是否有可能终止一个正在运行的线程而不设置/检查任何标志/信号/等等?


当前回答

如果您试图终止整个程序,您可以将线程设置为“守护进程”。看到 Thread.daemon

其他回答

实现一个线程是绝对可能的。方法,如下例代码所示:

import sys
import threading
import time


class StopThread(StopIteration):
    pass

threading.SystemExit = SystemExit, StopThread


class Thread2(threading.Thread):

    def stop(self):
        self.__stop = True

    def _bootstrap(self):
        if threading._trace_hook is not None:
            raise ValueError('Cannot run thread with tracing!')
        self.__stop = False
        sys.settrace(self.__trace)
        super()._bootstrap()

    def __trace(self, frame, event, arg):
        if self.__stop:
            raise StopThread()
        return self.__trace


class Thread3(threading.Thread):

    def _bootstrap(self, stop_thread=False):
        def stop():
            nonlocal stop_thread
            stop_thread = True
        self.stop = stop

        def tracer(*_):
            if stop_thread:
                raise StopThread()
            return tracer
        sys.settrace(tracer)
        super()._bootstrap()

###############################################################################


def main():
    test1 = Thread2(target=printer)
    test1.start()
    time.sleep(1)
    test1.stop()
    test1.join()
    test2 = Thread2(target=speed_test)
    test2.start()
    time.sleep(1)
    test2.stop()
    test2.join()
    test3 = Thread3(target=speed_test)
    test3.start()
    time.sleep(1)
    test3.stop()
    test3.join()


def printer():
    while True:
        print(time.time() % 1)
        time.sleep(0.1)


def speed_test(count=0):
    try:
        while True:
            count += 1
    except StopThread:
        print('Count =', count)

if __name__ == '__main__':
    main()

Thread3类运行代码的速度似乎比Thread2类快大约33%。

如果您试图终止整个程序,您可以将线程设置为“守护进程”。看到 Thread.daemon

我对这个游戏已经很晚了,但我一直在与一个类似的问题作斗争,下面的内容似乎为我完美地解决了这个问题,并且让我在守护子线程退出时做一些基本的线程状态检查和清理:

import threading
import time
import atexit

def do_work():

  i = 0
  @atexit.register
  def goodbye():
    print ("'CLEANLY' kill sub-thread with value: %s [THREAD: %s]" %
           (i, threading.currentThread().ident))

  while True:
    print i
    i += 1
    time.sleep(1)

t = threading.Thread(target=do_work)
t.daemon = True
t.start()

def after_timeout():
  print "KILL MAIN THREAD: %s" % threading.currentThread().ident
  raise SystemExit

threading.Timer(2, after_timeout).start()

收益率:

0
1
KILL MAIN THREAD: 140013208254208
'CLEANLY' kill sub-thread with value: 2 [THREAD: 140013674317568]

有一个图书馆就是为了这个目的而建的,别说了。尽管这里列出的一些注意事项仍然适用,但至少这个库提供了一种常规的、可重复的技术来实现既定的目标。

正如其他人所提到的,规范是设置一个停止标志。对于一些轻量级的东西(没有Thread的子类化,没有全局变量),lambda回调是一个选项。(注意if stop()中的括号。)

import threading
import time

def do_work(id, stop):
    print("I am thread", id)
    while True:
        print("I am thread {} doing something".format(id))
        if stop():
            print("  Exiting loop.")
            break
    print("Thread {}, signing off".format(id))


def main():
    stop_threads = False
    workers = []
    for id in range(0,3):
        tmp = threading.Thread(target=do_work, args=(id, lambda: stop_threads))
        workers.append(tmp)
        tmp.start()
    time.sleep(3)
    print('main: done sleeping; time to stop the threads.')
    stop_threads = True
    for worker in workers:
        worker.join()
    print('Finis.')

if __name__ == '__main__':
    main()

将print()替换为始终刷新的pr()函数(sys.stdout.flush())可以提高shell输出的精度。

(仅在Windows/Eclipse/Python3.3上测试)