如何在不使用 `return` 语句的情况下跳出 asyncio 协程? [英] How to break out of a asyncio coroutine without using the `return` statement?

问题描述

如果 except 语句被捕获，我在决定如何跳出这个协程时会遇到一些麻烦.通常，我只会使用:

I'm having a bit of trouble deciding how to break out of this coroutine if the except statement catches. Normally, I would just use:

def f(x):
    try:
        foo_var = next(a_volitile_generator(x))
    except Exception:
        print('it broked')
        return
    yield foo_var

但是！我将产生无数的这些生成器函数，使它们成为协程，最后 Future 将它们和化将它们加载到事件循环中.问题来了:

However! I'm going to spawn a gazillion of these generator funcs, make them coroutines, and finally Futureize them & load them into an event-loop. Here's the issue:

loop = asyncio.get_event_loop()
queue = asyncio.Queue(maxsize=100)

async def f(x, result_queue, *, loop=loop):
    while _some_condition() is True:
        try:
            foo_var = await a_volitile_async_generator(x)
        except Exception:
            print('it broked')
            # HELP BELOW
            raise StopAsyncIteration # <--ONLY OPTION????????
        await result_queue.put(foo_var)

将任何内容返回到事件循环在asyncio中是一个禁忌，因为你会破坏一切......即使那个值是None.

returning anything into the event-loop is a no-no in asyncio because you'll break everything...even if that value is None.

除此之外，我们还要说我想避免提高 StopAsyncIteration ......我试图弄清楚这是唯一的方法还是我有其他选择:

Aside from that, let's also say I'd like to avoid raising StopAsyncIteration...I'm trying to figure out if that's the only way to do this or if I have other options:

推荐答案

As 由@dirn 在评论中提到，您可以用简单的 break<替换 raise StopAsyncIteration 行/代码>:

As mentioned by @dirn in the comments, you can replace your raise StopAsyncIteration line with a simple break:

async def f(x, result_queue, *, loop=loop):
    while _some_condition() is True:
        try:
            foo_var = await a_volitile_async_generator(x)
        except Exception:
            print('it broked')
            break
        await result_queue.put(foo_var)

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