多重继承和访问不明确的元素

我试图通过多重继承和模板化来实现一些运行时类型信息技巧，但遇到了一个模糊性问题。

首先，我所说的一个非常基本的例子：

#include <memory>
struct A { };
struct B : public A { };
struct C : public B, public A { };
int main() {
std::shared_ptr<A> p0 = std::make_shared<B>(); // possible
std::shared_ptr<A> p1 = std::make_shared<C>(); // impossible
// Here the compiler obviously whines "‘A’ is an ambiguous base of ‘C’"
// but if we're especially naughty, we can circumvent it with a void* cast
// we still can't really use this pointer to anything though.
A* p2 = static_cast<A*>((void*)new C()); 
}

正如评论所暗示的那样，我可能特别顽皮，完全无视这种类型。这种恶作剧会让我直接进入开发人员的地狱，而且无论如何都不会让我随心所欲地使用指针，所以我实际上对std::shared_ptr<B>的基础很满意。但第二个(稍微接近现实(例子表明，这种妥协是不够的：

#include <iostream>
#include <memory>
int current_val = 0;
struct A {
int runtime_val;
A(int v) : runtime_val(v) 
{ 
std::cout << "ACtor with arg: " << v << std::endl; 
}
};
template<typename T>
struct B : public A {
static const int global_val;
B() : A(global_val) {}
};
template<typename T>
const int B<T>::global_val{current_val++};
struct X : public B<X> {};           // Base 
struct C : public X, public B<C> {}; // Leaf
struct D : public X, public B<D> {}; // Leaf
int main() {
std::cout << "Xval: " << B<X>::global_val << std::endl; // 0
std::cout << "Dval: " << B<D>::global_val << std::endl; // 1
std::cout << "Cval: " << B<C>::global_val << std::endl; // 2
// std::shared_ptr<A> px = std::make_shared<X>(); // Impossible because of ambiguity
std::shared_ptr<X> p0 = std::make_shared<X>(); // I'm fine with this, really
std::shared_ptr<X> p1 = std::make_shared<D>();
std::shared_ptr<X> p2 = std::make_shared<C>();
std::cout << "p0 val: " << p0->runtime_val << " (X)" << std::endl; // 0     :)
std::cout << "p1 val: " << p0->runtime_val << " (D)" << std::endl; // 0     :(
std::cout << "p2 val: " << p0->runtime_val << " (C)" << std::endl; // 0    >:(
}

在本例中，我使用B<T>::global_val作为给定类型<T>的运行时类型信息。当程序运行时，我得到以下输出(为了清晰起见，我添加了一些额外的注释(：

Xval: 0
Dval: 1
Cval: 2
ACtor with arg: 0  -- (p0 initialization)
ACtor with arg: 0  -- (p1 initialization (X base))
ACtor with arg: 1  -- (p1 initialization (D base))
ACtor with arg: 0  -- (p2 initialization (X base))
ACtor with arg: 2  -- (p2 initialization (C base))
p0 val: 0 (X)
p1 val: 0 (D)
p2 val: 0 (C)

v表似乎只想指向我的D和C类的X基。如何确保D和C实例的runtime_val将指向继承树的叶，而不是基？

附言：我试过让X基类纯粹是虚拟的，但没有成功。