如何限制函数可以运行的时间量（添加超时）？

Question

如何设置函数可以运行的最大时间限制？ 例如，使用time.sleep作为占位符函数，如何限制时间量time.sleep可以运行最多5分钟（300秒）？

import time 

try: 
    # As noted above `time.sleep` is a placeholder for a function 
    # which takes 10 minutes to complete. 
    time.sleep(600) 
except: 
    print('took too long') 

即，如何能够time.sleep(600)上述限制和300秒后中断？

Answer 1

在POSIX上，您可以在signal模块中获得一个简单而干净的解决方案。

import signal 
import time 

class Timeout(Exception): 
    pass 

def handler(sig, frame): 
    raise Timeout 

signal.signal(signal.SIGALRM, handler) # register interest in SIGALRM events 

signal.alarm(2) # timeout in 2 seconds 
try: 
    time.sleep(60) 
except Timeout: 
    print('took too long') 

注意事项：

• 不能在所有平台上，例如工作视窗。
• 不适用于线程应用程序，仅适用于主线程。

对于其他读者，上面的警告是一个交易断路器，你将需要一个更重量级的方法。最好的选择通常是在单独的进程（或可能是一个线程）中运行代码，如果时间过长，则终止该进程。例如，请参阅multiprocessing模块。

Answer 2

一个目前很可能优先选项来完成你想要的是蟒蛇的使用

多（特别是它的proc.join（timeoutTime）法）

模块（见tutorial）

只需复制/粘贴下面的代码示例并运行它。这是你以后的样子吗？

def beBusyFor(noOfSeconds): 
    import time 
    print(" beBusyFor() message: going to rest for", noOfSeconds, "seconds") 
    time.sleep(noOfSeconds) 
    print(" beBusyFor() message: was resting", noOfSeconds, "seconds, now AWAKE") 

import multiprocessing 

noOfSecondsBusy = 5; timeoutTime = 3 
print("--- noOfSecondsBusy = 5; timeoutTime = 3 ---") 
proc = multiprocessing.Process(target=beBusyFor, args=(noOfSecondsBusy,)) 
print("Start beBusyFor()") 
proc.start() 
print("beBusyFor() is running") 
proc.join(timeoutTime) 
if proc.is_alive(): 
    print(timeoutTime, "seconds passed, beBusyFor() still running, terminate()") 
    proc.terminate() 
else: 
    print("OK, beBusyFor() has finished its work in time.") 
#:if  

print() 

noOfSecondsBusy = 2; timeoutTime = 3 
print("--- noOfSecondsBusy = 2; timeoutTime = 3 ---") 
proc = multiprocessing.Process(target=beBusyFor, args=(noOfSecondsBusy,)) 
print("Start beBusyFor()") 
proc.start() 
print("beBusyFor() started") 
proc.join(timeoutTime) 
if proc.is_alive(): 
    print(timeoutTime, "seconds passed, beBusyFor() still running, terminate()") 
    proc.terminate() 
else: 
    print("OK, beBusyFor() has finished its work in time.") 
#:if  

它输出：

--- noOfSecondsBusy = 5; timeoutTime = 3 --- 
Start beBusyFor() 
beBusyFor() is running 
    beBusyFor() message: going to rest for 5 seconds 
3 seconds passed, beBusyFor() still running, terminate() 

--- noOfSecondsBusy = 2; timeoutTime = 3 --- 
Start beBusyFor() 
beBusyFor() started 
    beBusyFor() message: going to rest for 2 seconds 
    beBusyFor() message: was resting 2 seconds, now AWAKE 
OK, beBusyFor() has finished its work in time. 

另一种已知的对我的选择是使用一个

装饰功能和信号模块

结帐我在这里提供的web page with origin of the code（只有一个小的调整是必要的，以使其上的Python 3.6运行）：

import signal 

class TimeoutError(Exception): 
    def __init__(self, value = "Timed Out"): 
     self.value = value 
    def __str__(self): 
     return repr(self.value) 

def timeout(seconds_before_timeout): 
    def decorate(f): 
     def handler(signum, frame): 
      raise TimeoutError() 
     def new_f(*args, **kwargs): 
      old = signal.signal(signal.SIGALRM, handler) 
      signal.alarm(seconds_before_timeout) 
      try: 
       result = f(*args, **kwargs) 
      finally: 
       signal.signal(signal.SIGALRM, old) 
      signal.alarm(0) 
      return result 
     # new_f.func_name = f.func_name 
     new_f.__name__ = f.__name__ 
     return new_f 
    return decorate 

# Try it out: 

import time 

@timeout(5) 
def mytest(): 
    print("mytest() message: Started") 
    for i in range(1,10): 
     time.sleep(1) 
     print("mytest() message: %d seconds have passed" % i) 

try: 
    mytest() 
except TimeoutError as e: 
    print("stopped executing mytest() because it", e) 

print("continuing script execution past call of mytest()") 

上述输出的代码：

mytest() message: Started 
mytest() message: 1 seconds have passed 
mytest() message: 2 seconds have passed 
mytest() message: 3 seconds have passed 
mytest() message: 4 seconds have passed 
stopped executing mytest() because it 'Timed Out' 
continuing script execution past call of mytest()