Type must be a reference Type Error When Calling Generic Method

Question

I get a compile error

The type must be a reference type in order to use it as parameter 'T' in the generic type or method

on the 'Derived' class below. How to resolve this?

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

You get this error if you have constrained T to being a class

Answer 3

If you put constrains on a generic class or method, every other generic class or method that is using it need to have "at least" those constrains.