Question

我有一个基类和两个派生类，我需要将一个指向派生类对象的指针复制到另一个类中，就像示例一样。

class Base
{
public:
    Base(const Base& other);
}

class Derived1 :public Base
{
public:
    Derived1(const Derived& other): Base(other){...};
}

class Derived2: public Base
{
public:
    Derived2(const Derived& other): Base(other){...};
}

main()
{
    Derived 1 d1;
    Derived2 d2(d1)
}

我尝试从Derived 1 ti base（允许向上转换）传递，然后传递给* dynamic_cast * Base到Derived2并调用复制构造函数，但它不起作用。我只需要在两个派生对象之间复制两个对象的Base部分。

Answer 1

如果您的目的是复制Base类部分，请创建一个接收基类的构造函数。

Derived2(const Base& other): Base(other){...};

Derived1(const Base& other): Base(other){...};

两个派生类的C ++副本

1 个答案: