如何理解这段C++代码中的动态绑定？

Question

这是我的代码。我需要帮助使这个动态绑定工作。

#include<iostream>
#include<conio.h>
using namespace std;

class Shape {
    protected:
    double x,y;
    public:
    void get_data(double a,double b=3.14)   {
        x=a;
        y=b;
    }   
    virtual void display_area(){
    };
};

class Rectangle : public Shape  {
    public:
    void display_area() {
        cout<<"\nArea of Rectangle : "<<x*y<<" units.\n";
    }
};
class Circle : public Shape {
    public:
    void display_area() {
        cout<<"\nArea of Circle : "<<y*x*x<<" units.\n";
    }
};
class Triangle : public Shape   {
    public:
    void display_area() {
        cout<<"\nArea of Triangle : "<<0.5*x*y<<" units.\n";
    }
};

main()  {
    Shape *ptr;
    char opt,wait;
    double a,b;
    do{
        system("cls");
        cout<<"\n1.Area of Rectangle\n2.Area of Circle\n3.Area of Triangle\n4.Exit\n\nEnter your Option : ";
        opt=getche();
        switch(opt) {
            case '1':
                *ptr = new Rectangle;
                cout<<"\nEnter Length : ";
                cin>>a;
                cout<<"\nEnter Width  : ";
                cin>>b;
                ptr->get_data(a,b);
                ptr->display_area();
                break;
            case '2':
                *ptr = new Circle;
                cout<<"\nEnter Radius : ";
                cin>>a;
                ptr->get_data(a);
                ptr->display_area();
                break;
            case '3':
                *ptr = new Triangle;
                cout<<"\nEnter Base   : ";
                cin>>a;
                cout<<"\nEnter Height : ";
                cin>>b;
                ptr->get_data(a,b);
                ptr->display_area();
                break;
            case '4':
                cout<<"\n\nGoodBye ! Have a Nice Day !\n";
                exit(0);
            default:
                cout<<"\n\nEnter a Valid Option !\n";
        }
        cout<<"\n\nPress Any Key to Continue....";
        wait=getche();
    }while(opt!='4');
}

我得到的错误是

5   7  [Note] Shape& Shape::operator=(const Shape&)

5   7  [Note] no known conversion for argument 1 from 'Rectangle*' to 'const Shape&'

55  10 [Error] no match for 'operator=' (operand types are 'Shape' and 'Circle*')

55  10 [Note] candidate is:

5   7  [Note] Shape& Shape::operator=(const Shape&)

5   7  [Note] no known conversion for argument 1 from 'Circle*' to 'const Shape&'

62  10 [Error] no match for 'operator=' (operand types are 'Shape' and 'Triangle*')

62  10 [Note] candidate is:

5   7  [Note] Shape& Shape::operator=(const Shape&)

5   7  [Note] no known conversion for argument 1 from 'Triangle*' to 'const Shape&'

Answer 1

你正在分配 ptr 不正确。当你将一个对象分配给它时，你不应该对它进行减引用。

改变这个

*ptr = ...;

对此

ptr = ...;

同时，你也泄露了你用 new. 你要打电话 delete ptr; 当你使用完每个对象时。这也就意味着你需要给 Shape 以便正确调用正确的子代析构器。

class Shape {
... 
public:
    virtual ~Shape() {}
...
};

Answer 2

我想补充一下 不要使用原始指针.如果你使用了智能庞特。std::unique_ptr 那么你就可以避免错误。而且你不需要考虑删除对象。

std::unique_ptr<Shape> ptr;
ptr = std::make_unique<Rectangle>();