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From: Peter Otten <__peter__@web.de>
Subject: Re: What happens when you 'break' a generator?
Date: Tue, 29 Jul 2014 09:31:58 +0200
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Xref: csiph.com comp.lang.python:75340

Frank Millman wrote:

> Hi all
> 
> Python 3.4.1
> 
> Here is a simple generator -
> 
> def test():
>     print('start')
>     for i in range(5):
>         yield i
>     print('done')
> 
> x = test()
> for j in x:
>     print(j)
> 
> As expected, the output is -
> 
> start
> 0
> 1
> 2
> 3
> 4
> done
> 
> Here I break the loop -
> 
> x = test()
> for j in x:
>     print(j)
>     if j == 2:
>         break
> 
> Now the output is -
> 
> start
> 0
> 1
> 2
> 
> 'done' does not appear, so the generator does not actually terminate. What
> happens to it?
> 
> My guess is that normal scoping rules apply. Using my example, the
> generator is referenced by 'x', so when 'x' goes out of scope, the
> generator is garbage collected, even though it never completed.
> 
> Is this correct?

Yes. In newer Pythons try...finally works, so you can see for yourself:

>>> def f():
...     try:
...             for c in "abc": yield c
...     finally:
...             print("done")
... 
>>> g = f()
>>> for c in g:
...     print(c)
...     if c == "b": break
... 
a
b
>>> del g
done

Also:

>>> g = f()
>>> next(g)
'a'
>>> g.throw(GeneratorExit)
done
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in f
GeneratorExit