完全有可能实现 Thread.stop 方法,如以下示例代码所示:

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()

这个 Thread3 类运行代码的速度似乎比 Thread2 班级。

下面的代码演示了一种可以尝试的方法:

>>> variables = {}
>>> for n in range(11):
    variables[f'm{n}'] = n


>>> for key, value in variables.items():
    print(f'{key} = {value}')


m0 = 0
m1 = 1
m2 = 2
m3 = 3
m4 = 4
m5 = 5
m6 = 6
m7 = 7
m8 = 8
m9 = 9
m10 = 10
>>>