Why does the val_n function return an older value of self.__n? Seems to me that if obj.__n has been updated to 3, calling self.val__n should return 3 and not 2.
class myClass: def __init__(self,n=1): self.__n=n def val_n(self): return self.__n #create an instance, __n is 2 obj=myClass(2) #update __n to 3 obj.__n=3 #verify obj.__n has changed from 2 to 3 print(obj.__n) #why does this still return 2? print(obj.val_n()) This is based on Python’s name mangling feature. Using a leading double underscore for class attributes means Python will use name mangling.
class X: __attr = None x = X() x.__attr # raises AttributeError x._X__attr # None Python turns the name __attr in class X into _X__attr so that when subclassing, this attribute is less likely to be overwritten if a subclass also defines a __attr but the parent class’ attribute is needed.
When you do x.__n = 3, you’re not overwriting the attribute defined in the class since this has become x._myClass__n to any outside accessor. You are instead creating a new attribute (and Python doesn’t really check x.__n before assigning it to see if it’s already an attribute of the object - which is why no AttributeError is raised).
obj.__n = 3seems to change the value. The answer is that this assignment, as it does not occur inside a method definition, creates a new attribute whose name really is__n, as opposed to_myClass__n. Usedir(obj)to show that attributes by both names exist after the assignment.