Messages in this thread | | | Date | Tue, 22 Nov 2022 10:17:39 -1000 | From | Tejun Heo <> | Subject | Re: [PATCH v5 3/5] workqueue: Make too_many_workers() return the worker excess |
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Hello,
On Tue, Nov 22, 2022 at 07:29:35PM +0000, Valentin Schneider wrote: ... > The function currently returns true when > (nr_idle - 2) * MAX_IDLE_WORKERS_RATIO >= nr_busy > thus, the desired number of idle workers is expressed by > (nr_idle - 2) * MAX_IDLE_WORKERS_RATIO == nr_busy - 1 > IOW > nr_idle == ((nr_busy - 1) / MAX_IDLE_WORKERS_RATIO) + 2 > +/* How many idle workers should we get rid of, if any? */ > +static unsigned int worker_cull_count(struct worker_pool *pool)
Can we name it nr_workers_to_cull()?
> { > bool managing = pool->flags & POOL_MANAGER_ACTIVE; > int nr_idle = pool->nr_idle + managing; /* manager is considered idle */ > int nr_busy = pool->nr_workers - nr_idle; > > - return nr_idle > 2 && (nr_idle - 2) * MAX_IDLE_WORKERS_RATIO >= nr_busy; > + lockdep_assert_held(&pool->lock); > + > + /* > + * We keep at least 2 spare idle workers, but overall aim to keep at > + * most (1 / MAX_IDLE_WORKERS_RATIO) workers idle. > + */ > + return max(0, nr_idle - 2 - ((nr_busy - 1) / MAX_IDLE_WORKERS_RATIO));
I think we can do away with the subtraction on nr_busy. I don't think it'd make any material difference, so maybe we can do:
return max(0, nr_idle - 2 - nr_busy / MAX_IDLE_WORKERS_RATIO);
Thanks.
-- tejun
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