linux-kernel - Re: [PATCH] sched: Wake up all non-exclusive waiters in __wake_up

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Date:   Fri, 10 Mar 2017 16:00:27 +0900
From:   Byungchul Park <byungchul.park@....com>
To:     <peterz@...radead.org>, <mingo@...nel.org>
CC:     <linux-kernel@...r.kernel.org>, <kernel-team@....com>
Subject: Re: [PATCH] sched: Wake up all non-exclusive waiters in
 __wake_up_common()

On Wed, Mar 08, 2017 at 09:21:52AM +0900, Byungchul Park wrote:
> __wake_up_common() should wake up all non-exclusive waiters and
> exclusive waiters as many as nr_exclusive, but currently it does not.
> 
> Consider a wait queue like the following for example:
> 
>    A(exclusive) -> B(non-exclusive) -> C(non-exclusive)
> 
> Current code will wake up only A when nr_exclusive = 1, but has to wake
> up A, B and C. Make it do as we expect.

I'm wondering if I was wrong. Am I wrong?

> 
> Signed-off-by: Byungchul Park <byungchul.park@....com>
> ---
>  kernel/sched/wait.c | 15 +++++++++++++--
>  1 file changed, 13 insertions(+), 2 deletions(-)
> 
> diff --git a/kernel/sched/wait.c b/kernel/sched/wait.c
> index 9453efe..0ea1083 100644
> --- a/kernel/sched/wait.c
> +++ b/kernel/sched/wait.c
> @@ -67,12 +67,23 @@ static void __wake_up_common(wait_queue_head_t *q, unsigned int mode,
>  {
>  	wait_queue_t *curr, *next;
>  
> +	/*
> +	 * We use nr_exclusive = 0 to wake up all no matter whether
> +	 * WQ_FLAG_EXCLUSIVE is set. However, we have to distinguish
> +	 * between the case and having finished all exclusive wake-up.
> +	 * So make nr_exclusive non-zero in advance in the former case.
> +	 */
> +	nr_exclusive = nr_exclusive ?: -1;
> +
>  	list_for_each_entry_safe(curr, next, &q->task_list, task_list) {
>  		unsigned flags = curr->flags;
>  
> +		if ((flags & WQ_FLAG_EXCLUSIVE) && !nr_exclusive)
> +			continue;
> +
>  		if (curr->func(curr, mode, wake_flags, key) &&
> -				(flags & WQ_FLAG_EXCLUSIVE) && !--nr_exclusive)
> -			break;
> +		    (flags & WQ_FLAG_EXCLUSIVE))
> +			nr_exclusive--;
>  	}
>  }
>  
> -- 
> 1.9.1