How to multithread using threading

Question

The queueLock.acquire() line hangs a code that tries to take advantage of threading and Queue modules with following output:

run() on <newThread(Thread-1, started 4344102912)>

run() on <newThread(Thread-2, started 4348309504)>

...working delay: 1

Why?

import threading, thread, Queue, time

class newThread (threading.Thread):
    def __init__(self, delay=0, workQueue=None):
        self.delay=delay
        self.workQueue=workQueue
        threading.Thread.__init__(self)

    def run(self):
        print '\nrun() on %s'%self
        print_time(self.delay, self.workQueue)
        print '\nrun() exit %s'%self

def print_time(delay=None, workQueue=None):
    def onExit():
        if not workQueue.empty():
            data = workQueue.get()
            queueLock.release()
        else:
            queueLock.release()

    counter=0
    while True:
        queueLock.acquire()
        time.sleep(delay)
        print '\t...working delay: %s'%delay   

        if counter>5:
            onExit()

        counter=counter + 1



queueLock = threading.Lock()
workQueue = Queue.Queue(10)
threads = []

thread1 = newThread(delay=1, workQueue=workQueue)
thread1.start()
threads.append(thread1)

thread2 = newThread(delay=2, workQueue=workQueue)
thread2.start()
threads.append(thread2)

queueLock.acquire()
print '1. workQueue.empty():',workQueue.empty()
workQueue.put('One')
print '2. workQueue.empty():',workQueue.empty()
workQueue.put('Two')

queueLock.release()

#Wait for queue to empty
while not workQueue.empty():
    print '...passing while not workQueue.empty()'
    pass

for thread in threads:
    thread.join()

print '...completed'

Error message and stack trace would be very helpful. It would also help to show what the output is (i.e. which of your print lines actually executed) — Oliver Dain
– Oliver Dain, Commented Oct 11, 2015 at 2:20
Are you running standard C-based Python? If so, you can't do multithreading in the traditional sense. Look at the GIL — T.D. Smith
– T.D. Smith, Commented Oct 11, 2015 at 2:44
Some suggestions for your code: You don't have to derive from Thread, you can simply run any function with a plain Thread instance. It would also help if you didn't specify default parameters for functions where they may no sense at all, like passing None as queue to print_time(). Also, upgrade to Python 3! — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Oct 11, 2015 at 8:15
Please post some example on how to run any function with a plain Thread instance. — alphanumeric
– alphanumeric, Commented Oct 14, 2015 at 3:58

jfs · Accepted Answer · 2015-10-11 04:30:40Z

2

queueLock.acquire() blocks until queueLock.release() is called if queueLock is already acquired. counter > 5 never happens because even if queueLock is available on the first iteration of the loop; nothing releases it on the 2nd iteration.

Queue() has its own locks. Drop queueLock completely.

answered Oct 11, 2015 at 4:30

jfs

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1 Comment

Ulrich Eckhardt Over a year ago

I'd like to add that in order to not forget releasing a lock, you can use it as context manager, i.e. with some_lock.acquire(): ....

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