Difference in inference between [F <: List[A], A] and [F[_] <: List[A], A]

user_5205022 0
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Consider the difference in inference of type argument A to type constructors in the following two type parameter clauses

$ scala3-repl
scala> def f[F <: List[A], A](as: F) = as
def f[F <: List[A], A](as: F): F

scala> f(List(42))
val res0: List[Int] = List(42)

scala> def f[F[_] <: List[A], A](as: F[A]) = as
def f[F[_$1] <: List[A], A](as: F[A]): F[A]

scala> f(List(42))
val res1: List[Any] = List(42)

Why is type argument A to type constructor F inferred as Any in second case?

type-constructor scala-3 type-inference scala
2 Answers

Based on my interpretation of your definition in the 2nd case, F[_] is a List type constructor, but List[A] has to be an upper bound on any list F[_] can construct, so A has to be Any.

Probably what you were going for is this:

def f[F[_] <: List[_], A](as: F[A]) = as

Or

def f[F[x] <: List[x], A](as: F[A]) = as

especially for cases where x needs to be fixed to multiple constraint parameters (for an example please refer to @user comments below)

In the 1st case F is a concrete type so List[A] isn't an upper-bound on all list, but only on list F, so A doesn't have to be Any and the narrowest inferable type would be Int.

user_120664680

Not a full-fledged answer, just some food for thought: I've attempted to construct a counter-example, but couldn't quite come up with anything that would actually result in unsoundness under the assumption that A would be inferred as the narrowest type. Still, maybe you find it interesting.

Here is a function h with similar constraints, but instead of List, we take slightly different type constructors.

The main idea is that Cc has two separate type parameters:

  • The first is what is meant by _ in F[_]
  • The second one is the one that interacts with A in the <: Lst[A]-constraint

Note that this would not compile if the A was inferred to be the narrowest type (Nothing):

(run in 3.0.0-RC2)

scala> trait Lst[+X]
// defined trait Lst

scala> case class Cc[+I, +X](i: I) extends Lst[X]
// defined case class Cc

scala> type T[+I] = Cc[I, Nothing]
// defined alias type T[+I] = Cc[I, Nothing]

scala> def h[F[_] <: Lst[A], A](as: F[A]) = as
def h[F[_$1] <: Lst[A], A](as: F[A]): F[A]

scala> val xs: T[Int] = Cc(42)
val xs: T[Int] = Cc(42)

scala> h(xs)                                                                                             
val res9: Cc[Int, Nothing] = Cc(42)

Had A been inferred as the narrowest possible type satisfying the constraint of <: Lst[A], then A would be Nothing, and the argument would have to be of type T[Nothing] = Cc[Nothing, Nothing], which is uninhabited.

I think it's interesting, but I don't see why it would actually be bad if it didn't compile.

user_27077920
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