如何为包含指向(抽象)基类的指针的类定义拷贝构造函数或复制操作符

Table of Contents
    
    #include <iostream>
    using namespace std;
    
    // 如何为包含指向(抽象)基类的指针的类定义拷贝构造函数或复制操作符
    class Shape
    {
    public:
    	Shape()
    	{
    		cout << "Shape::Shape()" << endl;
    	}
    	virtual ~Shape()
    	{
    		cout << "Shape::~Shape()" << endl;
    	}
    	virtual Shape* clone() const = 0;
    };
    
    class Circle : public Shape
    {
    public:
    	Circle()
    	{
    		cout << "Circle::Circle()" << endl;
    	}
    	~Circle()
    	{
    		cout << "Circle::~Circle()" << endl;
    	}
    	virtual Shape* clone() const
    	{
    		return new Circle(*this);
    	}
    };
    
    class Square : public Shape
    {
    public:
    	Square()
    	{
    		cout << "Square::Square()" << endl;
    	}
    	~Square()
    	{
    		cout << "Square::~Square()" << endl;
    	}
    	virtual Shape* clone() const
    	{
    		return new Square(*this);
    	}
    };
    
    class Fred
    {
    public:
    	Fred(Shape* p) : p_(p)
    	{
    		cout << "Fred::Fred()" << endl;
    	}
    	~Fred()
    	{
    		cout << "Fred::~Fred()" << endl;
    		delete p_;
    	}
    	Fred(const Fred& f) : p_(f.p_->clone())
    	{
    	}
    	Fred& operator =(const Fred& f)
    	{
    		if (this != &f)
    		{
    			Shape* p2 = f.p_->clone();
    			delete p_;
    			p_ = p2;
    		}
    		return *this;
    	}
    
    private:
    	Shape* p_;
    };
    
    int _tmain(int argc, _TCHAR* argv[])
    {
    	Shape *circle = new Square();
    	Fred *fred = new Fred(circle);
    
    	delete fred;
    	//delete circle;
    
    	system("pause");
    	return 0;
    }