Why does the child class does not inherit the method from parent class in python in this example?

Question

I have the following code.

class Foo(object):
     def __init__(self):
         self.__baz = 40
     def foo(self):
         print self.__baz

class Bar(Foo):
     def __init__(self):
         #super(Bar, self).__init__()
         self.__baz = 21
     def bar(self):
         print self.__baz

x = Bar()
x.foo()
x.bar()

I get this error:

Traceback (most recent call last):
  File "classes.py", line 15, in <module>
    x.foo()
  File "classes.py", line 5, in foo
    print self.__baz
AttributeError: 'Bar' object has no attribute '_Foo__baz'

why is the foo method not inherited in Bar.

EDIT: It works fine, if you call super which is commented out.

Not sure, but isn't it because __variablename is a special variable? http://stackoverflow.com/a/1301369/2537322 — meyer9, Sep 27 '13 at 04:07
Not an answer, but adds more detail - if you change it from `__baz` to `baz` both calls print 21. If you call `super()` and leave as `__baz` they print `40` and `21` respectively. — , Sep 27 '13 at 04:09

mgilson · Accepted Answer · 2013-09-27T04:17:34.777

10

The double underscore attributes have their names mangled based on the current/containing namespace. In the function foo, the current namespace is Foo so when python looks up self.__baz, it will actually look for self._Foo__baz due to the name mangling scheme. Since nowhere in Foo have you actually set an __baz attribute, the class has no _Foo__baz attribute (it has a _Bar__baz attribute since you set self.__baz in a method within Bar).

Of course, as you've probably noticed, if you call Foo.__init__(self) in Baz.__init__ (directly or via super), you'll see the problem go away because Foo.__init__ sets __baz (i.e. _Foo__baz).

edited Sep 27 '13 at 04:17

answered Sep 27 '13 at 04:11

mgilson

300,191
65
633
696

so the foo method is not inherited correct? I am a little confused as to what is happening here... – brain storm Sep 27 '13 at 04:21
@user1988876 -- the `foo` method *is* inherited. But when you call it, the attribute that it is trying to access is *missing*. When python sees `self.__baz` inside the definition of the `Bar` class, it changes it to `self._Bar__baz`. When it sees `self.__baz` inside the definition of the `Foo` class, it changes it to `self._Foo__baz`. The whole idea here is so that `self.__baz` inside `Foo` can be distinct from `self.__baz` inside `Bar`. – mgilson Sep 27 '13 at 04:23
So actually when he does call `super(Bar, self).__init__()` child class Bar's object has **two** attributes one is `_Bar__baz` (because he do `self.__baz = 21` in child class init method) and other is `_Foo__baz` from parent class **?** Whereas in error case (if he doesn't call `super(Bar, self).__init__()`) Bar's object has only one attribute that is `_Bar__baz` **?** – Grijesh Chauhan Sep 27 '13 at 04:35
1

@GrijeshChauhan -- Exactly. – mgilson Sep 27 '13 at 04:40
Thanks, Yesterday only I learned inheritance in Python I don't understand a word "*`current/containing namespace`*" .. what does it mean. – Grijesh Chauhan Sep 27 '13 at 04:43
1

It means that `Foo.__init__` is in the `Foo` namespace. `Bar.__init__` is in the `Bar` namespace. When python is creating a class, it creates a new namespace (effectively with the name of the class). All of the functions and variables in that class belong to that namespace. So, when you're in `Foo.__init__` the "current namespace" is `Foo`. – mgilson Sep 27 '13 at 04:47

score 3 · Answer 2 · edited Sep 27 '13 at 04:14

3

When you name variables with double underscore like that in python the member name will be obfuscated. Declaring __baz gives you a member _Bar__baz.

class Bar(Foo):
 def __init__(self):
     #super(Bar, self).__init__()
     self.__baz = 21
 def bar(self):
     print self._Bar__baz

x = Bar()
x.bar()
>>> 21

edited Sep 27 '13 at 04:14

mgilson

300,191
65
633
696

answered Sep 27 '13 at 04:10

Aesthete

18,622
6
36
45

you have changed the definition of function `bar`. I would like to know why it works when the super call is made, as I made in my edit – brain storm Sep 27 '13 at 04:15
+ as you added a trick that is helpful for me, but is it not bad practice to use `_Bar__baz` explicitly **?** – Grijesh Chauhan Sep 27 '13 at 04:39

score 2 · Answer 3 · answered Sep 27 '13 at 04:11

By using the initial double underscores on __baz you requested "name mangling" to make a "private" variable. It's documented here:

http://docs.python.org/2/tutorial/classes.html#private-variables-and-class-local-references

If you change the name from __baz to just baz your code will work as shown.

score 1 · Answer 4 · answered Jul 09 '20 at 18:30

Since python 3.6, we can now use the __init_subclass__ function, which is called automatically before __init__ of the Child.

class Foo(object):
     def __init_subclass__(self):
         self.__baz = 40
     def foo(self):
         print(self.__baz)

class Bar(Foo):
     def __init__(self):
         self.__baz = 21
     def bar(self):
         print(self.__baz)

x = Bar()
x.foo()
x.bar()

output

40
21

Why does the child class does not inherit the method from parent class in python in this example?

4 Answers4