在asyncio 中跳出正在执行的task

Posted 2021-02-15 ywhyme

需求描述

代码在asyncio的框架中运行, 但是一旦一个task出现了长时间的堵塞,我们要跳过这个task(代码可能是用户输入的,例如用户编写的插件)

代码如下

(其中大部分代码出自官方的 asyncio 以及 signal)
重点的函数在于asyncio.gather 详情查看上面的链接

import asyncio
import os
import signal
import time


async def factorial(name, number):
    f = 1
    for i in range(2, number + 1):
        print(f"Task {name}: Compute factorial({i})...")
        await asyncio.sleep(1)
        f *= i
    print(f"Task {name}: factorial({number}) = {f}")


async def test():
    for i in range(100):
        print("sleep--", i)
        time.sleep(1)


def handler(signum, frame):
    print('Signal handler called with signal', signum)
    raise OSError("Couldn't open device!")


signal.signal(signal.SIGTERM, handler)

print(os.getpid())


async def main():
    # Schedule three calls *concurrently*:
    await asyncio.gather(
        test(),
        factorial("A", 2),
        factorial("B", 3),
        factorial("C", 4),
        return_exceptions=True
    )


asyncio.run(main())

运行结果如下

18733
sleep-- 0
sleep-- 1
sleep-- 2
sleep-- 3
sleep-- 4
sleep-- 5
sleep-- 6
Signal handler called with signal 15
Task A: Compute factorial(2)...
Task B: Compute factorial(2)...
Task C: Compute factorial(2)...
Task A: factorial(2) = 2
Task B: Compute factorial(3)...
Task C: Compute factorial(3)...
Task B: factorial(3) = 6
Task C: Compute factorial(4)...
Task C: factorial(4) = 24

Process finished with exit code 0

这样就可以跳过被长时间堵塞的task,或者长时间执行的task了