返回实现相同接口的不同泛型 [英] return different generics that implement same interface

问题描述

考虑到一种方法

static IEnumerable<IComparable> q()
{
   return new List<string>();
}

我正在尝试实现相同的目标，但是在我自己的班级上，结果收到转换错误cs0266

I am trying to achieve the same but on my own classes and as a result i receive casting error cs0266

我试图以这种方式强制转换 return(Common< Message>)new A(); ，但结果为 InvalidCastException

I tried to cast this way return (Common<Message>)new A(); but it results InvalidCastException

interface Common<T> where T : Message
{
    T Source { get; }
    void Show();
}
interface Message
{
    string Message { get; }
}
class AMsg : Message
{
    public string Message => "A";
}
class A : Common<AMsg>
{
    public AMsg Source => new AMsg();
    public void Show() { Console.WriteLine(Source.Message); }
}
static Common<Message> test()
{
    return new A(); //CS0266
}

该方法如何返回实现相同接口的不同泛型?

How can the method return different generics that implement same interface?

推荐答案

IEnumerable 是

IEnumerable is covariant which is why the first block of code works. To do the same thing you need to make your T type paramater covariant by adding the out modifier:

interface Common<out T> where T : Message
{
    T Source { get; }
    void Show();
}

现在您可以编写如下代码:

Now you can write code like this:

Common<Message> x = new A();

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