How To Simulate Broadcast Message Passing Between Thread
I'm writing a small concurrent program using Python 3.6. I have a question: my program has a small Thread class (which simulates a thread); this class has within it 3 methods tha
Solution 1:
a good way to communicate between threads is using queue it is better to use a designated queue for every thread this is how you implement it in your code:
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()
output:
starting fun1
fun2 method saying hello
fun3 method saying hello
fun3 terminating
fun2 terminating
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