Default class inheritance access

Question

Suppose I have a base and derived class:

class Base
{
    public:
    virtual void Do();
}

class Derived:Base
{
    public:
    virtual void Do();
}

int main()
{
    Derived sth;
    sth.Do(); // calls Derived::Do OK
    sth.Base::Do(); // ERROR; not calls Based::Do 
}

as seen I wish to access Base::Do through Derived. I get a compile error as "class Base in inaccessible" however when I declare Derive as

class Derived: public Base

it works ok.

I have read default inheritance access is public, then why I need to explicitly declare public inheritance here?

Answer 1

From standard docs, 11.2.2

In the absence of an access-specifier for a base class, public is assumed when the derived class is defined with the class-key struct and private is assumed when the class is defined with the class-key class.

So, for structs the default is public and for classes, the default is private...

Examples from the standard docs itself,

class D3 : B { / ... / }; // B private by default

struct D6 : B { / ... / }; // B public by default

Answer 2

You might have read something incomplete or misleading. To quote Bjarne Stroustrup from "The C++ programming Language", fourth Ed., p. 602:

In a class, members are by default private; in a struct, members are by default public (§16.2.4).

This also holds for members inherited without access level specifier.

A widespread convention is, to use struct only for organization of pure data members. You correctly used a class to model and implement object behaviour.

Answer 3

The default inheritance level (in absence of an access-specifier for a base class )for class in C++ is private. [For struct it is public]

class Derived:Base

Base is privately inherited so you cannot do sth.Base::Do(); inside main() because Base::Do() is private inside Derived

Answer 4

The default type of the inheritance is private. In your code,

class B:A

is nothing but

class B: private A