c++ - 虚拟继承(钻石)——为什么我需要从派生类最多的类向上转换为基类

Question

考虑以下几点:

#include <iostream>
#include <string>
using namespace std;


class A {
public:
   A(const char* sName) //conversion constructor
           : _sName(sName) {cout<<"(1)"<<endl;} ;
   A(const A& s) {cout<<"(2)"<<endl;} //copy constructor
   virtual ~A() {cout<<"(3)"<<endl;} //destructor
    void f1() {cout<<"(4)"<<endl; f2();} //Notice two commands!
    virtual void f2() =0;
  private:
    string _sName;
  };



  class B1: virtual public A {
  public:
     B1(const char* sAName, const char* sSName)
             : _sName1(sAName), A(sSName) {cout<<"(5)"<<endl;}
     B1(const B1& b1) : A(b1) {cout<<"(6)"<<endl;}
     ~B1() {cout<<"(7)"<<endl;}
     virtual void f1() {cout<<"(8)"<<endl;}
     virtual void f2() {cout<<"(9)"<<endl; f3();}
     virtual void f3() {cout<<"(10)"<<endl;}
  private:
     string _sName1;
  };



  class B2: virtual public A {
  public:
     B2(const char* sAName, const char* sSName)
             : _sName2(sAName), A(sSName) {cout<<"(11)"<<endl;}
     B2(const B2& b2) : A(b2) {cout<<"(12)"<<endl;}
     ~B2() {cout<<"(13)"<<endl;}
     virtual void f3() {f1(); cout<<"(14)"<<endl;}
  private:
      string _sName2;
  };

  class C: public B1, public B2 {
  public:
         C () : A(" this is A ") , B1(" this is " , " B1 ") , B2 (" this is " , " B2 ") {}
         C (const C& c) :  A(c) , B1(c) , B2(c) {}
         ~C() {cout<<"(15)"<<endl;}
         virtual void f1() {A::f1(); cout<<"(16)"<<endl;}
         void f3 () {cout<<"(17)"<<endl;}
  };


  int main() {
      /* some code */
      return 0;
  }

如您所见，我在 class C 中添加了 C 的 Ctor(构造函数)的实现。我不清楚的是，如果 B1 在其 Ctor 中为我完成这项工作，为什么我还需要从 C 到 A 的向上转换？意思是，如果我将 C 的 Ctor 写为:

C () : A(" this is A ") , B1(" this is " , " B1 ") , B2 (" this is " , " B2 ") {}

为什么我不能写:

C () : B1(" this is " , " B1 ") , B2 (" this is " , " B2 ") {}

谢谢，罗嫩

Answer 1

虚拟基础的构造函数之前被称为非虚拟分析器。在您的示例中，B1 ctor 无法调用 C 的虚拟基构造函数，因为稍后将调用 B1 ctor 本身。