为什么c#是这样设计的?

根据我的理解,一个接口只描述行为,并且服务于描述实现接口的类的契约义务。

如果类希望在共享方法中实现这种行为,为什么不应该呢?

以下是我想到的一个例子:

// These items will be displayed in a list on the screen.
public interface IListItem {
  string ScreenName();
  ...
}

public class Animal: IListItem {
    // All animals will be called "Animal".
    public static string ScreenName() {
        return "Animal";
    }
....
}

public class Person: IListItem {

    private string name;

    // All persons will be called by their individual names.
    public string ScreenName() {
        return name;
    }

    ....

 }

当前回答

举一个例子,我既没有接口方法的静态实现,也没有Mark Brackett介绍的“所谓的类型方法”:

当从数据库存储中读取数据时,我们有一个通用的DataTable类来处理从任何结构的表中读取数据。所有特定于表的信息都放在每个表的一个类中,每个表还保存来自DB的一行的数据,并且必须实现IDataRow接口。IDataRow中包含了要从数据库中读取的表的结构描述。DataTable在从DB中读取数据之前必须向IDataRow请求数据结构。目前看起来是这样的:

interface IDataRow {
  string GetDataSTructre();  // How to read data from the DB
  void Read(IDBDataRow);     // How to populate this datarow from DB data
}

public class DataTable<T> : List<T> where T : IDataRow {

  public string GetDataStructure()
    // Desired: Static or Type method:
    // return (T.GetDataStructure());
    // Required: Instantiate a new class:
    return (new T().GetDataStructure());
  }

}

每个表只需要读取一次GetDataStructure,实例化一个实例的开销是最小的。但是,在这种情况下它会很好。

其他回答

I think the question is getting at the fact that C# needs another keyword, for precisely this sort of situation. You want a method whose return value depends only on the type on which it is called. You can't call it "static" if said type is unknown. But once the type becomes known, it will become static. "Unresolved static" is the idea -- it's not static yet, but once we know the receiving type, it will be. This is a perfectly good concept, which is why programmers keep asking for it. But it didn't quite fit into the way the designers thought about the language.

因为它不可用,所以我采用如下所示的方式使用非静态方法。虽然不太理想,但至少对我来说,我找不到比这更有意义的方法了。

public interface IZeroWrapper<TNumber> {
  TNumber Zero {get;}
}

public class DoubleWrapper: IZeroWrapper<double> {
  public double Zero { get { return 0; } }
}

因为接口是继承结构,静态方法继承不好。

从概念上讲,接口没有理由不能定义包含静态方法的契约。

对于当前的c#语言实现,限制是由于允许继承基类和接口。如果“类SomeBaseClass”实现了“接口ISomeInterface”和“类SomeDerivedClass: SomeBaseClass, ISomeInterface”也实现了接口,实现接口方法的静态方法将会编译失败,因为静态方法不能与实例方法具有相同的签名(实例方法将出现在实现接口的基类中)。

静态类在功能上与单例类相同,并且具有与单例类相同的目的,只是语法更简洁。因为单例可以实现接口,所以静态的接口实现在概念上是有效的。

因此,这可以简单地归结为c#的实例名称冲突和跨继承的同名静态方法的限制。c#没有理由不能“升级”以支持静态方法契约(接口)。

假设你在问为什么你不能这样做:

public interface IFoo {
    void Bar();
}

public class Foo: IFoo {
    public static void Bar() {}
}

This doesn't make sense to me, semantically. Methods specified on an interface should be there to specify the contract for interacting with an object. Static methods do not allow you to interact with an object - if you find yourself in the position where your implementation could be made static, you may need to ask yourself if that method really belongs in the interface. To implement your example, I would give Animal a const property, which would still allow it to be accessed from a static context, and return that value in the implementation.

public class Animal: IListItem {
    /* Can be tough to come up with a different, yet meaningful name!
     * A different casing convention, like Java has, would help here.
     */
    public const string AnimalScreenName = "Animal";
    public string ScreenName(){ return AnimalScreenName; }
}

对于更复杂的情况,您总是可以声明另一个静态方法并委托给它。在尝试给出一个例子的过程中,我想不出您在静态和实例上下文中都要做一些非平凡的事情的任何理由,因此我将给您一个FooBar blob,并将其作为它可能不是一个好主意的指示。

您可以将类的静态方法和非静态方法看作是不同的接口。调用时,静态方法解析为单例静态类对象,而非静态方法解析为所处理的类的实例。所以,如果你在一个接口中使用静态和非静态方法,你实际上是在声明两个接口,而实际上我们想要接口被用来访问一个内聚的东西。