python利用signal自定义函数超时机制
阅读原文时间:2023年07月08日阅读:2

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:
50 # task2 start
51 # task2 end