该类型必须是引用类型，才能在通用类型或方法中将其用作参数“ T”

Question

I'm getting deeper into generics and now have a situation I need help with. I get a compile error on the 'Derived' class below as shown in the subject title. I see many other posts similar to this one but I'm not seeing the relationship. Can someone tell me how to resolve this?

using System;
using System.Collections.Generic;


namespace Example
{
    public class ViewContext
    {
        ViewContext() { }
    }

    public interface IModel
    {
    }

    public interface IView<T> where T : IModel 
    {
        ViewContext ViewContext { get; set; }
    }

    public class SomeModel : IModel
    {
        public SomeModel() { }
        public int ID { get; set; }
    }

    public class Base<T> where T : IModel
    {

        public Base(IView<T> view)
        {
        }
    }

    public class Derived<SomeModel> : Base<SomeModel> where SomeModel : IModel
    {

        public Derived(IView<SomeModel> view)
            : base(view)
        {
            SomeModel m = (SomeModel)Activator.CreateInstance(typeof(SomeModel));
            Service<SomeModel> s = new Service<SomeModel>();
            s.Work(m);
        }
    }

    public class Service<SomeModel> where SomeModel : IModel
    {
        public Service()
        {
        }

        public void Work(SomeModel m)
        {

        }
    }
}

Answer 1

I can't repro, but I suspect that in your actual code there is a constraint somewhere that T : class - you need to propagate that to make the compiler happy, for example (hard to say for sure without a repro example):

public class Derived<SomeModel> : Base<SomeModel> where SomeModel : class, IModel
                                                                    ^^^^^
                                                                 see this bit

Answer 2

如果将T限制为一class则会出现此错误

Answer 3

如果将约束放在通用类或方法上，则使用它的所有其他通用类或方法都必须“至少”具有这些约束。