Coroutines in python

Question

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.

What result were you expecting?

Adam Wagner
– Adam Wagner

2011-10-26 03:10:15 +00:00
Commented Oct 26, 2011 at 3:10 — Adam Wagner
– Adam Wagner, Commented Oct 26, 2011 at 3:10

Tavis Rudd · Accepted Answer · 2011-10-26 04:06:45Z

2

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.

answered Oct 26, 2011 at 4:06

Tavis Rudd

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1 Comment

Trantor Liu Over a year ago

Thank you Rudd. Let me explain this with my understanding again. The call r = receiver() is equals to r = coroutune(receiver)(), which becomes r = start(), thus r finally is g. Is that right?

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