On Sun, May 15, 2011 at 11:05 AM, Ian Kelly <ian.g.ke...@gmail.com> wrote:
> Actually, this won't work, because the value of the "yield None" gets
> ignored. Thus if you try to call send() twice in a row, the generator
> the treats second send() as if it were a next(), and it is not
> possible to have more than one item in the queue.
You're right. It needs a while loop instead of the if (and some slight
reordering):
def ints():
i=0
queue=[]
while True:
if queue: # see other thread, this IS legal and pythonic and
quite sensible
sent=(yield queue.pop(0))
else:
sent=(yield i)
i+=1
while sent is not None:
queue.append(sent)
sent=(yield None) # This is the return value from gen.send()
That should work.
Chris Angelico
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