Why one child class alone flings error ? does order matter?

Question

I was reading about diamond inheritance problem in java . So wanted to know how other languages have solved and found that c++ has solved it . And I found this How does virtual inheritance solve the "diamond" (multiple inheritance) ambiguity? . Then I tried this code in online c++ compiler . I tried

#include <iostream>

using namespace std;
class A                     { public: void eat(){ cout<<"A";} }; 
class B: virtual public A   { public: void eat(){ cout<<"B";} }; 
class C: virtual public A   { public: void eat(){ cout<<"C";} }; 
class D: public         B,C { public: void eat(){ cout<<"D";} }; 

int main()
{
    A *a = new D(); 
    a->eat(); 

}

It worked . Then I wanted to invoke B's eat and changed A *a = new D(); to

 B *a = new D(); 

and this too worked .But when I tried

C *a = new D(); 

It threw and error

error: 'C' is an inaccessible base of 'D' C *a = new D();

Why with C alone this error ? It is even public . Then why its not accessible ?

So what I did was I changed the order of inheritance to

class D: public         C,B { public: void eat(){ cout<<"D";} }; 

And now C *a = new D(); passed but B *a = new D(); failed .

Why is it so ?

Is the order important ?

When I googled I found this stackoverflow link , but found that it is for constructors . I could not understand why this case fails .

Please forgive if my question is stupid , as I am damn new to C++ .I am used to only java where there is no multiple inheritance .

Answer 1

D inherits publicly from B , but privately from C . Use this if you want to inherit publicly from both:

class D: public B, public C { public: void eat(){ cout<<"D";} };