Generator object with yield

Question

I have found the below piece of code on the internet, with the actual output

def countdown(start):
   print start
   if start == 0:
      yield 0
   else:
      yield countdown(start -1)

g = countdown(3)
g.next()
g.next()

the output of the code is

3 StopIteration

Can anyone break this code down for me, Is next() skipping the sequence of the generator? I can't seem to understand the flow here

possible duplicate of [python yield and stopiteration in one loop?](http://stackoverflow.com/questions/6784934/python-yield-and-stopiteration-in-one-loop) — rafaelc, May 14 '15 at 18:16
A link to the function: http://www.usrsb.in/Bouncing-Python-s-Generators-With-A-Trampoline.html — wwii, May 14 '15 at 18:49

Stefan Pochmann · Accepted Answer · 2015-05-14T19:07:38.530

I don't think you found that code on the internet. You seem to be missing the entire point of that blog post, and if you use g = g.next() instead of g.next() like it's shown there, then it makes sense and it works.

But to answer your question, why it goes wrong the way it goes wrong when you do it like you did:

countdown yields exactly one thing, either 0 or another generator. g.next() fetches that one thing, and since you don't do anything with it, it's lost. And then, since g is still the original generator, and it already gave you the one thing it had to give, you get that StopIteration when you ask it for a second thing.

If instead you assign that second generator back to g like you should, then you can ask that generator to give you its one item. And so on.

score 1 · Answer 2 · answered May 14 '15 at 18:22

1

Is this what you want?

def countdown(start):
  if start >= 0:
    yield start
    for i in countdown(start -1):
      yield i

g = countdown(3)
g.next()  #3
g.next()  #2
g.next()  #1
g.next()  #0
g.next()  #StopIteration

problem

yield countdown(start -1) return a generator and iteration then stop

answered May 14 '15 at 18:22

Jose Ricardo Bustos M.

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Actually i did not write the code, I found it on the internet with the answer `3 StopIteration`, so just wanted to understand the flow of the code – Bad_Coder May 14 '15 at 18:27
@Bad_Coder `3` is printed in first instruction, as `start` isn't equal to 3 return `countdown(start -1)` and stop iteration – Jose Ricardo Bustos M. May 14 '15 at 18:30
@Bad_Coder Can you give us the link? – Stefan Pochmann May 14 '15 at 18:39
@StefanPochmann: http://www.usrsb.in/Bouncing-Python-s-Generators-With-A-Trampoline.html – wwii May 14 '15 at 18:53
@wwii Yeah, thanks, I found it as well (easier than I had thought) and wrote an answer myself. – Stefan Pochmann May 14 '15 at 18:54

Generator object with yield

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