c++ - 菱形继承(钻石问题)

Question

我正在研究菱形继承(钻石问题)，并认为它适用于各种场景。这是我正在研究的其中之一。

#include <iostream> 
using namespace std;
class MainBase{

    public:
    int mainbase;
    MainBase(int i):mainbase(i){}
    void geta()
    {
        cout<<"mainbase"<<mainbase<<endl;
    }
};
class Derived1: public MainBase{

    public:
    int derived1;
    int mainbase;
    Derived1(int i):MainBase(i),derived1(i) {mainbase = 1;}
    public:
    void getderived1()
    {
        cout<<"derived1"<<derived1<<endl;
    }

};
class Derived2: public MainBase{

    public:
    int derived2;
    int mainbase;
    Derived2(int i):MainBase(i),derived2(i){mainbase = 2;}
    public:
    void getderived2()
    {
        cout<<"derived2"<<derived2<<endl;
    }
};
class Diamond: public Derived1, public Derived2{

    public:
    int diamond;
    int mainbase;
    Diamond(int i,int j, int x):Derived1(j),Derived2(x),diamond(i){mainbase=3;}
    public:
    void getdiamond()
    {
        cout<<"diamond"<<diamond<<endl;
    }
};
int main()
{
    Diamond d(4,5,6);
//    cout<< d.MainBase::mainbase;
    cout<<"tested"<<endl;
    cout<<d.mainbase;
    cout<<d.Derived2::mainbase<<endl;
    cout<<d.Derived1::mainbase<<endl;
    /*cout<<d.Derived2::MainBase::mainbase<<endl;
    cout<<d.Derived1::MainBase::mainbase<<endl;*/
}

我现在想知道如何访问 MainBase 类 mainbase 变量？任何输入。

Answer 1

你做你在那里做的事:

cout<<d.Derived2::MainBase::mainbase<<endl;
cout<<d.Derived1::MainBase::mainbase<<endl;

但是，它可能无法实现您想要实现的目标。也许，您应该使用 virtual 继承？你所拥有的意味着你的对象中将有两个 MainBase 成员拷贝，每个继承轨道一个。

(来自 MSDN)。

When a base class is specified as a virtual base, it can act as an indirect base more than once without duplication of its data members. A single copy of its data members is shared by all the base classes that use it as a virtual base.

可能这样的东西更适合你:

class Derived1: virtual public MainBase{