Reputation: 2595
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)
{
}
}
}
Upvotes: 222
Views: 132753
Reputation: 5158
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.
Upvotes: 10
Reputation: 1064324
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
Upvotes: 498