python3 asyncio等待回调-Java 学习之路

我现在成功了 . 我的想法是为行动的 done_callback 添加一个asyncio未来，但未来似乎永远不会完成 .

代码在这里：

def _generate_action_executor(self, action):

    async def run_action(goal):
        action_done = asyncio.Future()

        def done_callback(goal_status, result, future):
            status = ActionLibGoalStatus(goal_status)
            print('Action Done: {}'.format(status))
            future.set_result(result)

        action.send_goal(goal,
                         lambda x, y: done_callback(x,
                                                    y,
                                                    action_done))
        try:
            result = await action_done
            #problem future never done
        except asyncio.CancelledError as exc:
            action.cancel()
            raise exc

        return result

    return run_action

async def do_some_other_stuff(action):
    //do stuff
    my_goal = MyActionRequest('just do it')

    run_action = self._generate_action_executor(action)

    response = await run_action(my_goal)

    return response


if __name__ == "__main__":
    action = actionlib.SimpleActionClient('my_action',
                                           MyAction)

    try:
        loop = asyncio.get_event_loop()

        loop.run_until_complete(do_some_other_stuff(action))
    finally:
        loop.close()

2 回答

请记住，asyncio意味着在单个线程中运行 .

如果程序需要与其他线程交互，则必须使用其中一个专用函数：

这是一个简化的例子：

async def run_with_non_asyncio_lib(action, arg):
    future = asyncio.Future()
    loop = asyncio.get_event_loop()

    def callback(*args):
        loop.call_soon_threasafe(future.set_result, args)

    non_asyncio_lib.register_callback(action, arg, callback)
    callback_args = await future
    return process(*callback_args)

或者，loop.run_in_executor提供了一种通过在自己的线程中运行给定函数来与非asyncio库交互的方法：

async def run_with_non_asyncio_lib(action, arg):
    loop = asyncio.get_event_loop()
    future = loop.run_in_executor(None, non_asyncio_lib.run, action, arg)
    result = await future
    return result

回复于 2024-04-26T05:06:52+08:00

随着文森特的想法，

我实际上找到了解决问题的方法：

def _generate_action_executor(action):
    async def run_action(goal):
        loop = asyncio.get_event_loop()
        action_done = loop.create_future()

        def done_callback(goal_status, result, future, loop):
            status = ActionLibGoalStatus(goal_status)
            print('Action Done: {}'.format(status))
            loop.call_soon_threadsafe(future.set_result(result))

        action.send_goal(goal,partial(done_callback, future=action_done, loop=loop))
        try:
            await action_done
        except asyncio.CancelledError as exc:
            action.cancel()
            raise exc

        return action_done.result()

    return run_action

如果有人知道如何以更智能的方式实现它，请与使用者分享这些知识 .

最好的曼努埃尔

回复于 2024-04-26T05:06:52+08:00

python3 asyncio等待回调

2 回答

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