Python method name with double-underscore is overridden?

Question

Take a look at this.
Note that class B overrides A‘s a() method.

In [1]: class A(object):
   ...:     def __init__(self):
   ...:         self.a()
   ...:     def a(self):
   ...:         print "A.a()"
   ...:         
   ...:         

In [2]: class B(A):
   ...:     def __init__(self):
   ...:         super(B, self).__init__()
   ...:     def a(self):
   ...:         print "B.a()"
   ...:         
   ...:         

In [3]: b = B()
B.a()

No surprises there.

Now, take a look at this.
Note that the method now being overridden is __a().

In [7]: class A(object):
   ...:     def __init__(self):
   ...:         self.__a()
   ...:     def __a(self):
   ...:         print "A.__a()"
   ...:         
   ...:         

In [8]: class B(A):
   ...:     def __init__(self):
   ...:         super(B, self).__init__()
   ...:     def __a(self):
   ...:         print "B.__a()"
   ...:         
   ...:         

In [9]: b = B()
A.__a()

This behaviour surprised me.

Can anyone explain why A.__a() is called instead of B.__a()?

Anything __special__ about __a?

Update:
After reading Sean’s answer I wanted to see if I could override the name mangled method and got this result:

In [11]: class B(A):
   ....:     def __init__(self):
   ....:         super(B, self).__init__()
   ....:     def _A__a(self):
   ....:         print "B._A__a()"
   ....:         
   ....:         

In [12]: b = B()
B._A__a()

Asked By: Jason A

||

Source

Answer 1

keywords with a pattern of __* are class private names.

http://docs.python.org/reference/lexical_analysis.html#reserved-classes-of-identifiers

Quoting:

Names in this category, when used within the context of a class definition, are re-written to use a mangled form to help avoid name clashes between “private” attributes of base and derived classes

Private name mangling (emphasis added):

Private name mangling: When an identifier that textually occurs in a class definition begins with two or more underscore characters and does not end in two or more underscores, it is considered a private name of that class. Private names are transformed to a longer form before code is generated for them. The transformation inserts the class name in front of the name, with leading underscores removed, and a single underscore inserted in front of the class name. For example, the identifier __spam occurring in a class named Ham will be transformed to _Ham__spam. This transformation is independent of the syntactical context in which the identifier is used. If the transformed name is extremely long (longer than 255 characters), implementation defined truncation may happen. If the class name consists only of underscores, no transformation is done.

http://docs.python.org/reference/expressions.html#atom-identifiers

This means that behind the scenes, B.__a() is transformed to something like B._B__a()

Answered By: Sean

Answer 2

In [1]: class A(object):
...:     def __init__(self):
...:         self.a()
...:     def a(self):
...:         print "A.a()"
...:
...:     __str__ = a
...:         

In [2]: class B(A):
...:     def __init__(self):
...:         super(B, self).__init__()
...:     def a(self):
...:         print "B.a()"
...:         
...:         

In [3]: b = B()
        print str(b)
   A.a()

You will need to declare __str__ again in B.

Answered By: bnhelp

Python method name with double-underscore is overridden?

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