所以我有以下代码:

 /// <summary>
/// The 'Prototype' abstract class
/// </summary>
abstract class ColorPrototype
{
    public abstract ColorPrototype Clone();
}

/// <summary>
/// The 'ConcretePrototype' class
/// </summary>
class Color : ColorPrototype
{
    private int _red;
    private int _green;
    private int _blue;

    // Constructor
    public Color(int red, int green, int blue)
    {
        this._red = red;
        this._green = green;
        this._blue = blue;
    }

    // Create a shallow copy
    public override ColorPrototype Clone()
    {
        Console.WriteLine(
          "Cloning color RGB: {0,3},{1,3},{2,3}",
          _red, _green, _blue);

        return this.MemberwiseClone() as ColorPrototype;
    }
}

/// <summary>
/// Prototype manager
/// </summary>
class ColorManager
{
    private Dictionary<string, ColorPrototype> _colors =
      new Dictionary<string, ColorPrototype>();

    // Indexer
    public ColorPrototype this[string key]
    {
        get { return _colors[key]; }
        set { _colors.Add(key, value); }
    }
}

在主要方法中,我有:
ColorManager colormanager = new ColorManager();

// Initialize with standard colors
colormanager["red"] = new Color(255, 0, 0);

// User clones selected colors
ColorPrototype p1 = colormanager["red"].Clone();
bool isColor = p1 is Prototype.Color;
Color color1 = p1 as Color;

我的问题是,如果在克隆方法中将结果转换为Color(isColor),为什么变量p1的类型为ColorPrototype(return this.MemberwiseClone() as ColorPrototype;为true)?

引用:http://www.dofactory.com/net/prototype-design-pattern

最佳答案



因为即使将p1转换为ColorPrototypep1的运行时类型仍然是Color类型。强制转换引用类型使您可以将T类型视为U类型,但它不会更改基础运行时类型。

10-07 14:46