CODEWITHSUNDEEP
scalatypes
Sim
Sim

Reputation: 13528

AnyRef equivalent for an AnyVal type

Given a type T, is there an idiomatic Scala way to describe the AnyRef equivalent of T (let's call it ARE[T]). For example,

The purpose of the question is to allow implementing many methods such as:

def foo[A](...): ARE[A]

while avoiding the naive def foo[A <: AnyRef](...): A + overloading for Boolean, Byte, Char, Double, Float, Int, Long and Short.

Upvotes: 1

Views: 310

Answers (1)

Kolmar
Kolmar

Reputation: 14224

The standard way to implement a type computation like this is to create a typeclass:

sealed trait Box[T] {
  type Out
  def apply(t: T): Out
}

object Box {
  type Aux[T, ARE] = Box[T] { type Out = ARE }

  def make[T, ARE](f: T => ARE): Box.Aux[T, ARE] = new Box[T] {
    type Out = ARE
    def apply(t: T) = f(t)
  }

  implicit val int: Box.Aux[Int, java.lang.Integer] = make(Int.box)
  implicit val long: Box.Aux[Long, java.lang.Long] = make(Long.box)
  implicit val short: Box.Aux[Short, java.lang.Short] = make(Short.box)
  implicit val byte: Box.Aux[Byte, java.lang.Byte] = make(Byte.box)
  implicit val char: Box.Aux[Char, java.lang.Character] = make(Char.box)
  implicit val float: Box.Aux[Float, java.lang.Float] = make(Float.box)
  implicit val double: Box.Aux[Double, java.lang.Double] = make(Double.box)
  implicit val boolean: Box.Aux[Boolean, java.lang.Boolean] = make(Boolean.box)
  implicit val unit: Box.Aux[Unit, scala.runtime.BoxedUnit] = make(Unit.box)
  implicit def anyRef[T <: AnyRef]: Box.Aux[T, T] = make(identity)

  def box[T](t: T)(implicit are: Box[T]): are.Out = are(t)
}

This can be used like any other typeclass. For example, you can compute the type ARE with the help of Box.Aux in your own functions:

def box2[T, ARE](t: T)(implicit box: Box.Aux[T, ARE]): ARE = box(t)

Scala accepts the output of Box.box when AnyRef is expected:

scala> def foo[T <: AnyRef](anyRef: T): T = anyRef
foo: [T <: AnyRef](anyRef: T)T

scala> foo(10)
<console>:13: error: inferred type arguments [Int] do not conform to method foo's type parameter bounds [T <: AnyRef]
       foo(10)
       ^
<console>:13: error: type mismatch;
 found   : Int(10)
 required: T
       foo(10)
           ^

scala> foo(Box.box(10))
res1: Box.int.Out = 10

And Scala also knows the exact ARE type returned from Box.box:

scala> def bar[T](t: T)(implicit ev: T =:= java.lang.Integer) = ev
bar: [T](t: T)(implicit ev: =:=[T,Integer])=:=[T,Integer]

scala> bar(10)
<console>:13: error: Cannot prove that Int =:= Integer.
       bar(10)
          ^

scala> bar(Box.box(10))
res2: =:=[Box.int.Out,Integer] = <function1>

Upvotes: 2

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