No error for Python private variable

Question

I encountered a strange issue while using python. consider the below code:

class Foo:
    __bar=None

foo=Foo()
foo.bar=10
print(foo.bar)

I beleive I should get an error here. But I get the output as 10. I get the same output even if I change the code as:

foo=Foo()
foo.__bar=10
print(foo.__bar)

can someone explain why I am not getting an error here? Also what should I do to get an error? Edited for additional clarification:

class Foo:
    def __init__(self):
        self.__bar=10
    def get_bar(self):
        return self.__bar

foo=Foo()
foo.__bar=20
print(foo.__bar)
print(foo.get_bar())

Can someone explain why I am getting the answer as 20 and 10? I thought I will get an error for trying to access a private instance variable outside the class. Thanks for the help.

Why do you believe that you should get an error? What are you doing wrong, in your opinion? — Hashmush, Feb 18 '15 at 05:46
Underscores for "private" variables is merely a convention. Nothing more. Also, `bar` and `__bar` are not the same property. Nothing in Python says that this should trigger anything. — deceze, Feb 18 '15 at 05:46
This question is not a duplicate, becuase it is not asking about leading double-underscore in general, but about a specific usage, and none of the answers in the other question cover the answer to this question. — shx2, Feb 18 '15 at 06:03

score 4 · Accepted Answer · answered Feb 18 '15 at 06:10

First of all, you should read the answers to the related quesion.

Second of all, don't confuse between class-members and instance members. Your definition of the class Foo includes a class member named __bar, but assigning to foo.__bar adds an instance-member which overshadows it (and not overwrites it). Those are different things.

Now, when you assign (or in general, access) a leading-double-underscore attribute from "outside" (in your example, using foo.__bar), the name does not get mangaled, and everything works without extra magic. However, the standard way of accessing leading-double-underscore attributes is from within methods of the class (using self.__bar), and that is when the magic happens.

To demonstrate:

class Foo(object): pass

f = Foo()
foo.__bar = 10
foo.__bar
=> 10
foo.__dict__.keys()
=> dict_keys(['__bar'])

class Foo(object):
    def __init__(self):
       self.__bar = 10

foo = Foo()
foo.__dict__.keys()
=> dict_keys(['_Foo__bar'])  # mangaled!
foo.__bar  # cannot access from "outside"
=> AttributeError

Thanks @shx2. Can you kindly calify what do you mean by "class member" and "instance variable"? Do you mean that class member like "static variable" as in Java? — codingsplash, Feb 18 '15 at 08:57
I have edited my question to add a new scenario..can you look into it please? — codingsplash, Feb 18 '15 at 11:35
Just take a look at `foo.__dict__`. That should make it clear. — shx2, Feb 18 '15 at 12:21

No error for Python private variable

1 Answers1