Python 多线程：错误<var_name>已在变量增加上定义

Question

我正在使用Python

3.12.7

。 我有多个服务器，它们在网络环境中以多线程方式运行。 我想增加

self.msg_id

函数中的变量

handle_message

。 其他一切都运行良好。但是，当我执行

self.msg_id += 1

行时，我收到错误：

[ERROR] Attribute msg_id already defined. Plugin conflict?

从变量中读取值效果很好。 请注意，这是大学的一项任务，因此在代码中存在一些奇怪的注释。

Server类的init

class Server(Bottle):

    def __init__(self, ID, IP, server_list):
        super(Server, self).__init__()
        self.id = int(ID)
        self.ip = str(IP)
        self.server_list = server_list
        self.msg_id = 1

        self.status = {
            "crashed": False,
            "notes": "",
            "num_entries": 0,  # we use this to generate ids for the entries
        }

        self.lock = threading.RLock()  # use reentry lock for the server
        self.msg_lock = threading.RLock()

        # Handle CORS
        self.route("/<:re:.*>", method="OPTIONS", callback=self.add_cors_headers)
        self.add_hook("after_request", self.add_cors_headers)

        # Those two http calls simulate crashes, i.e., unavailability of the server
        self.post("/crash", callback=self.crash_request)
        self.post("/recover", callback=self.recover_request)
        self.get("/status", callback=self.status_request)

        # Define REST URIs for the frontend (note that we define multiple update and delete routes right now)
        self.post("/entries", callback=self.create_entry_request)
        self.get("/entries", callback=self.list_entries_request)
        self.put("/entries/<entry_id>", callback=self.update_entry_request)
        self.delete("/entries/<entry_id>", callback=self.delete_entry_request)

        # REST URIs for the later algorithms
        self.post("/message", callback=self.message_request)

        self.board = Board()

相关功能

    def handle_message(self, message):
        # Note that you might need to use the lock
        print("Received message: ", message)

        if "type" in message:
            type = message["type"]

            if type == "add_entry":

                assert (
                    self.id == 0
                )  # ID 0 is coordinator only this server should receive add_entry messages right now
                with self.msg_lock:
                    entry_value = message["entry_value"]

                    # We can safely propagate here since we always have a single frontend client, right? So no need to lock, right? Right?!
                    for other in self.server_list:
                        # TODO: Send message to other servers concurrently?
                        self.send_message(
                            other,
                            {
                                "type": "propagate",
                                "entry_value": entry_value,
                                "origin_id": self.id,
                                "message_id": self.msg_id,
                            },
                        )
                    self.msg_id += 1

            elif type == "propagate":
                entry_value = message["entry_value"]
                msg_id = message["message_id"]
                # Let's hope this is from the coordinator
                with self.lock:
                    if msg_id in self.board.indexed_entries:
                        return {}
                    self.status["num_entries"] += 1
                    entry = Entry(msg_id, entry_value)
                    self.board.add_entry(entry)
            else:
                print("Received weird message?")

        return {}

主要代码执行

# Sleep a bit to allow logging to be attached
time.sleep(2)

# the server_list contains all server ips of the distributed blackboard
server_list = os.getenv("SERVER_LIST").split(",")
own_id = int(os.getenv("SERVER_ID"))
own_ip = server_list[own_id]

server = Server(own_id, own_ip, server_list)

NUM_THREADS = 10
print("#### Starting Server {} with {} threads".format(str(own_id), NUM_THREADS))
httpserver.serve(
    server,
    host="0.0.0.0",
    port=80,
    threadpool_workers=NUM_THREADS,
    threadpool_options={"spawn_if_under": NUM_THREADS},
)

Answer 1

我没有想法我没有想法 我没有想法

我没有想法 我没有想法