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Re: local variable 'a' referenced b

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On 10/03/2012 01:54 AM, Demian Brecht wrote:
> On 12-10-02 07:26 PM, Dave Angel wrote:
>>
>> if you're stuck with Python2.x, you can use a mutable object for a, and
>> mutate it, rather than replace it.  For example,
>>
>>
>> def foo():
>>       a = [3]
>>       def bar():
>>           b=2
>>           a.append(b)   #this mutates a, but doesn't assign it
>>           print (a)
>>           a[0] += b  #likewise, for a number within the list
>>           print (a)
>>       bar()
>>
>> That should work in either 2.x or 3.2
>>
>
> Alternatively, you can restructure your code by simply adding a
> parameter to bar(). Nice thing about this is that if you ever move
> bar() out into another module, then you don't have to worry about
> documenting the side effects on 'a' so users (including yourself)
> aren't confused later:
>
> >>> def foo():
> ...     a = 1
> ...     def bar(n):
> ...             b = 2
> ...             return n + b
> ...     a = bar(a)
> ...     print a
> ...
> >>> foo()
> 3
>
>

One problem with short examples is they mask the reason for the code to
be structured that way.  I assumed that the OP was really talking about
a closure, and that sharing that variable was deliberate.  I seldom
write nested functions otherwise.

-- 

DaveA

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Re: local variable 'a' referenced b Dave Angel <d@davea.name> - 2012-10-03 02:08 -0400

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