python利用signal自定义函数超时机制

利用signal模块，可以给函数设置一个超时时间（精确到秒），结合装饰器实现：超时时间内函数运行完成正常退出，超时时间内函数未运行完成则抛出Timeout异常。

 1 # coding:utf8
 2 import time
 3 import signal
 4
 5
 6 # 自定义超时异常
 7 class TimeoutError(Exception): pass
 8
 9 # 超过超时时间，调用此函数
10 def handler(signum, frame):
11     raise TimeoutError()
12
13 # 函数超时装饰器
14 def time_out(interval, doc):
15     def decorator(func):
16         def wrapper(*args, **kwargs):
17             try:
18                 signal.signal(signal.SIGALRM, handler)
19                 signal.alarm(interval)       # interval秒后向进程发送SIGALRM信号
20                 result = func(*args, **kwargs)
21                 signal.alarm(0)              # 函数在规定时间执行完后关闭alarm闹钟
22                 return result
23             except TimeoutError as e:
24                 # 捕获到超时异常，要做的事情
25                 print("The function failed to run due to timeout, func:<%s>" % doc)
26         return wrapper
27     return decorator
28
29 @time_out(1, "demo.py中task1函数")
30 def task1():
31     print("task1 start")
32     time.sleep(1.2)
33     print("task1 end")
34
35
36 @time_out(1, "demo.py中task2函数")
37 def task2():
38     print("task2 start")
39     time.sleep(0.8)
40     print("task2 end")
41
42
43 if __name__ == "__main__":
44     task1()
45     task2()
46
47
48 # task1 start
49 # The function failed to run due to timeout, func:<demo.py中task1函数>
50 # task2 start
51 # task2 end