Why do managed attributes just work for class attributes and not for instance attributes in python?

Question

To illustrate the question check the following code:

class MyDescriptor(object):
  def __get__(self, obj, type=None):
    print "get", self, obj, type
    return self._v
  def __set__(self, obj, value):
    self._v = value
    print "set", self, obj, value
    return None

class SomeClass1(object):
  m = MyDescriptor()

class SomeClass2(object):
  def __init__(self):
    self.m = MyDescriptor()

x1 = SomeClass1()
x2 = SomeClass2()

x1.m = 1000
# ->  set <__main__.MyDescriptor object at 0xb787c7ec> <__main__.SomeClass1 object at 0xb787cc8c> 10000
x2.m = 1000 # I guess that this overwrites the function. But why?
# ->
print x1.m
# -> get <__main__.MyDescriptor object at 0xb787c7ec> <__main__.SomeClass1 object at 0xb787cc8c> <class '__main__.SomeClass1'> 10000
print x2.m
# -> 10000

1. Why doesn't x2.m = 1000 not call the __set__-function? It seems that this overwrites the function. But why?
2. Where is _v in x1? It is not in x1._v

Answer 1

To answer your second question, where is _v?

Your version of the descriptor keeps _v in the descriptor itself. Each instance of the descriptor (the class-level instance SomeClass1, and all of the object-level instances in objects of class SomeClass2 will have distinct values of _v.

Look at this version. This version updates the object associated with the descriptor. This means the object (SomeClass1 or x2) will contain the attribute _v.

class MyDescriptor(object):
  def __get__(self, obj, type=None):
    print "get", self, obj, type
    return obj._v
  def __set__(self, obj, value):
    obj._v = value
    print "set", self, obj, value

Answer 2

You should read this and this.

It overwrites the function because you didn't overload the __set__ and __get__ functions of SomeClass but of MyDescriptor class. Maybe you wanted for SomeClass to inherit MyDescriptor? SomeClass1 prints the "get" and "set" output because it's a static method AFAIK. For details read the upper links.

Answer 3

I found _v of x1: It is in SomeClass1.__dict__['m']._v

For the version suggested by S.Lott within the other answer: _v is in x1._v