Re: [PATCH v5 3/5] workqueue: Make too_many_workers() return the worker excess

[Tejun Heo <tj@kernel.org>]

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