6

收到特定消息“退出”后shutdown()如何调用?SocketServer据我所知,调用serve_forever()将阻塞服务器。

谢谢!

4

2 回答 2

6

使用来源,卢克!

摘自 SocketServer.py:

   def serve_forever(self, poll_interval=0.5):
        """Handle one request at a time until shutdown.

        Polls for shutdown every poll_interval seconds. Ignores
        self.timeout. If you need to do periodic tasks, do them in
        another thread.
        """
        self.__is_shut_down.clear()
        try:
            while not self.__shutdown_request:
                # XXX: Consider using another file descriptor or
                # connecting to the socket to wake this up instead of
                # polling. Polling reduces our responsiveness to a
                # shutdown request and wastes cpu at all other times.
                r, w, e = select.select([self], [], [], poll_interval)
                if self in r:
                    self._handle_request_noblock()
        finally:
            self.__shutdown_request = False
            self.__is_shut_down.set()

    def shutdown(self):
        """Stops the serve_forever loop.

        Blocks until the loop has finished. This must be called while
        serve_forever() is running in another thread, or it will
        deadlock.
        """
        self.__shutdown_request = True
        self.__is_shut_down.wait()
于 2010-10-05T14:37:56.950 回答
4

不,serve_forever是定期检查标志(默认为 0.5 秒)。调用 shutdown 将引发此标志并导致 serve_forever 结束。

于 2010-10-05T12:20:31.653 回答