python-descriptors

In Python 3 class A(object): attr = SomeDescriptor() ... def somewhere(self): # need to check is type of self.attr is SomeDescriptor() desc = self.__class__.__dict__[attr_name] return isinstance(desc, SomeDescriptor) Is there better way to do it? I don't like this self.__class__.__dict__ stuff A.attr causes Python to call SomeDescriptor().__get__(None, A) so if you have SomeDescriptor.__get__ return self when inst is None , then A.attr will return the descriptor: class SomeDescriptor(): def __get__(self, inst, instcls): if inst is None: # instance attribute accessed on class, return self

According to Python's documentation , Data descriptors with __set__() and __get__() defined always override a redefinition in an instance dictionary. I have no problem understanding this sentence, but can someone clarify for me why such a rule is in place? After all, if I want to override an attribute in an instance dictionary, I already need to do that explicitely ( inst.__dict__["attr"] = val ), as a naive inst.attr = val would call the descriptor's __set__ method, which would (usually) not override the attribute in the instance dictionary. edit: just to make it clear, I understand what is

Take this code for example: class SomeClass(): def a_method(self): pass print(SomeClass.a_method is SomeClass.a_method) # Example 1: False print(SomeClass.a_method == SomeClass.a_method) # Example 2: True print(SomeClass().a_method is SomeClass().a_method) # Example 3: False print(SomeClass().a_method == SomeClass().a_method) # Example 4: False Example 1: I would have guessed that they were the same object. Does Python make a copy of the method each time it is referenced? Example 2: Expected. Example 3: Expected, since they are different objects. Example 4: Why doesn't this output match