Re: [PATCH] sched/debug: avoid executing show_state and causing rcu stall warning

[Liu Song <liusong@linux.alibaba.com>]

> * Liu Song <liusong@linux.alibaba.com> wrote:
>
>>> * Liu Song <liusong@linux.alibaba.com> wrote:
>>>
>>>> From: Liu Song <liusong@linux.alibaba.com>
>>>>
>>>> If the number of CPUs is large, "sysrq_sched_debug_show" will execute for
>>>> a long time. Every time I execute "echo t > /proc/sysrq-trigger" on my
>>>> 128-core machine, the rcu stall warning will be triggered. Moreover,
>>>> sysrq_sched_debug_show does not need to be protected by rcu_read_lock,
>>>     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
>>>> and no rcu stall warning will appear after adjustment.
>>>>
>>> That doesn't mean it doesn't have to be protected by *any* lock - which
>>> your patch implements AFAICS.
>>>
>>> There's a couple of lines such as:
>>>
>>>           for_each_online_cpu(cpu) {
>> Hi,
>>
>> Here I refer to the implementation of "sysrq_timer_list_show", and I don't
>> see any lock.
>>
>> Maybe there is a problem with the implementation of "sysrq_timer_list_show".
> But we are talking about sysrq_sched_debug_show(), which your patch tries
> to relax the RCU locking of.

Hi,

I'm not sure for_each_online_cpu && print_cpu must need a lock to 
protect, so I refer to other codes

under kernel that reference the implementation. It looks like some 
places use "get_online_cpus" to prevent

cpu hotplug, but many places don't have obvious protection, so I'm also 
confused if protection is necessarily

required.


Thanks

>
> Thanks,
>
> 	Ingo