字典更新函数内部但不在函数外部

Question

我有一个线程 (new_client_thread) 可以更新字典并打印它。 当我试图再次打印字典但从主线程时，字典没有改变。当线程 (new_client_thread) 再次更改字典时，所有更改也都保存了，但仅在此线程内而不在主线程内。

这是我的 server.py

from threading import Thread
from threading import Lock


def add_client_to_dict(my_id, connect):
    my_socket_list[int(my_id)] = connect
    print my_socket_list


def print_my_socket_list():
    print my_socket_list
    if len(my_socket_list) == 0:
        print "dictionary is empty"
        return None
    for item in my_socket_list:
        print item


def connection_failed(connecting):
    print "deleting"
    for key in my_socket_list.keys():
        if my_socket_list[key] == connecting:
            my_socket_list.pop(key)
            print "deleted"
            connecting.close()
            break


def new_clients():
    from Class import Socket_control
    while True:
        connect = Socket_control.open_socket()
        thread_listener = Thread(target=Socket_control.recv_msg, args=(connect,))
        thread_listener.start()


def menu():
    choice = 1
    while choice != 0:
        choice = raw_input("What would you like to do?\n[0] exit.\n[1] print list of computers.")
        if choice == "1":
            print_my_socket_list()
            if len(my_socket_list) == 0:
                continue
            computer = raw_input("Choose a computer :")
            msg = raw_input("Send message: ")
            my_socket_list[computer].send_message("MSG:" + msg)


def main():

    new_client_thread = Thread(target=new_clients)
    new_client_thread.start()
    menu()

my_socket_list = {}
my_socket_list["0"] = None

if __name__ == "__main__":
    main()

这是我的套接字控件

import socket
import datetime
import server
from mutex import mutex
def open_socket():
    global connect
    try:
        sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        sock.bind(("0.0.0.0", 8200))
    except socket.error:
        print "Socket Error"

    sock.listen(1)
    connect, addr = sock.accept()
    return connect


def recv_msg(connecting):
    while True:
        try:
            data = connecting.recv(1024)
            if len(data) == 0:
                server.connection_failed(connect)
                return None
            print "recv " + data
            data = data.decode('utf-8')
            if data[:4] == "MYID":
                handle_client_id(data[4:])
        except:
            server.connection_failed(connecting)
            print "Recv Error"
            return None




def handle_client_id(data):
    my_id = data
    if my_id == "0":
        my_id = get_last_number()+1
        send_message("MYID" + str(my_id))
    file_writing_new_client(my_id)
    server.add_client_to_dict(my_id,connect)


def send_message(data):
    try:
        print "sent " + data
        connect.send(data)
    except:
        print "Sent Error"
        server.connection_failed(connect)


def get_last_number():
    with open("jennifer_clients_list.txt") as my_file:
        my_file.seek(0)
        first_char = my_file.read(1)
        my_file.seek(0)
        if not first_char:
            return 0
        else:
            lines = my_file.readlines()
            print "text: "
            print lines
            data = lines[-1].split(" ")[0]
            return int(data)

Answer 1

在 CPython 中，您应该可以轻松地在一个线程中更新 dict，并在另一个线程中访问它（参见 Using a global dictionary with threads in Python）。

问题的近因是您试图使用错误的键来引用字典。当您将 ID 存储在字典中时，您首先将其转换为 int，但是当您从主线程访问它时，您使用的是用户提供的 str 输入；即int('88') 与'88' 的键不相同。

如果换行：

my_socket_list[computer].send_message("MSG:" + msg)

到：

my_socket_list[int(computer)].send_message("MSG:" + msg)

您将进入下一个错误（即socket 对象没有send_message 方法，但如果您将其更改为.send，它将起作用）。

回到最初的主题，让您的代码依赖于诸如 CPython 的全局解释器锁之类的实现细节通常是个坏主意，因此我建议将您的字典访问与Lock 同步。