当其中一个线程中存在未捕获的异常时，Python 多线程程序不会退出

Question

下面的代码产生 100 个线程并随机生成一个异常。即使所有线程都执行完毕（同时产生了一些异常），主程序仍然没有退出。难道我做错了什么？需要修改什么才能让其中一个线程发生异常，主线程仍然退出？

from __future__ import print_function

from threading import Thread

import sys

import random

from queue import Queue

__author__ = 'aanush'

"""

Testing if threading exits the python script gracefully

"""

class NoException(Exception):

    pass

class ThreadFail(Thread):

    """

    Class which helps us in doing multi-threading which improves performance of the script

    """

    def __init__(self, name, counter, queue_):

        Thread.__init__(self)

        self.queue = queue_

        self.threadID = counter

        self.name = name

        self.counter = counter

    def run(self):

        while True:

            # Expand the tuple from queue and pass it to the target function

            some_random_num = self.queue.get()

            func_test(some_random_num)

            self.queue.task_done()

def func_test(random_num):

    if random_num <= 10:

        print("Sleep time - {} greater than 10. Not raising exception".format(random_num))

    else:

        print('sleep time less than 10 : Raising exception')

        raise NoException

queue = Queue()

for thread_num in range(100):

    worker = ThreadFail('Thread-{}'.format(thread_num), thread_num, queue)

    worker.daemon = True

    worker.start()

for x in range(1000):

    queue.put(random.randrange(1, 15))

queue.join()

Answer 1

您在这里遇到了僵局。由于异常而终止的线程不会释放对共享资源的持有锁，因此queue会被破坏。您需要捕获线程内的异常并让它们优雅地退出。def run(self):&nbsp; &nbsp; while True:&nbsp; &nbsp; &nbsp; &nbsp; # Expand the tuple from queue and pass it to the target function&nbsp; &nbsp; &nbsp; &nbsp; some_random_num = self.queue.get()&nbsp; &nbsp; &nbsp; &nbsp; try:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; func_test(some_random_num)&nbsp; &nbsp; &nbsp; &nbsp; except NoException:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; pass&nbsp; &nbsp; &nbsp; &nbsp; finally:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.queue.task_done()