Question

我正在使用Python 3.6编写一个小型并发程序。我有个问题：我的程序有一个小的Thread类（可模拟线程）；此类具有3个作为子线程执行的方法：

class myThread(Thread):
  def __init__(self, identifier):
    super(myThread, self).__init__() 

  def fun1(self):
    # broadcasts messages

  def fun2(self):
    # event that occurs when a message arrives
    # do something

  def fun3(self):
    # event that occurs when a message arrives
    # do something

  def run(self):
    t1 = Thread(target = self.fun1)
    t2 = Thread(target = self.fun2)
    t3 = Thread(target = self.fun3)
    t1.start()
    t2.start()
    t3.start()

如您所见，fun1()发送其他2个线程必须接收的广播消息（他发送对象）。如何在Python中轻松实现此功能？我已经看到最简单的方法是使用Queue，但是我有一些疑问...我应该把这个队列放在哪里？常规方法如何在不清空此队列的情况下使用提交的对象（因为“广播”对象必须由其他方法使用）？每次将新对象添加到队列中（就像是事件一样）时，该方法如何执行其主体？

非常感谢大家！

Answer 1

在线程之间进行通信的一种好方法是使用队列最好为每个线程使用指定的队列这是在代码中实现它的方式：

from queue import Queue
from threading import Thread
import time

# define some queues
fun2_q = Queue()
fun3_q = Queue()

class myThread(Thread):
    def __init__(self, identifier):
        super(myThread, self).__init__() 

    def fun1(self):
        print('starting fun1')

        # broadcasts messages
        fun2_q.put('say something')
        fun3_q.put('say something')

        fun2_q.put('quit')
        fun3_q.put('quit')



    def fun2(self):
        # event that occurs when a message arrives
        # as a listener we should use infinite loop to monitor messages 
        # we will use non blocking way to read the queue using "if", also we can use fun2_q.get_nowait()
        # instead of "if fun2_q.qsize() > 0:" statement

        while True:
            if fun2_q.qsize() > 0:
                msg = fun2_q.get()
                if msg == 'say something':
                    print('fun2 method saying hello')
                elif msg == 'quit':
                    break  # quit thread

            # do other stuff below if no messages coming

            time.sleep(0.1)  # to stop while loop from abusing processor

        print('fun2 terminating')


    def fun3(self):
        # event that occurs when a message arrives
        # we will use a blocking way to read the queue
        while True:

            msg = fun3_q.get() # it will block here waiting for a message to come
            if msg == 'say something':
                print('fun3 method saying hello')
            elif msg == 'quit':
                    break  # quit thread

            # can't do other stuff below if no messages coming, the loop will stuck waiting new message

            # time.sleep(0.1)  # no need for it since the loop will wait anyway

        print('fun3 terminating')

    def run(self):
        t1 = Thread(target = self.fun1)
        t2 = Thread(target = self.fun2)
        t3 = Thread(target = self.fun3)
        t1.start()
        t2.start()
        t3.start()

my_thread = myThread(1)
my_thread.run()

输出：

starting fun1
fun2 method saying hello
fun3 method saying hello
fun3 terminating
fun2 terminating

Python：如何模拟Thread之间的广播消息传递？

1 个答案: