needed to be covered under the comment,
after all, it won't have any peers to flaunt it's share...
Regards,
Michael Wang
>
>> + init_cfs_bandwidth(&root_task_group.cfs_bandwidth);
>> +#endif
>> +
>> for_each_possible_cpu(i) {
>> struct rq *rq;
>>
>> @@ -6966,28 +6991,7 @@ void __init sched_init(void)
>> init_cfs_rq(&rq->cfs);
>> init_rt_rq(&rq->rt, rq);
>> #ifdef CONFIG_FAIR_GROUP_SCHED
>> - root_task_group.shares = ROOT_TASK_GROUP_LOAD;
>> INIT_LIST_HEAD(&rq->leaf_cfs_rq_list);
>> - /*
>> - * How much cpu bandwidth does root_task_group get?
>> - *
>> - * In case of task-groups formed thr' the cgroup filesystem, it
>> - * gets 100% of the cpu resources in the system. This overall
>> - * system cpu resource is divided among the tasks of
>> - * root_task_group and its child task-groups in a fair manner,
>> - * based on each entity's (task or task-group's) weight
>> - * (se->load.weight).
>> - *
>> - * In other words, if root_task_group has 10 tasks of weight
>> - * 1024) and two child groups A0 and A1 (of weight 1024 each),
>> - * then A0's share of the cpu resource is:
>> - *
>> - * A0's bandwidth = 1024 / (10*1024 + 1024 + 1024) = 8.33%
>> - *
>> - * We achieve this by letting root_task_group's tasks sit
>> - * directly in rq->cfs (i.e root_task_group->se[] = NULL).
>> - */
>> - init_cfs_bandwidth(&root_task_group.cfs_bandwidth);
>> init_tg_cfs_entry(&root_task_group, &rq->cfs, NULL, i, NULL);
>> #endif /* CONFIG_FAIR_GROUP_SCHED */
>>
>> --
>> 1.7.4.1
>>
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