On Dec 3, 11:18 pm, [EMAIL PROTECTED] wrote:
> I just came acrosshttp://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
>
> def make_counter(start_num):
> start = start_num
> def counter():
> start += 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.
>
> How can I do it in Python?
In Python 3.0, you can do this:
def make_counter(start):
def counter():
nonlocal start
start += 1
return start # I assume you forgot this above
return counter
In Python 2.x, there is no nonlocal statement, so the best you can do
is improvise something like this:
def make_counter(start_num):
start = [start_num]
def counter():
start[0] += 1
return start[0]
return counter
You can access variables from an enclosing scope starting from Python
2.2 (from 2.1 with from __future__ import nested_scopes), but you
could not rebind those variables until Python 3.0 came out today.
Carl Banks
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