在双钻石中到达场或遥远基类的方法是什么？

Question

在“经典”钻石问题中，如下所示（virtual前面没有public D的一个，在class C和class B后面），可以解决使用名称范围运算符::产生歧义，例如在类A的构造函数中：

/* 
 *   D                            D   D
 *  / \   which without 'virtual' |   |
 * B   C      is actually:        B   C
 *  \ /                            \ /
 *   A                              A
 */
#include <iostream>
using namespace std;
class D                      { public: char d = 'D';};
class C : public D           { public: char c = 'C';};
class B : public D           { public: char b = 'B';};
class A : public B, public C { public: char a = 'A'; A();};

A::A() {
    cout << B::d; //This works! B's d, inherited from D.
    cout << C::d; //This works! C's d, inherited from D.
    //cout << D::d;     //This doesn't work (ambiguous)
    //cout << B::D::d;  //Doesn't work either though.
    //cout << C::D::d;  //Doesn't work either though.
}

int main() {
    A a;
    cout << endl;
    return 0;
}

现在考虑像这样的双钻石：

/* 
 *   G                          G  G G  G
 *  / \                         |  | |  |
 * E   F                        E  F E  F
 *  \ /                          \ / \ /
 *   D                            D   D
 *  / \   which without 'virtual' |   |
 * B   C      is actually:        B   C
 *  \ /                            \ /
 *   A                              A
 */
#include <iostream>
using namespace std;
class G                      { public: char g = 'G';};
class E : public G           { public: char e = 'E';};
class F : public G           { public: char f = 'F';};
class D : public E, public F { public: char d = 'D';};
class C : public D           { public: char c = 'C';};
class B : public D           { public: char b = 'B';};
class A : public B, public C { public: char a = 'A'; A();};

A::A() {
    cout << /* How do I reach any of the two e's or f's 
               or any of the four g's?*/
}

int main() {
    A a;
    cout << endl;
    return 0;
}

[如何准确地到达从E，F和G继承的字段？对我而言，实际上最合乎逻辑的是以下内容。

cout << B::D::d;
cout << B::D::E::e; 
cout << B::D::F::e; 
cout << B::D::E::G::g; 
cout << B::D::F::G::g; 

cout << C::D::d;
cout << C::D::E::e; 
cout << C::D::F::e; 
cout << C::D::E::G::g; 
cout << C::D::F::G::g; 

但是（使用g ++）它们都产生'X' is an ambiguous base of 'A'.形式的错误。

有人可以解释为什么这行不通，什么是正确的方法？我想念什么？

Answer 1

让我们从您遇到的错误开始。 'X' is an ambiguous base of 'A'.

Answer 2

不起作用的原因：