Home/ Questions/Q 101123
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-11T00:42:38+00:00

To illustrate the question check the following code: class MyDescriptor(object): def __get__(self, obj, type=None):

  • 0

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
Share

Leave an answer

You must login to add an answer.

Need An Account,

1 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
    • 0