Question

Should’t this be valid C# code?

class A<T> where T : class {

    public void DoWork<K>() where K : T {

        var b = new B<K>(); // <- compile time error
    }
}

class B<U> where U : class {

}

The compiler spits this error:

error CS0452: The type ‘K’ must be a reference type in order to use it as parameter ‘U’ in the generic type or method ‘ConsoleApplication1.B’

Shouldn’t the compiler be able to figure out that K is constraint to be of type T or derived from T so it should obviously be a reference type (T is constrained to be a reference type)?

Answer 1

The constraints are applied when the type parameter is specified. A type has not been specified for K even though K is being specified for U. As U requires its type to be a reference type, the compiler is looking to confirm that K is indeed a reference type, but it cannot. Therefore, you need to explicitly state that it will be.

The specification states in section 4.4.4:

For each where clause, the type argument A that corresponds to the named type parameter is checked against each constraint…

and later:

A compile-time error occurs if one or more of a type parameter’s constraints are not satisfied by the given type arguments.

Since type parameters are not inherited, constraints are never inherited either.

This last point indicates that K will not inherit the constraints from T.

Update
While my conclusions appear correct, my evidence is a little shaky as was clarified in a now deleted response from Eric Lippert’s response. There, Eric stated that the correct part of the specification is:

A type parameter is known to be a
reference type if it has the reference
type constraint or its effective base
class is not object or
System.ValueType.