underscoreio/scala

Works

sealed trait Sum[+A, +B] {
  def flatMap[AA >: A, C](f: B => Sum[AA, C]): Sum[AA, C] =
    this match {
      case Failure(v) => Failure(v)
      case Success(v) => f(v)
    }
}
final case class Failure[A](value: A) extends Sum[A, Nothing]
final case class Success[B](value: B) extends Sum[Nothing, B]

Doesn't Work

sealed trait Sum[+A, +B] {
  def flatMap[C](f: B => Sum[A, C]): Sum[A, C] =
    this match {
      case Failure(v) => Failure(v)
      case Success(v) => f(v)
    }
}
final case class Failure[A](value: A) extends Sum[A, Nothing]
final case class Success[B](value: B) extends Sum[Nothing, B]

I don't get it. If I take f: B => Sum[A, C] and construct it from trait Function1[-I, +O]{ def apply(i: I): O} it seems B is the contravariant argument while Sum[A, C] is the covariant term in this function. How does it end up that A is the contravariant term also, what does this mean 👇 ?

"It is declared with type B => Sum[A, C] and thus a supertype is covariant in B and contravariant in A and C."

I expected it to be B is the contraviant term, why? it's the argument to function f,meaning it's B that has to be type bounded not A and Sum[A, C] to be covariant why? it's the result type of f. But I'm clearly wrong, what I'm I missing here?

I didn't really get the argument for contravariant terms in a function why supertypes?

Source:

"Back to flatMap, the function f is a parameter, and thus in a contravariant position. This means we accept supertypes of f. It is declared with type B => Sum[A, C] and thus a supertype is covariant in B and contravariant in A and C. B is declared as covariant, so that is fine. C is invariant, so that is fine as well. A on the other hand is covariant but in a contravariant position. Thus we have to apply the same solution we did for Box above."

Excerpt From: Noel Welsh and Dave Gurnell. “Essential Scala”.