python 中的协程
我从一本书上读到了以下代码，并对此有一些疑问。
def coroutine(func):
def start(*args, **kwargs):
g = func(*args, **kwargs)
g.next()
return g
return start
@coroutine
def receiver():
print("Ready to receive")
while True:
n = (yield)
print("Got %s" % n)
r = receiver()
r.send("hello, world")
通过使用协程，不需要初始.next()。我的理解是，如果r = receive()，则r = start，所以当我调用r.send()时，它等于start.send()，然后我再次调用.next()，对吗？但结果并不是我所期望的。
I read the following code from a book, and have some questions about it.
def coroutine(func):
def start(*args, **kwargs):
g = func(*args, **kwargs)
g.next()
return g
return start
@coroutine
def receiver():
print("Ready to receive")
while True:
n = (yield)
print("Got %s" % n)
r = receiver()
r.send("hello, world")
By using coroutine, no initial .next() is needed. My understanding is, if r = receiver(), then r = start, so when I call r.send(), it equals to start.send(), then I call .next() again, right? But the result is not what I expected.
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评论（1）
你的问题不是协程。您误解了函数装饰器。在
r = receive()之后，r不是start而是g。阅读一下函数装饰，你就会明白发生了什么。Your problem isn't the coroutine. You're misunderstanding the function decorator. After
r = receiver(), r is not start but g. Read up on function decoration and you'll understand what is going on.