c# - 无法在 C# 中将具体类型转换为其接口(interface)的泛型版本

Question

我有以下界面:

public interface INotificationHandler<T>
{
    Task<T> Handle(string msg);
}

还有几个像这样愉快地实现它的类:

public class FooHandler : INotificationHandler<Foo>
{
    public Task<Foo> Handle(string msg) { return Task.FromResult<Foo>(new Foo()); }
}

public class BarHandler : INotificationHandler<Bar>
{
    public Task<Bar> Handle(string msg) { return Task.FromResult<Bar>(new Bar()); }
}

我想在一个集合中保留一组 INotificationHandler 实例，当我收到消息“foo”时，使用 FooHandler，“bar”获取 BarHandler，等等...

var notificationHandlers = new Dictionary<string, INotificationHandler<object>>();
notificationHandlers["foo"] = new FooHandler();
notificationHandlers["bar"] = new BarHandler();
...
public void MessageReceived(string type, string msg)
{
    INotificationHandler<object> handler = notificationHandlers[type];
    handler.Notify(msg).ContinueWith((result) => /* do stuff with a plain object */)
}

但是这无法编译，因为我的泛型没有公共(public)基类型，这是设计使然。任何对象都应该能够从 MessageReceived 中的 INotificationHandler 返回.

Cannot implicitly convert type FooHandler to INotificationHandler<object>. An explicit conversion exists (are you missing a cast?)

我如何使用 INotificationHandler<T>这样我就不需要关心其具体实现的泛型类型？

Answer 1

如果您需要类型安全，您可以使用以下层次结构。

public interface INotificationHandler
{
    Task<object> Handle(string msg);
}

public abstract BaseHandler<T> : INotificationHandler
{
    Task<object> INotificationHandler.Handle(string msg)
    {
        return Handle(msg);
    }

    public abstract Task<T> Handle(string msg);
}

public class FooHandler : BaseHandler<Foo>
{
    public override Task<Foo> Handle(string msg) { return Task.FromResult<Foo>(new Foo()); }
}

public class BarHandler : BaseHandler<Bar>
{
    public override Task<Bar> Handle(string msg) { return Task.FromResult<Bar>(new Bar()); }
}

var notificationHandlers = new Dictionary<string, INotificationHandler>();
notificationHandlers["foo"] = new FooHandler();
notificationHandlers["bar"] = new BarHandler();
...
public void MessageReceived(string type, string msg)
{
    INotificationHandler handler = notificationHandlers[type];
    handler.Notify(msg).ContinueWith((result) => /* do stuff with a plain object */)
}