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如何正确设置 sys.excepthook

如何正确设置 sys.excepthook

米脂 2023-03-16 09:54:59
我编写了以下代码来理解多进程环境中的 sys.excepthook。我正在使用 python 3。我创建了 2 个进程,它们将打印并等待获取 ctrl+c。from multiprocessing import Processimport multiprocessingimport sysfrom time import sleepclass foo:    def f(self, name):        try:            raise ValueError("test value error")        except ValueError as e:            print(e)        print('hello', name)        while True:            passdef myexcepthook(exctype, value, traceback):    print("Value: {}".format(value))    for p in multiprocessing.active_children():        p.terminate()def main(name):    a = foo()    a.f(name)sys.excepthook = myexcepthookif __name__ == '__main__':    for i in range(2):        p = Process(target=main, args=('bob', ))        p.start()当我按下时,我期待以下结果ctrl+Cpython /home/test/test.pytest value errorhello bobtest value errorhello bobValue: <KeyboardInterrupt>但不幸的是,我得到了以下结果。/home/test/venvPython3/bin/python /home/test/test.pytest value errorhello bobtest value errorhello bobError in atexit._run_exitfuncs:Traceback (most recent call last):Process Process-1:  File "/usr/lib/python3.6/multiprocessing/popen_fork.py", line 28, in poll    pid, sts = os.waitpid(self.pid, flag)KeyboardInterruptTraceback (most recent call last):  File "/usr/lib/python3.6/multiprocessing/process.py", line 258, in _bootstrap    self.run()  File "/usr/lib/python3.6/multiprocessing/process.py", line 93, in run    self._target(*self._args, **self._kwargs)  File "/home/test/test.py", line 26, in main    a.f(name)  File "/home/test/test.py", line 15, in f    passKeyboardInterruptProcess Process-2:Traceback (most recent call last):  File "/usr/lib/python3.6/multiprocessing/process.py", line 258, in _bootstrap    self.run()  File "/usr/lib/python3.6/multiprocessing/process.py", line 93, in run    self._target(*self._args, **self._kwargs)  File "/home/test/test.py", line 26, in main    a.f(name)  File "/home/test/test.py", line 15, in f    passKeyboardInterruptProcess finished with exit code 0如果有人能指出我做错了什么,那将是一个很大的帮助。另外,请让我知道如何获得预期的输出。
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蝴蝶不菲

TA贡献1810条经验 获得超4个赞

你几乎做到了。首先,使用exctype打印:


def myexcepthook(exctype, value, traceback):

    print("Value: {}".format(exctype))

    for p in multiprocessing.active_children():

        p.terminate()

并join()创建流程,防止过早退出


if __name__ == '__main__':

    pr = []

    for i in range(2):

        p = Process(target=main, args=('bob', ))

        p.start()

        pr.append(p)

    for p in pr:

        p.join()


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反对 回复 2023-03-16
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