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Date: Sun, 15 May 2011 10:47:09 +1000
Subject: Re: I don't understand generator.send()
From: Chris Angelico <rosuav@gmail.com>
To: python-list@python.org
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On Sun, May 15, 2011 at 10:08 AM, Victor Eijkhout <see@sig.for.address> wro=
te:
> =A0 =A0 =A0 =A0yield i

> =A0 =A0 =A0 =A0r =3D gen.send(2)

When you send() something to a generator, it becomes the return value
of the yield expression. See the example here:
http://docs.python.org/whatsnew/2.5.html#pep-342-new-generator-features

For what you're doing, there's a little complexity. If I understand,
you want send() to be like an ungetc call... you could do that like
this:


def ints():
   i=3D0
   while True:
       sent=3D(yield i)
       if sent is not None:
          yield None  # This becomes the return value from gen.send()
          yield sent  # This is the next value yielded
       i +=3D 1

This lets you insert at most one value per iteration. Supporting more
than one insertion is more complicated, but changing the loop
structure entirely may help:

def ints():
    i=3D0
    queue=3D[]
    while True:
        if queue:  # see other thread, this IS legal and pythonic and
quite sensible
            sent=3D(yield queue.pop(0))
        else:
            sent=3D(yield i)
            i+=3D1
        if sent is not None:
            yield None  # This is the return value from gen.send()
            queue.append(sent)

With this generator, you maintain a queue of sent values (if you want
it to be a LIFO stack rather than a FIFO queue, just change the pop(0)
to just pop()), and if the queue's empty, it produces sequential
integers. (Incidentally, the sent values don't have to be integers. I
leave it to you to decide whether that's any use or not.)

Hope that helps!

Chris Angelico