Why must virtual base classes be constructed by the most derived class?

Question

The following code won't compile:

class A { public:     A(int) {} };  class B: virtual public A { public:     B(): A(0) {} };  // most derived class class C: public B { public:     C() {} // wrong!!! }; 

If I call A's constructor in C's constructor initialization list, that is:

// most derived class class C: public B { public:     C(): A(0) {} // OK!!! }; 

it does work.

Apparently, the reason is because virtual base classes must always be constructed by the most derived classes.

I don't understand the reason behind this limitation.

Answer 1

Because it avoids this:

class A { public:     A(int) {} };  class B0: virtual public A { public:     B0(): A(0) {} };  class B1: virtual public A { public:     B1(): A(1) {} };  class C: public B0, public B1 { public:     C() {} // How is A constructed? A(0) from B0 or A(1) from B1? }; 

Answer 2

Because in the class hierarchy having a virtually inherited base class, the base class would/may be shared by multiple classes (in diamond inheritance for example, where the same base class is inherited by multiple classes). It means, there would be only one copy of the virtually-inherited base class. It essentially means, the base class must be constructed first. It eventually means the derived class must instantiate the given base class.

For example:

class A; class B1 : virtual A; class B2 : virtual A; class C: B1,B2 // A is shared, and would have one copy only.