scala - Why is parameter in contravariant position?

Question

I'm trying to use a covariant type parameter inside a trait to construct a case-class like so:

trait MyTrait[+T] {
  private case class MyClass(c: T)
}

compiler says:

error: covariant type T occurs in contravariant position in type T of value c

I then tried the following but it also didn't work:

trait MyTrait[+T] {
  private case class MyClass[U <: T](c: U)
}

the error this time is:

error: covariant type T occurs in contravariant position in type >: Nothing <: T of type U

Could somebody explain why the T is in a covariant position here and suggest a solution for this problem? Thx!

Answer 1

This is a fundamental feature of object-oriented programming that doesn't get as much attention as it deserves.

Suppose you have a collection C[+T]. What the +T means is that if U <: T, then C[U] <: C[T]. Fair enough. But what does it mean to be a subclass? It means that every method should work that worked on the original class. So, suppose you have a method m(t: T). This says you can take any t and do something with it. But C[U] can only do things with U, which might not be all of T! So you have immediately contradicted your claim that C[U] is a subclass of C[T]. It's not. There are things you can do with a C[T] that you can't do with a C[U].

Now, how do you get around this?

One option is to make the class invariant (drop the +). Another option is that if you take a method parameter, to allow any superclass as well: m[S >: T](s: S). Now if T changes to U, it's no big deal: a superclass of T is also a superclass of U, and the method will work. (However, you then have to change your method to be able to handle such things.)

With a case class, it's even harder to get it right unless you make it invariant. I recommend doing that, and pushing the generics and variance elsewhere. But I'd need to see more details to be sure that this would work for your use case.