有没有办法在异步 python 中等待值 [关闭]答案

【问题标题】：Is there a way to wait for value in async python [closed]有没有办法在异步 python 中等待值 [关闭]
【发布时间】：2020-10-22 06:40:15
【问题描述】：

基本上我正在尝试在 Python 中使用异步代码。一开始，我想我会做一个斐波那契计算器。但是我无法使用它，这是我的代码

import asyncio as a
cache = {}

async def fib(n):

    if n < 10:
        a.sleep(0.001)
    else:
        if n in cache:
            result = cache[n]
        else:
            result = await fib(n - 2) + await fib(n - 1)
    cache[n] = result
    return result

async def main():
    res = await fib(100)
    print(res)

a.run(main())

我得到的错误是

Warning (from warnings module):
  File "<pyshell#3>", line 4
RuntimeWarning: coroutine 'sleep' was never awaited
Traceback (most recent call last):
  File "<pyshell#6>", line 1, in <module>
    a.run(main())
  File "C:\Users\Owner\AppData\Local\Programs\Python\Python38-32\lib\asyncio\runners.py", line 43, in run
    return loop.run_until_complete(main)
  File "C:\Users\Owner\AppData\Local\Programs\Python\Python38-32\lib\asyncio\base_events.py", line 616, in run_until_complete
    return future.result()
  File "<pyshell#5>", line 2, in main
    res = await fib(100)
  File "<pyshell#3>", line 9, in fib
    result = await fib(n - 2) + await fib(n - 1)
  File "<pyshell#3>", line 9, in fib
    result = await fib(n - 2) + await fib(n - 1)
  File "<pyshell#3>", line 9, in fib
    result = await fib(n - 2) + await fib(n - 1)
  [Previous line repeated 43 more times]
  File "<pyshell#3>", line 11, in fib
    return result
UnboundLocalError: local variable 'result' referenced before assignment

谢谢。

【问题讨论】：

await a.sleep(0)
if n < 10 "result" 未设置但已使用。
什么意思？我退货了。
您的代码实际上是同步的，因为您只使用事件循环安排一件事 (main)。

标签： python python-3.x async-await python-asyncio

【解决方案1】：

你错过了初始化值。它们被预先添加到下面的缓存中：

import asyncio as a
cache = {0: 0, 1: 1}

async def fib(n):
    if n in cache:
        result = cache[n]
    else:
        result = await fib(n - 2) + await fib(n - 1)
    cache[n] = result
    return result

async def main():
    res = await fib(100)
    print(res)

a.run(main())

【讨论】：