Python 套接字服务器

发布于 2024-09-26 14:57:07 字数 134 浏览 14 评论 0原文

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

谢谢!

How can I call shutdown() in a SocketServer after receiving a certain message "exit"? As I know, the call to serve_forever() will block the server.

Thanks!

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圈圈圆圆圈圈 2024-10-03 14:57:07

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()

Use the source, Luke!

Excerpt from 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()
何处潇湘 2024-10-03 14:57:07

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

No the serve_forever is checking a flag on a regular basis (by default 0.5 sec). Calling shutdown will raise this flag and cause the serve_forever to end.

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