`asyncio.run()` 不等待协程完成

Question

我在 Python 3.7.3 中运行此代码

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__")

并得到这个输出：

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

我还期待看到：

done waiting for 1
done waiting for 2

即，为什么期望asyncio.run(X)会等待协程完成后再继续。

Answer 1

如果您想等待由 生成的所有任务的完成create_task，那么您需要显式地执行此操作，例如，仅await依次执行它们或使用 asyncio 工具，如gather 或（此处wait描述了差异）。否则，当退出主协程时，它们将被取消，并被传递给。asyncio.runasyncio.run

例子：

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__