Question

我正在尝试设置3个线程并在队列中执行5个任务。这个想法是线程将首先同时运行前三个任务，然后两个线程将完成其余两个任务。但是该程序似乎冻结了。我无法发现任何问题。

from multiprocessing import Manager
import threading
import time
global exitFlag 
exitFlag = 0


class myThread(threading.Thread):
    def __init__(self, threadID, name, q):
        threading.Thread.__init__(self)
        self.threadID = threadID
        self.name = name
        self.q = q

    def run(self):
        print("Starting " + self.name)
        process_data(self.name, self.q)
        print("Exiting " + self.name)


def process_data(threadName, q):
    global exitFlag
    while not exitFlag:
        if not workQueue.empty():
            data = q.get()
            print("%s processing %s" % (threadName, data))
        else:
            pass
        time.sleep(1)
    print('Nothing to Process')


threadList = ["Thread-1", "Thread-2", "Thread-3"]
nameList = ["One", "Two", "Three", "Four", "Five"]
queueLock = threading.Lock()
workQueue = Manager().Queue(10)
threads = []
threadID = 1

# create thread
for tName in threadList:
    thread = myThread(threadID, tName, workQueue)
    thread.start()
    threads.append(thread)
    threadID += 1

# fill up queue
queueLock.acquire()
for word in nameList:
    workQueue.put(word)
queueLock.release()

# wait queue clear
while not workQueue.empty():
    pass
# notify thread exit
exitFlag = 1
# wait for all threads to finish
for t in threads:
    t.join()
print("Exiting Main Thread")

我不知道到底发生了什么，但是在删除join()部分之后，该程序可以很好地运行。我不明白的是，当队列清空时，exitFlag应该已经发出了信号。因此，似乎以某种方式未通过process_data（）检测到该信号

Answer 1

您的代码存在多个问题。首先，由于全局解释器锁定（GIL），CPython中的线程不会“同时”运行Python代码。线程必须持有GIL才能执行Python字节码。默认情况下，如果线程没有阻止它，因为它确实阻塞了I / O，所以它在GIL中最多保留5毫秒（Python 3.2+）。要并行执行Python代码，您必须使用multiprocessing。

您也不必使用Manager.Queue而不是queue.Queue。 Manager.Queue是单独管理程序中的queue.Queue。您在此处引入了绕过IPC和内存复制的弯路，这毫无用处。

您陷入僵局的原因是您在这里有比赛条件：

    if not workQueue.empty():
        data = q.get()

这不是原子操作。线程可以检查workQueue.empty()，然后删除GIL，让另一个线程耗尽队列，然后继续进行data = q.get()，如果您不将某些内容再次放入队列，它将永远阻塞。 Queue.empty()检查是一种常规的反模式，不需要使用它。使用有毒药丸（前哨值）来代替获取循环并让工人知道他们应该退出。您需要与工作人员一样多的哨兵值。查找有关iter(callabel, sentinel) here的更多信息。

import time
from queue import Queue
from datetime import datetime
from threading import Thread, current_thread


SENTINEL = 'SENTINEL'


class myThread(Thread):

    def __init__(self, func, inqueue):
        super().__init__()
        self.func = func
        self._inqueue = inqueue

    def run(self):
        print(f"{datetime.now()} {current_thread().name} starting")
        self.func(self._inqueue)
        print(f"{datetime.now()} {current_thread().name} exiting")


def process_data(_inqueue):
    for data in iter(_inqueue.get, SENTINEL):
        print(f"{datetime.now()} {current_thread().name} "
              f"processing {data}")
        time.sleep(1)


if __name__ == '__main__':


    N_WORKERS = 3

    inqueue = Queue()
    input_data = ["One", "Two", "Three", "Four", "Five"]

    sentinels = [SENTINEL] * N_WORKERS # one sentinel value per worker
    # enqueue input and sentinels
    for word in input_data +  sentinels:
        inqueue.put(word)

    threads = [myThread(process_data, inqueue) for _ in range(N_WORKERS)]

    for t in threads:
        t.start()
    for t in threads:
        t.join()

    print(f"{datetime.now()} {current_thread().name} exiting")

示例输出：

2019-02-14 17:58:18.265208 Thread-1 starting
2019-02-14 17:58:18.265277 Thread-1 processing One
2019-02-14 17:58:18.265472 Thread-2 starting
2019-02-14 17:58:18.265542 Thread-2 processing Two
2019-02-14 17:58:18.265691 Thread-3 starting
2019-02-14 17:58:18.265793 Thread-3 processing Three
2019-02-14 17:58:19.266417 Thread-1 processing Four
2019-02-14 17:58:19.266632 Thread-2 processing Five
2019-02-14 17:58:19.266767 Thread-3 exiting
2019-02-14 17:58:20.267588 Thread-1 exiting
2019-02-14 17:58:20.267861 Thread-2 exiting
2019-02-14 17:58:20.267994 MainThread exiting

Process finished with exit code 0

如果您不坚持要继承Thread的子类，则也可以只使用multiprocessing.pool.ThreadPool multiprocessing.dummy.Pool，它会在后台为您完成工作。

使用`thread.join（）`时多线程冻结

1 个答案: