Default inheritance access specifier

Question

If I have for example two classes A and B, such that class B inherits A as follows:

class B: public A

Answer 1

The "type" of inheritance depends on how the class is defined. There are default access specifiers applied to inheritance. From the C++ standard:

[class.access.base]/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. [ Example:

class B { /* ... */ };
class D1 : private B { /* ... */ };
class D2 : public B { /* ... */ };
class D3 : B { /* ... */ };             // B private by default
struct D4 : public B { /* ... */ };
struct D5 : private B { /* ... */ };
struct D6 : B { /* ... */ };            // B public by default
class D7 : protected B { /* ... */ };
struct D8 : protected B { /* ... */ };

Here B is a public base of D2, D4, and D6, a private base of D1, D3, and D5, and a protected base of D7 and D8.  — end example ]