call private class function in exec() python

Question

I am having an unexpected issue. This is the simplified code:

class test():
    def __init__(self,name):
        self.name = name

    def __private(self):
        print(self.name)

    def public(self):
        exec("self.__private()")

obj = test('John')
obj.public()

does anyone know how I can make this code work without getting rid of the exec statement?

That's not a private method. There is no such thing as "private" in Python. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Sep 18, 2017 at 16:35
@Ignacio Vazquez-Abrams it is the closest one can get to a private function if you called it like obj.__private() it would return an error — Daan
– Daan, Commented Sep 18, 2017 at 16:40
It is not any sort of private. It is not intended for making things private. It is for preventing attribute name conflicts when using multiple inheritance. — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Sep 18, 2017 at 16:42
Why do you need exec at all? This seems like an XY problem to me. — Kevin
– Kevin, Commented Sep 18, 2017 at 16:43

vaultah · Accepted Answer · 2017-09-18 17:09:29Z

2

Names of attributes starting with two underscores and not ending with two underscores are subject to name mangling.

When you do

def __private(self):
    print(self.name)

in a class definition, you actually create a method called _<ClassName>__private.

You can still access it normally as __private within the class body (using a proper qualifier, such as self, if applicable), but in order to call it via exec or eval, or outside of the class body, you'll have to use its full name:

def public(self):
    exec('self._test__private()')

edited Sep 18, 2017 at 17:09

answered Sep 18, 2017 at 16:43

vaultah

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