Asyncio create task without await.md

created	2026-04-02 18:42:02 +0800
modified	2026-04-02 18:42:18 +0800

Asyncio create task without await

If you create an asyncio task with asyncio.create_task() and then sleep (e.g., await asyncio.sleep(3)) without awaiting the task itself, the task will still run concurrently during that sleep. The event loop will switch to the task, execute its code (including the print statement), and then resume the sleep.

So in your scenario—where the task only contains a print—you will see the printed output even without awaiting the task. The side effect (printing) happens as soon as the task runs.

However, if you meant “get the result” as the return value of the coroutine (e.g., if it returned a string or number), you cannot obtain that without either:

• await task (which would also wait for completion), or
• calling task.result() after the task is done (which requires checking task.done() or using a callback).

Important notes:

• The event loop must keep running long enough for the task to finish. In your example, asyncio.sleep(3) gives it that time.
• If you don’t keep a reference to the task and the program ends immediately after the sleep, the task may be cancelled or garbage‑collected before completing. But as long as you await the sleep (or otherwise keep the loop alive), the task runs fine.

Example that prints the output without awaiting the task:

import asyncio

async def my_task():
    print("Hello from task")
    # No return value, just print

async def main():
    task = asyncio.create_task(my_task())
    await asyncio.sleep(3)   # task runs during this sleep
    # task not awaited – but print already happened

asyncio.run(main())

Output: Hello from task

If the task returned a value (e.g., return 42), you would not get 42 without awaiting task or calling task.result() after it finishes.