`asyncio.run（）`不等待协同程序完成

Question

我在Python3.7.3中运行这段代码

import asyncio

async def fun(time):
    print(f"will wait for {time}")
    await asyncio.sleep(time)
    print(f"done waiting for {time}")

async def async_cenas():
    t1 = asyncio.create_task(fun(1))
    print("after 1")
    t2 = asyncio.create_task(fun(2))
    print("after 2")

def main():
    t1 = asyncio.run(async_cenas())
    print("ok main")
    print(t1)

if __name__ == '__main__':
    main()
    print("finished __name__")

并获得此输出：

after 1
after 2
will wait for 1
will wait for 2
ok main
None
finished __name__

我还希望看到：

done waiting for 1
done waiting for 2

也就是说，为什么期望asyncio.run(X)在继续之前会等待协同路由完成

Answer 1

如果您想等待create_task生成的所有任务完成，那么您需要显式地执行，例如，只需依次为它们执行await，或者使用异步IO工具，如^{}或^{}（区别在here中描述）。否则，当退出传递给asyncio.run的主协程时，asyncio.run将取消它们

例如：

import asyncio

async def fun(time):
    print(f"will wait for {time}")
    await asyncio.sleep(time)
    print(f"done waiting for {time}")

async def async_cenas():
    t1 = asyncio.create_task(fun(1))
    print("after 1")
    t2 = asyncio.create_task(fun(2))
    print("after 2")
    await asyncio.wait({t1, t2}, return_when=asyncio.ALL_COMPLETED)
    # or just
    # await t1
    # await t2

def main():
    t1 = asyncio.run(async_cenas())
    print("ok main")
    print(t1)

if __name__ == '__main__':
    main()
    print("finished __name__")

after 1
after 2
will wait for 1
will wait for 2
done waiting for 1
done waiting for 2
ok main
None
finished __name__