Messages in this thread | | | Date | Wed, 3 Aug 2022 17:25:49 +0800 | Subject | Re: [PATCH] sched/debug: avoid executing show_state and causing rcu stall warning | From | Liu Song <> |
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> * 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
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