default method parameter behavior

[Primoz Skale]

<jianbing.chen at gmail.com> wrote in message 
news:879d72da-825f-4b2b-8244-87367647fb6e at f63g2000hsf.googlegroups.com...
>I ran into a similar situation like the following (ipython session).
> Can anyone please explain why the behavior?
> Thanks in advance.
>
> In [11]: def foo(b=[]):
>   ....:     b.append(3)
>   ....:     return b
>   ....:
>
> In [12]: foo()
> Out[12]: [3]
>
> In [13]: foo()
> Out[13]: [3, 3]
>
> In [14]: foo([])
> Out[14]: [3]
>
> In [15]: foo([])
> Out[15]: [3]

I think it has something to do with foo.func_defaults thing :) If parameter 
that is assigned a default value is mutable then foo.func_defaults is 
changed everytime there is no parameter passed to the function.

For example:

>>> def f(b=[]):
 b.append(3)
 return b

>>> f.func_defaults   #default is [], because function was not yet called
([],)
>>> f()                     #no args
[3]
>>> f.func_defaults  #default is now b=[3], because b is mutable object
([3],)
>>> f([])                  #empty; default still == [3]
[3]
>>> f.func_defaults  #as we can see here
([3],)
>>> f()                   #again no args; default is changed to [3,3]
[3, 3]
>>> f.func_defaults
([3, 3],)
>>> f([])                 #no args
[3]
>>> f.func_defaults
([3, 3],)


As *I* understand it, it goes something like this. Because mutable objects 
change globaly if not passed correctly, and because [] is mutable, python 
creates a *def global* object, which is only seen by function that created 
it, with a name b to which it assigns a default value of []. But when this 
default [] is changed in function, it becomes [3], and so on.....

Correct me if I am wrong...

P.