Date: Mon, 21 Nov 2016 13:23:44 +0100
From: Oleg Nesterov <oleg@redhat.com>
To: Davidlohr Bueso <dave@stgolabs.net>
Cc: mingo@kernel.org, peterz@infradead.org, john.stultz@linaro.org,
        dimitrysh@google.com, linux-kernel@vger.kernel.org,
        Davidlohr Bueso <dbueso@suse.de>
Subject: Re: [PATCH 3/3] locking/percpu-rwsem: Avoid unnecessary writer
        wakeups
Message-ID: <20161121122343.GA635@redhat.com>
References: <1479495277-9075-1-git-send-email-dave@stgolabs.net> <1479495277-9075-4-git-send-email-dave@stgolabs.net>
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On 11/18, Davidlohr Bueso wrote:
>
> +static bool __readers_active_check(struct percpu_rw_semaphore *sem)
> +{
> +	return !(per_cpu_sum(*sem->read_count) !=0);
> +}

Hmm,

	return per_cpu_sum(*sem->read_count) == 0;

looks more clear, but this is minor,

>  int __percpu_init_rwsem(struct percpu_rw_semaphore *sem,
>  			const char *name, struct lock_class_key *rwsem_key)
>  {
> @@ -103,41 +141,11 @@ void __percpu_up_read(struct percpu_rw_semaphore *sem)
>  	__this_cpu_dec(*sem->read_count);
>  
>  	/* Prod writer to recheck readers_active */
> -	swake_up(&sem->writer);
> +	if (__readers_active_check(sem))
> +		swake_up(&sem->writer);

Suppose we have 2 active readers which call __percpu_up_read() at the same
time and the pending writer sleeps.

What guarantees that one of these readers will observe per_cpu_sum() == 0 ?
They both can read the old value of the remote per-cpu counter, no?

Oleg.