Question

刚刚找到LinFu - 看起来非常令人印象深刻，但我不能相当看看如何做我想做的事情 - 这是mixin的多重继承（我说的组合/授权）我的VB5 / 6天 - 当我有一个工具来生成繁琐的重复委托代码时 - 它正在寻找一个C＃等价物，我找到了LinFu）。

进一步编辑：澄清组合/授权和mixin的含义。

public class Person : NEOtherBase, IName, IAge 
{

    public Person()
    {

    }

    public Person(string name, int age)
    {
        Name = name;
        Age = age;
    }

    //Name "Mixin" - you'd need this code in any object that wanted to
    //use the NameObject to implement IName
    private NameObject _nameObj = new NameObject();

    public string Name
    {
        get { return _nameObj.Name; }
        set { _nameObj.Name = value; }
    }
    //--------------------

    //Age "Mixin" you'd need this code in any object that wanted to
    //use the AgeObject to implement IAge
    private AgeObject _ageObj = new AgeObject();

    public int Age
    {
        get { return _ageObj.Age; }
        set { _ageObj.Age = value; }
    }
    //------------------

}

public interface IName
{
    string Name { get; set; }
}

public class NameObject : IName
{
    public NameObject()
    {}

    public NameObject(string name)
    {
        _name = name;
    }

    private string _name;
    public string Name { get { return _name; } set { _name = value; } }

}

public interface IAge 
{
    int Age { get; set; }
}

public class AgeObject : IAge
{
    public AgeObject()
    {}

    public AgeObject(int age)
    {
        _age = age;
    }

    private int _age;
    public int Age { get { return _age; } set { _age = value; } }

}

想象一下具有更多属性的对象，在更多的“子类”中使用，你开始看到单调乏味。代码生成工具实际上只是很好 ......

所以，林富...... 下面的mixin示例很好，但我想要一个真正的Person class （如上所述） - LinFu-esque的做法是什么？或者我错过了重点？

编辑：我需要能够使用已经子类化的类来完成此任务。

DynamicObject dynamic = new DynamicObject();

IPerson person = null;

// This will return false
bool isPerson = dynamic.LooksLike<IPerson>();

// Implement IPerson
dynamic.MixWith(new HasAge(18));
dynamic.MixWith(new Nameable("Me"));

// Now that it’s implemented, this
// will be true
isPerson = dynamic.LooksLike<IPerson>();

if (isPerson)
    person = dynamic.CreateDuck<IPerson>();

// This will return “Me”
string name = person.Name;

// This will return ‘18’
int age = person.Age;

林富 - 不太明白我该怎么做

0 个答案: