python协程之await等待过程理解

目标

用一个等待过程理解，能够实现一致的方便，70行代码

例子说明

某方需要2，执行req2需要执行这个时间，最终执行时间是3秒，大约1秒，同时执行；如果最终时间是秒，如果执行的话，具体请参看，友情提示是,yield from 方法后面如果是它的工具，可以进入____

import time
from collections import deque

_delay = deque()


class FutureX:
def __init__(self, coro=None, delay_second=None):
self.coro = coro
if delay_second:
self.start = delay_second + time.time()

def step(self):
coro = self.coro
try:
result = coro.send(None)
except StopIteration as e:
print(e.value)
pass
else:
if isinstance(result, FutureX):
_delay.append((self._wakeup, result))
else:
pass

def _wakeup(self):
self.step()

def __iter__(self):
yield self
return None


def coroutine(func):
co = func.__code__
func.__code__ = co.replace(co_flags=co.co_flags | 0x100)
return func


@coroutine
def sleep0(seconds):

future = FutureX(delay_second=seconds)
b = yield from future
return seconds


async def req1(delay_seconds):
resp_time = await sleep0(delay_seconds)
return resp_time


async def req2(delay_seconds):
resp_time = await sleep0(delay_seconds)
return resp_time


t1 = time.time()
f1, f2 = FutureX(req1(2)), FutureX(req2(1))
f1.step()

f2.step()


while _delay:
callback, args = _delay.popleft()
start = args.start
if not start:
continue
while True:
end = time.time()
if start <= end:
try:
callback()
except StopIteration as e:
pass
break

print(f'花费的时间：{round(time.time() - t1,1)}')

'''
结果:
2
1
花费的时间：2.0
'''



审核编辑：刘清