Basic list/dictionary question

[Mick Krippendorf]

Ralax wrote:
> On Nov 11, 8:58 pm, Chris Rebert <c... at rebertia.com> wrote:
>> In [2]: def foo(z, a=[]):
>>    ...:     a.append(z)
>>    ...:     return a
>>    ...:
>>
>> In [3]: foo(1)
>> Out[3]: [1]
>>
>> In [4]: foo(2)
>> Out[4]: [1, 2]
>>
>> In [5]: foo(2)
>> Out[5]: [1, 2, 2]
>>
>> In [6]: foo(3)
>> Out[6]: [1, 2, 2, 3]
>>
>> In [7]: foo(4,[])
>> Out[7]: [4]
>>
>> In [8]: foo(5)
>> Out[8]: [1, 2, 2, 3, 5]
> 
> Usefull!

I think Chris post was meant as a warning, but yes, it is useful
sometimes. Here's a less error-prone version:

>>> def bar():
...     def foo(z, a=[]):
...         a.append(z)
...         return a
...     return foo
...
>>> f = bar()
>>> f(1)
[1]
>>> f(2)
[1, 2]
>>> g = bar()
>>> g(3)
[3]
>>> g(4)
[3, 4]

Regards,
Mick.