How to publicly inherit from a base class but make some of public methods from the base class private in the derived class?

Question

For example, class Base has two public methods: foo() and bar(). Class Derived is inherited from class Base. In class Derived, I want to make foo() public but bar() private. Is the following code the correct and natural way to do this?

class Base {    public:      void foo();      void bar(); };  class Derived : public Base {    private:      void bar(); }; 

Answer 1

Section 11.3 of the C++ '03 standard describes this ability:

11.3 Access declarations
The access of a member of a base class can be changed in the derived class by mentioning its qualified-id in the derived class declaration. Such mention is called an access declaration. The effect of an access declaration qualified-id ; is defined to be equivalent to the declaration using qualified-id

So there are 2 ways you can do it.

Note: As of ISO C++ '11, access-declarations (Base::bar;) are prohibited as noted in the comments. A using-declaration (using Base::bar;) should be used instead.

1) You can use public inheritance and then make bar private:

class Base { public:     void foo(){}     void bar(){} };  class Derived : public Base { private:     using Base::bar; }; 

2) You can use private inheritance and then make foo public:

class Base { public:     void foo(){}     void bar(){} };  class Derived : private Base { public:     using Base::foo; }; 

Note: If you have a pointer or reference of type Base which contains an object of type Derived then the user will still be able to call the member.

Answer 2

There is really no way to do what you want because if you derive publicly from Base, a user of the class will always be able to:

Derived d; Base& b = d; b.bar(); 

It isn't "correct or natural" to make public base class functions inaccessible in a class derived publicy from the base class; instead, the base class interface should be refactored such that those functions are not public or are split into a separate class.