线程 - Python如何实现让一个函数超时退出?

Question

需求: 多个子线程同时调用a函数处理数据,如何实现让a函数超时之后return或者raise?

1 signal不能在子线程中使用,pass
2 把函数扔到进程里,开销太大,pass
3 把函数扔到单独线程里,改编进程类另其可接受终止信号,再开另外一个线程做监控,不易实现,且个人认为杀掉进程对系统资源的风险很大.

有没有什么好的解决方案,或者已有类库里面的timeout参数都是如何实现的呢?gevent有没有什么好的线程终止方案?

参考:
http://stackoverflow.com/ques...

Answer 1

Do you mean that the function is always doing CPU-intensive computing tasks? You can split the task, and after executing a small task, check whether it times out, and return if it times out, otherwise continue.

Supplement:
Let’s do this

# coding=utf-8
import datetime
import time


def run(n):
    s = 0
    for i in range(0, n):
        for j in range(0, n):
            s += 1
    return s


def run_within_time(n, time_in_millisecond):
    s = 0

    start_time = long(time.time() * 1000)
    for i in range(0, n):
        for j in range(0, n):
            s += 1
        elapsed = long(time.time() * 1000) - start_time
        # 做完一部分任务后,判断是否超时
        if elapsed >= time_in_millisecond:
            s = -1
            break
    return s


num = 10000
print long(time.time() * 1000)
print datetime.datetime.now()
print run(num)
print 'run :'
print datetime.datetime.now()
print run_within_time(num, 200)
print 'run_within_time:'
print datetime.datetime.now()

My output here is:

2016-07-28 22:25:33.271503
100000000
run :
2016-07-28 22:25:37.473611
-1
run_within_time:
2016-07-28 22:25:37.673276

I don’t know if the scenario you are facing is like this. The specific problem needs to be analyzed in detail.

Answer 2

signal module:

Reference http://blog.sina.com.cn/s/blo...

Answer 3

First look at the possible blocking places in the function. If there is a timeout, add a timeout.