Python bug with [] as default value in functions?

[Dinu Gherman]

On Thu, 12 Apr 2001 13:55:04 +0100, Robin Becker
<robin at jessikat.fsnet.co.uk> wrote:

>Hi I think you are getting involved with one of the standard 'features'
>of python.
>
>eg
>>>> def dingo(n,L=[]):
>...     if n: L.append(n)
>...     return L
>... 
>>>> dingo(1)
>[1]
>>>> dingo(0)
>[1]
>>>>
>
>ie the first call of dingo modified the default argument. Since the
>empty list is mutable that particular version of it gets modified by the
>first call and changes the default for the second.
>
>Hope your tooth ache is better. 


I guess your example shouldn't use 0 (although the result does not 
change, in this case). So you still get this:

>>> def pingo(n, L=[]):
... 	if n: L.append(n)
... 	return L
... 
>>> 
>>> pingo(1)
[1]
>>> pingo(0)
[1]
>>> pingo(2)
[1, 2]
>>> 

Instead, I claim one should get the following: 

>>> pingo(1)
[1]
>>> pingo(0)
[]
>>> pingo(2)
[2]
>>> 

I can work around that 'feature', but I'm pretty sure I don't like it!
It makes a very good candidate for a "Python Obscurities" collection!

This is entirely different from a recursive context, where I *do*
expect the list to be shared across different calls on different lev-
els! And effectively it is, but only when I provide the default value
in the top-level call (adding to the *obscurity* list as well ;-).

Not-much-enlighted'ly,

Dinu


-- 
Dinu C. Gherman
dinu at reportlab dot com
http://www.reportlab.com
................................................................
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