[Tutor] beginner: using optional agument in __init__ breaks my code

[Karl Pflästerer]

On 25 Jun 2006, brb.shneider at yahoo.de wrote:

[...]
> This code works as intended. Now my idea is to provide
> an optional argument to the constructor. So I change
> it to:
>
>     def __init__(self, q =[]):
>         self.queue = q
>
> Now, something very strange happens: 
>
>>>> a = Queue()
>>>> b = Queue()
>>>> a.insert(12)
>>>> print b
> [12]
>>>> 
>
> Why do a and b share the same data? "self.queue" is
> supposed to be an instance variable. What does may
> change of the __init__ method do here?

The values of optional arguments are only once evaluated (when Python
reads the definition). If you place there mutable objects like e.g. a
list most of the time the effect you see is not what you want. So you
have to write it a bit different.

     def __init__(self, q = None):
         if not q: q = []
         self.queue = q

or

     def __init__(self, q = None):
         self.queue = q or []


Now you get a fresh list for each instance.



   Karl
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