如何对函数进行超时装饰?

时间:2016-10-27 06:31:26

标签: python time

我试图在某些特定功能上进行装饰。如果函数的工作时间超过五秒钟,则deco会引发错误或中断函数。

def time_limit(func):
    def time_out():
        raise AssertionError

    @wraps(func)
    def deco(*args, **kwargs):
        timer = Timer(5, time_out)
        timer.start()
        res = func(*args, **kwargs)
        return res

    return deco

然而即使装饰工作,该功能仍可正常工作:

In [69]: @time_limit
...: def f():
...:     time.sleep(6)
...:     print 'aaa'
...:     

In [70]: f()
Exception in thread Thread-2764:
Traceback (most recent call last):
File "/usr/local/Cellar/python/2.7.12/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 801, in __bootstrap_inner
self.run()
File      "/usr/local/Cellar/python/2.7.12/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 1073, in run
self.function(*self.args, **self.kwargs)
File "<ipython-input-68-cc74c901d8b8>", line 4, in time_out
raise AssertionError
AssertionError

aaa

如何解决问题? PS。我使用apscheduler来传播这个功能,所以超时装饰会更好。

0 个答案:

没有答案