How to change access level of inherited methods in Scala
scala
Solution
This would break the basic foundation of object-oriented programming - that you can use subclass everywhere where super class was expected (polymorphism).
If it was allowed to narrow down the visibility of a method (e.g. from `public` to `private`) client code receiving an instance of `Lib` would not be allowed to receive `A extends Lib`. Client code expects `someLibFunction()` to be accessible and subclass cannot change that contract.
That being said neither Scala nor any object-oriented language is allowed to narrow down the visibility of any method when subclassing. Note that widening the visibility (e.g. from `protected` to `public` is perfectly possible).
In other words you are not extending an existing API (library). You are creating a completely different API (library) that has a different contract.
Final example: you have a `Vehicle` class that has a `capacity()` and can `drive()`. `Car` can extend `Vehicle` by adding some new capabilities like `refuel()`. But `Container` cannot extend `Vehicle` and hide driving capability. `Container` can contain `Vehicle` (or vice-versa), also `Container` and `Vehicle` might have common parent like `CanHoldCargo`.
Problem
I am extending an existing library (lets call it `Lib`) that I cannot modify directly. Is there any way to make some methods from `Lib` private so that they cannot be accessed outside of subclasses? For example if I have: ``` class A extends Lib { def fun(i: int) = someLibFunction(i) //someLibFunction is inherited from Lib } ``` how can I make `someLibFunction` private in A, despite it being public/protected in Lib?