How can i make a method available only from within the class

Question

Good evening, i need an advice, googling i couldn't find a proper direction. I need to make a method available only within the class (i.e other methods or functions), if called from the program as a method of the object referring to the class i want:

the method to be invisible/not available to the intellisense
if i'm stubborn, and code it anyway, must raise an error.

Attaching a screenshot to make it more clear. Any advice is appreciated, Thank you.

Screenshot of the problem

Python doesn't let you absolutely hide a function from a developer. But the closest thing to it would be a "dunder" method. It will hide it from intellisense and then make it harder for a developer to see it. — Rashid 'Lee' Ibrahim, Dec 23 '21 at 18:36
afaik dunder methods refer to double leading _and_ trailing underscores, stuff like `__init__`. "Private" methods are marked with only leading double underscores. — Puff, Dec 23 '21 at 18:39

score 2 · Accepted Answer · answered Dec 23 '21 at 18:36

2

There's no private methods in python. Common usage dictates to precede a method that's only supposed to be used internally with one or two underscores, depending on the case. See here: What is the meaning of single and double underscore before an object name?

answered Dec 23 '21 at 18:36

Puff

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OK solved, you put me on the rightgood path, i hid to pycharm's intellisense the method via the double underscore (VS Code still see it though, but meh...) this way the method is still callable as myobj._myclass__hismethod() but i could forbid it's execution via the accessify module as someone suggested me. – Alexandro Dec 23 '21 at 22:35

score 2 · Answer 2 · answered Dec 23 '21 at 18:58

As others have mentioned there are no private methods in Python. I also don't know how to make it invisible for intelisense (probably there is some setting), but what you could theoretically do is this:

import re


def make_private(func):
    def inner(*args, **kwargs):
        name = func.__name__
        pattern = re.compile(fr'(.*)\.{name}')
        with open(__file__) as file:
            for line in file:
                lst = pattern.findall(line)
                if (lst and not line.strip().startswith('#')
                   and not all(g.strip() == 'self' for g in lst)):
                    raise Exception()
        return func(*args, **kwargs)
    return inner


class MyClass:

    @make_private
    def some_method(self):
        pass

    def some_other_method(self):
        self.some_method()


m = MyClass()
# m.some_method()
m.some_other_method()

It (make_private) is a decorator which basically when you call the function it is decorating, it first reads the entire file line by line and tries to find if in all of the file this method is called without being prefixed with self.. So if it is not then it is considered to be called from outside the class and an Exception is raised (probably add some message to it tho).

Issues could start once you have multiple files and this wouldn't entirely prevent someone from calling it if they really wanted for example if they did it like this:

self = MyClass()
self.some_method()

But mostly this would raise an exception.

this is a hacky atrocity, I love it +1 – Fynn Dec 23 '21 at 19:22 — Fynn, Dec 23 '21 at 19:22

score 1 · Answer 3 · answered Dec 23 '21 at 22:42

OK Solved, to hide the method to the ide's Intellisense i added the double underscore (works fine with pycharm, not with vscode) then i used the accessify module to prevent forced execution calling myobj._myclass__somemethod()

from accessify import private

class myclass:
    @private
    def __somemethod(self)

How can i make a method available only from within the class

3 Answers3