Python异步队列未显示任何异常

Question

1. 如果我运行此代码，它将挂起而不会抛出ZeroDivisionError.
2. 如果我在await asyncio.gather(*tasks, return_exceptions=True) 上面移动await queue.join()，它最终会抛出ZeroDivisionError并停止。
3. 如果我然后注释掉1 / 0并运行，它将执行所有内容，但最终会挂起。

现在的问题是，我怎样才能同时实现：

1. 能够像上面的情况 2 一样看到意外的异常，并且......
2. 当队列中的所有任务完成时实际停止

.

import asyncio
import random
import time


async def worker(name, queue):
    while True:
        print('Get a "work item" out of the queue.')
        sleep_for = await queue.get()

        print('Sleep for the "sleep_for" seconds.')
        await asyncio.sleep(sleep_for)

        # Error on purpose
        1 / 0

        print('Notify the queue that the "work item" has been processed.')
        queue.task_done()

        print(f'{name} has slept for {sleep_for:.2f} seconds')

async def main():
    print('Create a queue that we will use to store our "workload".')
    queue = asyncio.Queue()

    print('Generate random timings and put them into the queue.')
    total_sleep_time = 0
    for _ in range(20):
        sleep_for = random.uniform(0.05, 1.0)
        total_sleep_time += sleep_for
        queue.put_nowait(sleep_for)

    print('Create three worker tasks to process the queue concurrently.')
    tasks = []
    for i in range(3):
        task = asyncio.create_task(worker(f'worker-{i}', queue))
        tasks.append(task)

    print('Wait until the queue is fully processed.')
    started_at = time.monotonic()

    print('Joining queue')
    await queue.join()
    total_slept_for = time.monotonic() - started_at

    print('Cancel our worker tasks.')
    for task in tasks:
        task.cancel()

    print('Wait until all worker tasks are cancelled.')
    await asyncio.gather(*tasks, return_exceptions=True)

    print('====')
    print(f'3 workers slept in parallel for {total_slept_for:.2f} seconds')
    print(f'total expected sleep time: {total_sleep_time:.2f} seconds')

asyncio.run(main())

Answer 1

有几种方法可以解决这个问题，但中心思想是在 asyncio 中，与经典线程不同，一次等待多个事情很简单。

例如，您可以 awaitqueue.join()和工作任务，以先完成者为准。由于工作人员没有正常完成（您稍后取消它们），因此工作人员完成意味着它已经提出。

# convert queue.join() to a full-fledged task, so we can test
# whether it's done
queue_complete = asyncio.create_task(queue.join())

# wait for the queue to complete or one of the workers to exit
await asyncio.wait([queue_complete, *tasks], return_when=asyncio.FIRST_COMPLETED)

if not queue_complete.done():
    # If the queue hasn't completed, it means one of the workers has
    # raised - find it and propagate the exception.  You can also
    # use t.exception() to get the exception object. Canceling other
    # tasks is another possibility.
    for t in tasks:
        if t.done():
            t.result()  # this will raise