标签：为什么 its pwa _each events accept one 测试 next
这里的问题就是，在第5步的时候，会有多少个线程被唤醒而从epoll_wait()调用返回？答案是不一定，可能只有一个，也可能有部分，也可能是全部。当然在多个线程都唤醒的情况下，只会有一个线程accept()调用会成功。
为何如此？从内核代码分析，原因如下：
static void __wake_up_common(wait_queue_head_t *q, unsigned int mode,
int nr_exclusive, int wake_flags, void *key)
{
wait_queue_t *curr, *next;
list_for_each_entry_safe(curr, next, &q->task_list, task_list) {
unsigned flags = curr->flags;
if (curr->func(curr, mode, wake_flags, key) &&
(flags & WQ_FLAG_EXCLUSIVE) && !--nr_exclusive)
break;
}
}
因为__wake_up_common()的调用是从wake_up_locked()开始的，__wake_up_common的各个参数值为：
if (!list_empty(&ep->rdllist)) {
/*
* Wake up (if active) both the eventpoll wait list and
* the ->poll() wait list (delayed after we release the lock).
*/
if (waitqueue_active(&ep->wq))
wake_up_locked(&ep->wq);
if (waitqueue_active(&ep->poll_wait))
pwake++;
}
if (epi->event.events & EPOLLONESHOT)
epi->event.events &= EP_PRIVATE_BITS;
else if (!(epi->event.events & EPOLLET)) {
/*
* If this file has been added with Level
* Trigger mode, we need to insert back inside
* the ready list, so that the next call to
* epoll_wait() will check again the events
* availability. At this point, no one can insert
* into ep->rdllist besides us. The epoll_ctl()
* callers are locked out by
* ep_scan_ready_list() holding "mtx" and the
* poll callback will queue them in ep->ovflist.
*/
list_add_tail(&epi->rdllink, &ep->rdllist);
ep_pm_stay_awake(epi);
}
标签：为什么 its pwa _each events accept one 测试 next
原文地址：http://www.cnblogs.com/sduzh/p/6810469.html
