Question

这是一个示例代码，显示了如何在C ++中使用装饰设计模式的示例，虽然代码看起来很简单，但它让我感到困惑的是它为什么正常工作..

代码示例：

#include <iostream>
using namespace std;

// 1. "lowest common denom"
class Widget
{
  public:
    virtual void draw() = 0;
};

class TextField: public Widget
{
    // 3. "Core" class & "is a"
    int width, height;
  public:
    TextField(int w, int h)
    {
        width = w;
        height = h;
    }

    /*virtual*/
    void draw()
    {
        cout << "TextField: " << width << ", " << height << '\n';
    }
};

// 2. 2nd level base class
class Decorator: public Widget  // 4. "is a" relationship
{
    Widget *wid; // 4. "has a" relationship
  public:
    Decorator(Widget *w)
    {
        wid = w;
    }

    /*virtual*/
    void draw()
    {
        wid->draw(); // 5. Delegation
    }
};

class BorderDecorator: public Decorator
{
  public:
    // 6. Optional embellishment
    BorderDecorator(Widget *w): Decorator(w){}

    /*virtual*/
    void draw()
    {
        // 7. Delegate to base class and add extra stuff
        Decorator::draw();
        cout << "   BorderDecorator" << '\n';
    }
};

class ScrollDecorator: public Decorator
{
  public:
    // 6. Optional embellishment
    ScrollDecorator(Widget *w): Decorator(w){}

    /*virtual*/
    void draw()
    {
        // 7. Delegate to base class and add extra stuff
        Decorator::draw();
        cout << "   ScrollDecorator" << '\n';
    }
};

int main()
{
  // 8. Client has the responsibility to compose desired configurations
  Widget *aWidget = new BorderDecorator(new BorderDecorator(new ScrollDecorator(new TextField(80, 24))));
  aWidget->draw();
}

输出产生：

TextField: 80, 24
   ScrollDecorator
   BorderDecorator
   BorderDecorator

我不能得到的部分是，据我所知，代码永远不会到达

cout << "   BorderDecorator" << '\n';
cout << "   ScrollDecorator" << '\n';

的作用是什么

Decorator::draw();

看起来像递归电话..如果我的问题在这里有点混乱，请道歉。

Answer 1

在从Decorator派生的类中，对Decorator::draw()的调用是对基类方法的显式调用，即使它被声明为virtual。它根本不是一个递归调用，而是一种调用基类功能，然后在之后（或之前）添加更多功能的方法。

C ++装饰设计模式 - 代码有效，但不知道为什么

1 个答案: