Can C# 4.0 variance help me call a base class constructor with an upcast?

Question

I was reading a bit on generic variance and I don't have a full understanding of it yet but I'd like to know if it makes something like the following possible?

class A<T> { }

class B { }

class C : B { }

class My1  {
    public My1(A<B> lessDerivedTemplateParameter)
    {
    }
}

class My2 : My1 {
    public My2(A<C> moreDerivedTemplateParameter)
        : base(moreDerivedTemplateParameter) // <-- compile error here, cannot convert
    {
    }
}

Answer 1

No, because while C inherits from B, A<C> does not inherit from A<B>.

To understand why this is the case, imagine if A<T> were instead List<T>:

class B { }

class C : B { }

class D : B { }

class My1  {
    public My1(List<B> lessDerivedTemplateParameter)
    {
       // This is totally legal
       lessDerivedTemplateParameter.Add(new D());
    }
}

class My2 : My1 {
    public My2(List<C> moreDerivedTemplateParameter)
        // if this were allowed, then My1 could add a D to a list of Bs
        : base(moreDerivedTemplateParameter)
    {
    }
}

Now on the other hand, this is legal:

interface IA<out T> { 
    public T GetSome();
}

class B { }

class C : B { }

class D : B { }

class My1  {
    public My1(IA<B> lessDerivedTemplateParameter)
    {
       // This is totally legal
       var someB = lessDerivedTemplateParameter.GetSome();
    }
}

class My2 : My1 {
    public My2(IA<C> moreDerivedTemplateParameter)
        // This is allowed, because an A<C> only *produces* C's (which are also B's)
        // so the base class (which consumes B's, and doesnt care if they are C's) 
        // can use an IA<C>
        : base(moreDerivedTemplateParameter)
    {
    }
}