C#：重写返回类型

发布于 2024-07-25 17:37:27 字数 815 浏览 6 评论 0原文

有没有办法重写 C# 中的返回类型？如果是的话，如何进行？如果不是，为什么以及推荐的做法是什么？

我的情况是，我有一个带有抽象基类及其后代的接口。我想这样做（好吧，不是真的，但作为一个例子！）：

public interface Animal
{
   Poo Excrement { get; }
}

public class AnimalBase
{
   public virtual Poo Excrement { get { return new Poo(); } }
}

public class Dog
{
  // No override, just return normal poo like normal animal
}

public class Cat
{
  public override RadioactivePoo Excrement { get { return new RadioActivePoo(); } }
}

RadioactivePoo 当然继承自 Poo。

我想要这个的原因是，那些使用 Cat 对象的人可以使用 Excrement 属性，而不必将 Poo 转换为 RadioactivePoo 例如，猫 仍然可能是动物 列表的一部分，其中用户可能不一定知道或关心他们的放射性粪便。希望这是有道理的......

据我所知，编译器至少不允许这样做。所以我想这是不可能的。但您会推荐什么解决方案呢？

原文

Is there way to override return types in C#? If so how, and if not why and what is a recommended way of doing it?

My case is that I have an interface with an abstract base class and descendants of that. I would like to do this (ok not really, but as an example!) :

public interface Animal
{
   Poo Excrement { get; }
}

public class AnimalBase
{
   public virtual Poo Excrement { get { return new Poo(); } }
}

public class Dog
{
  // No override, just return normal poo like normal animal
}

public class Cat
{
  public override RadioactivePoo Excrement { get { return new RadioActivePoo(); } }
}

RadioactivePoo of course inherits from Poo.

My reason for wanting this is so that those who use Cat objects could use the Excrement property without having to cast the Poo into RadioactivePoo while for example the Cat could still be part of an Animal list where users may not necessarily be aware or care about their radioactive poo. Hope that made sense...

As far as I can see the compiler doesn't allow this at least. So I guess it is impossible. But what would you recommend as a solution to this?

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悲喜皆因你 2024-08-01 17:37:27

通用基类怎么样？

public class Poo { }
public class RadioactivePoo : Poo { }

public class BaseAnimal<PooType> 
    where PooType : Poo, new() {
    PooType Excrement {
        get { return new PooType(); }
    }
}

public class Dog : BaseAnimal<Poo> { }
public class Cat : BaseAnimal<RadioactivePoo> { }

编辑：一个新的解决方案，使用扩展方法和标记接口...

public class Poo { }
public class RadioactivePoo : Poo { }

// just a marker interface, to get the poo type
public interface IPooProvider<PooType> { }

// Extension method to get the correct type of excrement
public static class IPooProviderExtension {
    public static PooType StronglyTypedExcrement<PooType>(
        this IPooProvider<PooType> iPooProvider) 
        where PooType : Poo {
        BaseAnimal animal = iPooProvider as BaseAnimal;
        if (null == animal) {
            throw new InvalidArgumentException("iPooProvider must be a BaseAnimal.");
        }
        return (PooType)animal.Excrement;
    }
}

public class BaseAnimal {
    public virtual Poo Excrement {
        get { return new Poo(); }
    }
}

public class Dog : BaseAnimal, IPooProvider<Poo> { }

public class Cat : BaseAnimal, IPooProvider<RadioactivePoo> {
    public override Poo Excrement {
        get { return new RadioactivePoo(); }
    }
}

class Program { 
    static void Main(string[] args) {
        Dog dog = new Dog();
        Poo dogPoo = dog.Excrement;

        Cat cat = new Cat();
        RadioactivePoo catPoo = cat.StronglyTypedExcrement();
    }
}

这样 Dog 和 Cat 都继承自 Animal （正如评论中所述，我的第一个解决方案没有保留继承）。< br>
有必要使用标记接口显式标记类，这很痛苦，但这也许可以给您一些想法...

第二次编辑 @Svish：我修改了代码以明确显示扩展方法不以任何方式强制执行 iPooProvider 继承自 BaseAnimal 的事实。 “更强类型”是什么意思？

What about a generic base class?

public class Poo { }
public class RadioactivePoo : Poo { }

public class BaseAnimal<PooType> 
    where PooType : Poo, new() {
    PooType Excrement {
        get { return new PooType(); }
    }
}

public class Dog : BaseAnimal<Poo> { }
public class Cat : BaseAnimal<RadioactivePoo> { }

EDIT: A new solution, using extension methods and a marker interface...

public class Poo { }
public class RadioactivePoo : Poo { }

// just a marker interface, to get the poo type
public interface IPooProvider<PooType> { }

// Extension method to get the correct type of excrement
public static class IPooProviderExtension {
    public static PooType StronglyTypedExcrement<PooType>(
        this IPooProvider<PooType> iPooProvider) 
        where PooType : Poo {
        BaseAnimal animal = iPooProvider as BaseAnimal;
        if (null == animal) {
            throw new InvalidArgumentException("iPooProvider must be a BaseAnimal.");
        }
        return (PooType)animal.Excrement;
    }
}

public class BaseAnimal {
    public virtual Poo Excrement {
        get { return new Poo(); }
    }
}

public class Dog : BaseAnimal, IPooProvider<Poo> { }

public class Cat : BaseAnimal, IPooProvider<RadioactivePoo> {
    public override Poo Excrement {
        get { return new RadioactivePoo(); }
    }
}

class Program { 
    static void Main(string[] args) {
        Dog dog = new Dog();
        Poo dogPoo = dog.Excrement;

        Cat cat = new Cat();
        RadioactivePoo catPoo = cat.StronglyTypedExcrement();
    }
}

This way Dog and Cat both inherit from Animal (as remarked in the comments, my first solution did not preserve the inheritance).
It's necessary to mark explicitly the classes with the marker interface, which is painful, but maybe this could give you some ideas...

SECOND EDIT @Svish: I modified the code to show explitly that the extension method is not enforcing in any way the fact that iPooProvider inherits from BaseAnimal. What do you mean by "even more strongly-typed"?

C#：重写返回类型

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