c++ - 来自基类和派生类的多态赋值运算符

Question

我希望能够将派生对象或基对象复制到派生对象，并且我希望根据复制对象的类型以多态方式选择正确的运算符。

此代码无效，我希望 b1 = (A&)b2; 使用 B & operator= (B const &other) 因为 b2 是 B，但它使用 B & operator= (A const &other):

#include<iostream> 

using namespace std;

class A {
public:
    A & operator= (A const &other) {
        // Here copy A members...
        cout<<"A to A"<<endl;
        return *this;
    }
};

class B: public A {
public: 
    B & operator= (A const &other) {
        A::operator=(other); // Copy A members.
        cout<<"A to B"<<endl;
        return *this;
    }
    B & operator= (B const &other) {
        A::operator=(other); // Copy A members.
        // Here copy B members...
        cout<<"B to B"<<endl; 
        return *this;
    }
};

int main() 
{
    B b1, b2;
    A a2;
    b1 = b2;
    cout<<endl; 
    b1 = (A&)b2;
    cout<<endl; 
    b1 = a2;
    cout<<endl; 
    return 0; 
}

我想我必须做一些虚拟的，但我不知道怎么做。

Answer 1

我希望能够将派生对象或基础对象复制到派生对象，这是设计不佳的表现。

最好争取一种设计，其中类层次结构中只有叶级类是可实例化的。这允许您拥有干净的、非虚拟的、赋值运算符函数只处理正确类型的对象。

class A {
   public:

      // Make A uninstantiable.
      virtual ~A() = 0;

      A & operator= (A const &other) {
         // Here copy A members...
         cout<<"A to A"<<endl;
         return *this;
      }
};

class B: public A {
   public: 

      // Not necessary.
      // B & operator= (A const &other) { ... }

      B & operator= (B const &other) {
         A::operator=(other); // Copy A members.
         // Here copy B members...
         cout<<"B to B"<<endl; 
         return *this;
      }
};