Question

Code in test.py:

class Base(object):
    def __init__(self, l=[]):
        self.l = l

    def add(self, num):
        self.l.append(num)

    def remove(self, num):
        self.l.remove(num)

class Derived(Base):
    def __init__(self, l=[]):
        super(Derived, self).__init__(l)

Python shell session:

Python 2.6.5 (r265:79063, Apr  1 2010, 05:22:20) 
[GCC 4.4.3 20100316 (prerelease)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import test
>>> a = test.Derived()
>>> b = test.Derived()
>>> a.l
[]
>>> b.l
[]
>>> a.add(1)
>>> a.l
[1]
>>> b.l
[1]
>>> c = test.Derived()
>>> c.l
[1]

I was expecting “C++-like” behavior, in which each derived object contains its own instance of the base class. Is this still the case? Why does each object appear to share the same list instance?

Answer 1

You’re making a common Python newcomer mistake.

See my answer here:
How should I declare default values for instance variables in Python?

Briefly explained, Python interprets the class definitions only once. That means everything declared in the __init__() method is only created once. Or, in another words, your [] list default argument is only made once.

Then self.l = l assigns a reference to the same instance every time you create a new class, hence the behaviour you weren’t expecting.

The Pythonic way is this (partial code):

def __init__(self, arg=None):
    if arg is None:
        arg = []
    self.arg = arg

Also, you should consider using a better naming convention than l, which is hard to read and might be mistaken as 1 or |.