What is this Scala syntax?

Question

Take a look at this:

/** takes a Spellbook and returns a Spellbook guaranteeing 
  * that all spells have been loaded from the database. */
 def checkIfSpellsLoaded[S <: Spellbook](spellbook :S) :Option[S { type SpellsLoaded }] =
    if (spellbook.spellsLoaded) Some(spellbook.asInstanceOf[S { type SpellsLoaded }])
    else None
 def checkIfOwnerLoaded[S <: Spellbook](spellbook :S) :Option[S { type OwnerLoaded }] =
     if (spellbook.ownerLoaded) Some(spellbook.asInstanceOf[S { type OwnerLoaded }])
     else None

What is that { type X } doing as part of a type parameter?? What is going on here?

Abstract type member https://docs.scala-lang.org/tour/abstract-type-members.html used in type refinement — Mateusz Kubuszok, Jan 29 '23 at 02:41

Dmytro Mitin · Accepted Answer · 2023-01-29T08:09:01.393

In Scala class members can be def, val and (relevant for us) type

https://docs.scala-lang.org/tour/abstract-type-members.html

https://typelevel.org/blog/2015/07/13/type-members-parameters.html

Scala: Abstract types vs generics

How to work with abstract type members in Scala

Type members are used to create path-dependent types

What is meant by Scala's path-dependent types?

https://docs.scala-lang.org/scala3/book/types-dependent-function.html

If Spellbook has type members SpellsLoaded, OwnerLoaded

trait Spellbook {
  type SpellsLoaded
  type OwnerLoaded

  def spellsLoaded: Boolean
  def ownerLoaded: Boolean
}

then for S <: Spellbook the types S, S { type SpellsLoaded } and S { type OwnerLoaded } are the same

type S <: Spellbook

implicitly[(S { type SpellsLoaded }) =:= S] // compiles
implicitly[S =:= (S { type SpellsLoaded })] // compiles
implicitly[(S { type OwnerLoaded }) =:= S]  // compiles
implicitly[S =:= (S { type OwnerLoaded })]  // compiles

But if Spellbook doesn't have type members SpellsLoaded, OwnerLoaded

trait Spellbook {
  // no SpellsLoaded, OwnerLoaded

  def spellsLoaded: Boolean
  def ownerLoaded: Boolean
}

then the refined types S { type SpellsLoaded } and S { type OwnerLoaded } are just subtypes of S (having those type members)

implicitly[(S { type SpellsLoaded }) <:< S] // compiles
// implicitly[S <:< (S { type SpellsLoaded })] // doesn't compile
implicitly[(S { type OwnerLoaded }) <:< S] // compiles
// implicitly[S <:< (S { type OwnerLoaded })] // doesn't compile

and the refined types S { type SpellsLoaded = ... } and S { type OwnerLoaded = ... } in their turn are subtypes of the former refined types

implicitly[(S {type SpellsLoaded = String}) <:< (S {type SpellsLoaded})] // compiles
// implicitly[(S {type SpellsLoaded}) <:< (S {type SpellsLoaded = String})] // doesn't compile
implicitly[(S {type OwnerLoaded = Int}) <:< (S {type OwnerLoaded})] // compiles
// implicitly[(S {type OwnerLoaded}) <:< (S {type OwnerLoaded = Int})] // doesn't compile

S { type SpellsLoaded } and S { type OwnerLoaded } are shorthands for S { type SpellsLoaded >: Nothing <: Any } and S { type OwnerLoaded >: Nothing <: Any } while S { type SpellsLoaded = SL } and S { type OwnerLoaded = OL } are shorthands for S { type SpellsLoaded >: SL <: SL } and S { type OwnerLoaded >: OL <: OL }.

Casting .asInstanceOf[S { type SpellsLoaded }], .asInstanceOf[S { type OwnerLoaded }] looks like SpellsLoaded, OwnerLoaded are used as phantom types

https://books.underscore.io/shapeless-guide/shapeless-guide.html#sec:labelled-generic:type-tagging (5.2 Type tagging and phantom types)

So you seem to encode in types that the methods checkIfSpellsLoaded, checkIfOwnerLoaded were applied to S.

What is this Scala syntax?

1 Answers1