How to derive type parameters with inherited types?

Question

I have the following setup:

public abstract class super { }

public class sub : super { }

public static void Foo<T>(T element, Action<T> action)
    where T : new()
{ }

And I want to do this:

Action<super> superAction = (s) => { };
Foo(new sub(), superAction);

However, this fails because the second line is trying to call Foo<super>instead Foo<sub>. This will work:

Foo<sub>(new sub(), superAction);

• Is it necessary in any case to display the type parameter in this case?
• Why doesn't this infer the correct type in the first place?

EDIT:
The problem boils down to what is possible:

Action<sub> subAction = superAction;

But the compiler does not use this fact for output logic.
So, the answer to question 1:

Foo(new sub(), superAction as Action<sub>);

Question 2, why the compiler does not do this on its own, is still unresolved.

EDIT2:
2:
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Answer 1

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Answer 2

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:

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, :

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