Python-函数调用超时

Question

我正在Python中调用一个函数，该函数可能会停滞并迫使我重新启动脚本。

如何调用该函数或将其包装在其中，以便如果花费的时间超过5秒，脚本将取消该函数并执行其他操作？

Answer 1

如果在UNIX上运行，则可以使用信号包：

In [1]: import signal

# Register an handler for the timeout
In [2]: def handler(signum, frame):
   ...:     print("Forever is over!")
   ...:     raise Exception("end of time")
   ...: 

# This function *may* run for an indetermined time...
In [3]: def loop_forever():
   ...:     import time
   ...:     while 1:
   ...:         print("sec")
   ...:         time.sleep(1)
   ...:         
   ...:         

# Register the signal function handler
In [4]: signal.signal(signal.SIGALRM, handler)
Out[4]: 0

# Define a timeout for your function
In [5]: signal.alarm(10)
Out[5]: 0

In [6]: try:
   ...:     loop_forever()
   ...: except Exception, exc: 
   ...:     print(exc)
   ....: 
sec
sec
sec
sec
sec
sec
sec
sec
Forever is over!
end of time

# Cancel the timer if the function returned before timeout
# (ok, mine won't but yours maybe will :)
In [7]: signal.alarm(0)
Out[7]: 0

调用后10秒钟，将调用alarm.alarm(10)处理程序。这引发了一个异常，你可以从常规Python代码中拦截该异常。

该模块不能很好地与线程配合使用（但是，谁可以呢？）

请注意，由于发生超时时会引发异常，因此该异常可能最终在函数内部被捕获并被忽略，例如一个这样的函数：

def loop_forever():
    while 1:
        print('sec')
        try:
            time.sleep(10)
        except:
            continue