如何通过Python中的多个函数“双返”?

问题描述 投票:2回答:1

我有三个功能“滴入”并在另一个内部调用,如下所示:

async def start_swings(self, server, user):
    try:
        while self.jailed[server.id][user.id]['current'] != 0:
            await self.get_question(server, user)
        else:
            await self.escape_jail(server, user)
    except KeyError:  # User isn't jailed
        pass

async def get_question(self, server, user):
    rnd = random.choice(range(2))
    if rnd == 0:
        await self.swing_math(server, user)
    elif rnd == 1:
        await self.swing_reverse(server, user)

async def swing_math(self, server, user):
    num1 = random.randint(1, 1000)
    num2 = random.randint(1, 1000)
    ops = ['+', '-']
    op = random.choice(ops)
    ans = int(eval(str(num1) + op + str(num2)))
    await self.bot.send_message(user, "**\N{HAMMER} It's going to take you `{}` more swings to break free.**\n```What is {} {} {}?```\n*I might doze off if you take too long to answer. If you don't get a reply from me within a few seconds, type `>swing`. If you try to use `>swing` as a way to reset your current question, you'll get a cooldown penalty.*".format(self.jailed[server.id][user.id]['current'], num1, op, num2))
    resp = await self.bot.wait_for_message(author=user)
    resp = resp.clean_content
    if resp == str(ans):
        await self.bot.send_message(user, "Correct. \N{WHITE HEAVY CHECK MARK} Take a swing.")
        self.jailed[server.id][user.id]['current'] -= 1
        dataIO.save_json('data/jail/jailed.json', self.jailed)
    elif resp == ">swing":
        return
    else:
        await self.bot.send_message(user, "Wrong. \N{CROSS MARK} Let me put that brick back in place.")
        self.jailed[server.id][user.id]['current'] += 1
        dataIO.save_json('data/jail/jailed.json', self.jailed)

在第三个函数中,您将看到以下语句:

elif resp == ">swing":
    return

我怎样才能有效地“双重回报”,以便它返回start_swings而不是get_questions?我是Python的新手,并不知道该怎么称呼它。我正在尝试执行此脚本以完全停止在代码的这个区域:

elif resp == ">swing":
   return

...而不是让它返回到最后一个被调用的函数。

python python-3.x function return python-asyncio
1个回答
3
投票

您不能像过去那样手动“弹出”堆栈,但是您可以使用异常(即使异常未被抛出被认为是不好的做法)“:

从异常继承:

class DoubleReturnException(Exception):
     pass

安装处理程序:

    while self.jailed[server.id][user.id]['current'] != 0:
        try:
            await self.get_question(server, user)
        except DoubleReturnException:
            pass
    else:

然后,而不是返回,只是:

raise DoubleReturnException

返回到处理程序的except部分(无论调用的数量是多少)。

定义自己的异常是必要的,因为它确保程序因此而返回,而不是ValueError或程序可能抛出的任何其他异常。

这必须是一个例外情况。通过在各地抛出异常进行编程会导致“意大利面条代码”并以某种方式恢复邪恶的goto声明。你可能会发现在稳定产品的最后阶段很有趣,当你无法打破特定情况的所有(有缺陷的)设计时(但我个人从来没有使用过这样的技巧)。

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