Why invariant generic type parameterized with subclass of upper bounds fails to conform?

Question

Given:

class Invar[T]
trait ExtendsAnyref extends AnyRef
def f(a: Invar[ExtendsAnyref]) = {}

The following is erroneous

scala> val x: Function1[Invar[_ <: AnyRef], Unit] = f
<console>:13: error: type mismatch;
 found   : Invar[ExtendsAnyref] => Unit
 required: Invar[_ <: AnyRef] => Unit
       val x: Function1[Invar[_ <: AnyRef], Unit] = f
                                                    ^

Why?

I understand that in Scala, generic types have by default nonvariant subtyping. Thus, in the context of this example, instances of Invar with different type parameters would never be in a subtype relationship with each other. So an Invar[ExtendsAnyref] would not be usable as a Invar[AnyRef].

But I am confused about the meaning of _ <: AnyRef which I understood to mean "some type below AnyRef in the type hierarchy." ExtendsAnyref is some type below AnyRef in the type hierarchy, so I would expect Invar[ExtendsAnyref] to conform to Invar[_ <: AnyRef].

I understand that function objects are contravariant in their input-parameter types, but since I use Invar[_ <: AnyRef] rather than Invar[AnyRef] I understood, apparently incorrectly, the use of the upper bounds would have the meaning "Invar parameterized with Anyref or any extension thereof."

What am I missing?

"Invar parameterized with Anyref or any extension thereof." This is correct. And it is also the answer to your question: you want a function that takes `Invar[AnyRef]`, but instead are creating a function that takes what you wrote above. That's a different type. — Dima, Jun 12 '16 at 16:40
"you want a function that takes `Invar[AnyRef]`..." But I don't want that. I want a function that takes `Invar` parameterized with **any subclass of `AnyRef`**. You say that what I wrote excludes the "subclass of" part, but if the upper limit `_ <:` doesn't do that, what does it do? — Adam Mackler, Jun 12 '16 at 16:53
Ok, the other way around: you want a function " that takes Invar parameterized with any subclass of AnyRef", but are giving it a function that takes `Invar[AnyRef]` .That's a different type. The property of being a different type is commutative, so it does not matter: if A is not the same type as B, then B is not the same type as A. — Dima, Jun 12 '16 at 17:57
You lost me. Where am I "giving it a function that takes `Invar[AnyRef]`?" The function `f()` takes an argument of type `Invar[ExtendsAnyref]`, and in fact the error message says `found : Invar[ExtendsAnyref] => Unit`. What function are you talking about that takes `Invar[AnyRef]`? — Adam Mackler, Jun 12 '16 at 18:06
Yes, a typo, sorry. You give it `[ExtendsAnyRef]` it wants `[_ extends AnyRef]`. You keep missing the point though: doesn't matter what exact types they are, what matters is that they are different. — Dima, Jun 12 '16 at 18:08

Alexey Romanov · Accepted Answer · 2016-06-12T21:11:15.773

When you write

val x: Function1[Invar[_ <: AnyRef], Unit] = ...

it means x must accept any Invar[_ <: AnyRef]. That is, it must accept Invar[AnyRef], Invar[String], etc. f obviously doesn't: it only accepts Invar[ExtendsAnyref].

In other words, you need to combine your last two paragraphs: because functions are contravariant in argument types, for Function1[Invar[ExtendsAnyref], Unit] to conform to Function1[Invar[_ <: AnyRef], Unit] you'd need Invar[_ <: AnyRef] to conform to Invar[ExtendsAnyref], not vice versa.

If you

want a function that takes Invar parameterized with any subclass of AnyRef

this can be written as Function1[Invar[A], Unit] forSome { type A <: AnyRef }. However, I don't believe there is anything useful you could do with an object of this type, because 1) the only thing you can do with a function is to apply it to an argument, but 2) you don't know what arguments this function accepts.

Why invariant generic type parameterized with subclass of upper bounds fails to conform?

1 Answers1