How can I do this (from Perl) in Python? (closures)
Chris Rebert
clp at rebertia.com
Thu Dec 4 00:31:44 EST 2008
On Wed, Dec 3, 2008 at 9:18 PM, <excord80 at gmail.com> wrote:
> I just came across http://www.perl.com/pub/a/2002/05/29/closure.html
> and wanted to try the "canonical example of closures" in Python. I
> came up with the following, but it fails:
>
> #######################
> #!/usr/bin/env python
>
Depending on your version of Python, you need to do either (A) or (B).
(A) requires Python 3.0 IIRC.
> def make_counter(start_num):
> start = start_num
(B) replace prev line with: start = [start_num]
> def counter():
(A) add: nonlocal start
> start += 1
(B) replace prev line with: start[0] += 1
> return counter
>
> from_ten = make_counter(10)
> from_three = make_counter(3)
>
> print from_ten() # 10
> print from_ten() # 11
> print from_three() # 3
> print from_ten() # 12
> print from_three() # 4
> ####################
>
> The error message is: "UnboundLocalError: local variable 'start'
> referenced before assignment". The same thing happens if I omit start
> and just use start_num directly.
See http://www.python.org/dev/peps/pep-3104/ for more info.
Cheers,
Chris
--
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>
> How can I do it in Python?
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