Re: [PATCH 3/5] debugobjects: Don't start fill if there are remaining nodes locally

[Thomas Gleixner]

On Mon, Sep 02 2024 at 22:05, Zhen Lei wrote:

> If the conditions for starting fill are met, it means that all cores that
> call fill() later are blocked until the first core completes the fill
> operation. But obviously, for a core that has free nodes locally, it does
> not need to be blocked. This is good in stress situations.

Sure it's good, but is it correct? You need to explain why this can't
cause a pool depletion. The pool is filled opportunistically.

Aside of that the lock contention in fill_pool() is minimal. The heavy
lifting is the allocation of objects.

> diff --git a/lib/debugobjects.c b/lib/debugobjects.c
> index aba3e62a4315f51..fc8224f9f0eda8f 100644
> --- a/lib/debugobjects.c
> +++ b/lib/debugobjects.c
> @@ -130,10 +130,15 @@ static void fill_pool(void)
>  	gfp_t gfp = __GFP_HIGH | __GFP_NOWARN;
>  	struct debug_obj *obj;
>  	unsigned long flags;
> +	struct debug_percpu_free *percpu_pool;

Please keep variables in reverse fir tree order.

https://www.kernel.org/doc/html/latest/process/maintainer-tip.html#variable-declarations
  
>  	if (likely(READ_ONCE(obj_pool_free) >= debug_objects_pool_min_level))
>  		return;
>  
> +	percpu_pool = this_cpu_ptr(&percpu_obj_pool);

You don't need the pointer

> +	if (likely(obj_cache) && percpu_pool->obj_free > 0)

	if (likely(obj_cache) && this_cpu_read(percpu_pool.obj_free) > 0)

This lacks a comment explaining the rationale of this check.

Thanks,

        tglx