Class and Method level generic type constraints interaction

Question

Consider the following class:

public class DerivedClassPool<TBase> where TBase : class
{
    public TBase Get(Type componentType)
    {
        // Not important, but you get the idea
        return Activator.CreateInstance(componentType) as TBase;
    }

    public TDerived SomeMethod<TDerived>() where TDerived : TBase
    {
        return Get(typeof(TBase)) as TDerived;
    }
}

Note that I've restricted the TBase generic class argument to be a class: where TBase : class
I've also restricted the TDerived generic method argument to be TBase or something derived from that: where TDerived : TBase.

I get an error on the as TDerived line:

The type parameter 'TDerived' cannot be used with the 'as' operator because it does not have a class type constraint nor a 'class' constraint

I understand that to prevent the error I need to add the constraint class, so I'd get:

where TDerived : class, TBase

Why do I have to do this when TBase is already constrained to be a class and TDerived is constrained to be a TBase or derived from it?

Answer 1

UPDATE: This question was the subject of my blog on September 19th, 2011. Thanks for the great question!

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Why do i have to do this when TBase is already constrained to be a class and TDerived is constrained to be a TBase or derived from it?

Because a value type can be derived from a reference type. int is derived from reference types object and System.ValueType and implements a number of interfaces. That doesn't make int a reference type.

It is perfectly legal for you to call SomeMethod<int> on an instance of DerivedClassPool<object> because int is derived from object.

Now, there are cases where your criticism would be warranted. One can construct situations in which two type parameters are related in such a way that both of them logically can only be reference types, but only one of them is classified by the language as "known to be of reference type".

As an exercise to the reader: can you find one? It might be necessary to carefully read section 10.1.5 of the specification for a precise definition of "known to be a reference type".

Answer 2

Because TBase could be an interface and therefore TDerived could be a value type.