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__")

并获得此输出：

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

我也期待看到：

done waiting for 1
done waiting for 2

即，为什么期望 asyncio.run(X) 会在继续之前等待协程完成。

Answer 1

如果你想等待 create_task 产生的所有任务完成，那么你需要明确地完成它，例如，只需为它们依次 await 或 asyncio 设施，如 { {3}} 或 gather（差异描述为 wait）。否则，它们将在退出主协程时被 asyncio.run 取消，并传递给 asyncio.run。

示例：

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__

`asyncio.run()` 不等待协程完成

1 个答案: