C#中的存在类型？

Question

我目前在C#中面临一个问题，我认为可以使用存在类型来解决这个问题。然而，我真的不知道它们是否可以在C#中创建，或者模拟（使用其他构造）。

public interface MyInterface<T>
{
    T GetSomething();
    void DoSomething(T something);
}

public class MyIntClass : MyInterface<int>
{
    int GetSomething()
    {
        return 42;
    }

    void DoSomething(int something)
    {
        Console.Write(something);
    }
}

public class MyStringClass : MyInterface<string>
{
    string GetSomething()
    {
        return "Something";
    }

    void DoSomething(string something)
    {
        SomeStaticClass.DoSomethingWithString(something);
    }
}

public static void DoALotOfThingsTwice(){
    var listOfThings = new List<MyInterface<T>>(){
        new MyIntClass(),
        new MyStringClass();
    };

    foreach (MyInterface<T> thingDoer in listOfThings){
        T something = thingDoer.GetSomething();
        thingDoer.DoSomething(something);
        thingDoer.DoSomething(something);
    }
}

public static void DoALotOfThingsTwice(){
    var listOfThings = new List<∃T.MyInterface<T>>(){
        new MyIntClass(),
        new MyStringClass();
    };

    foreach (∃T.MyInterface<T> thingDoer in listOfThings){
        T something = thingDoer.GetSomething();
        thingDoer.DoSomething(something);
        thingDoer.DoSomething(something);
    }
}

Answer 1

由于doalotofthingstwice不依赖于t，因此可以将其包装在操作中，并将其存储在列表中。

public static Action DoSomethingTwice<T>(this MyInterface<T> i)
{
    return () =>
    {
        T something = i.GetSomething();
        i.DoSomething(something);
        i.DoSomething(something);
    };
}

然后

var listOfThings = new List<Action>() {
    new MyIntClass().DoSomethingTwice(),
    new MyStringClass().DoSomethingTwice()
};