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SubjectRe: [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@lge.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

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