C＃-接口通用方法约束以匹配派生类型

Question

Let's say I have an interface IA, containing a generic method called Foo. 假设我有一个接口IA，其中包含一个称为Foo的通用方法。

public interface IA {
    int Foo<T>(T otherType);
}

I want T to be the same type as the derived class: 我希望T与派生类具有相同的类型：

class A : IA {
    int Foo(A otherType)
    {
    }
}

I tried following (syntax error): 我尝试了以下操作（语法错误）：

public interface IA {
    int Foo<T>(T otherType) where T : this;
}

How does my constraint need to look like to achieve that? 我的约束看起来如何实现这一目标？

Answer 1

You would have to do it like this: 您必须这样做：

public interface IA<T>
{
    int Foo(T otherType);
}

class A : IA<A>
{
    public int Foo(A otherType)
    {
        return 42;
    }
}

It's the only way to enforce the interface member's generic type. 这是强制接口成员的泛型类型的唯一方法。