在Scala特征中使用self类型作为返回类型

Question

我想定义一个特征，它使用具体类的类型，在它的一个抽象方法中将其扩展为返回类型。这在Scala (2.13)中是可能的吗？例如，下面的代码无法编译，因为我找不到绑定ConcreteType的方法

trait Shape
trait Growable {
  def grow() : ConcreteType
}
case class Circle(x : Int, y : Int, size : Int) extends Shape with Growable {
  def grow() : Circle = this.copy(size = size + 1)
}
case class Square(x : Int, y : Int, size : Int) extends Shape with Growable {
  def grow() : Square = this.copy(size = size + 1)
}

我用下面的代码实现了一些类似的功能：

trait Shape
trait Growable[T <: Shape] {
  def grow() : T
}
case class Circle(x : Int, y : Int, size : Int) extends Shape with Growable[Circle] {
  def grow() : Circle = this.copy(size = size + 1)
}
case class Square(x : Int, y : Int, size : Int) extends Shape with Growable[Square] {
  def grow() : Square = this.copy(size = size + 1)
}

然后，此代码的用户将像这样使用它：

val circle : Circle = Circle(0, 0, 10).grow()
val square : Square = Square(0, 0, 10).grow()
// or
val shapes : Seq[Shape] = List(circle, square).map(_.grow())

我希望避免必须通过泛型传递类型，这似乎是多余的。对于如何实现这一点有什么想法吗？

Answer 1

考虑shapeless lenses方法

import shapeless._

sealed trait Shape
case class Circle(x : Int, y : Int, size : Int) extends Shape
case class Square(x : Int, y : Int, size : Int) extends Shape

implicit val circleLens = lens[Circle].size
implicit val squareLens = lens[Square].size

implicit class GrowableShape[T <: Shape](shape: T) {
  def grow()(implicit shapeLense: Lens[T, Int]): T =
    shapeLense.modify(shape)(_ + 1)
}

Circle(0, 0, 10).grow()
Square(0, 0, 10).grow()

哪种输出

res0: Circle = Circle(0,0,11)
res1: Square = Square(0,0,11)

或者，考虑使用vanilla scala的类型类解决方案

trait Growable[T <: Shape] {
  def grow(shape: T): T
}

sealed trait Shape
case class Circle(x : Int, y : Int, size : Int) extends Shape
case class Square(x : Int, y : Int, size : Int) extends Shape

implicit val circleGrowable = new Growable[Circle] {
  def grow(shape: Circle): Circle = shape.copy(size = shape.size + 1)
}

implicit val squareGrowable = new Growable[Square] {
  def grow(shape: Square): Square = shape.copy(size = shape.size + 1)
}

implicit class GrowableShape[T <: Shape](shape: T) {
  def grow()(implicit growable: Growable[T]): T =
    growable.grow(shape)
}

Circle(0, 0, 10).grow()
Square(0, 0, 10).grow()

哪种输出

res0: Circle = Circle(0,0,11)
res1: Square = Square(0,0,11)

Answer 2

在最简单的方式中，由于scala方法/函数结果类型本质上是协方差，即() => Growable是() => Circle或() => Square的超类型，您可以通过在实现中显式提供具体类型来简单地执行以下操作：

  trait Shape
  trait Growable {
    def grow() : Growable
  }
  case class Circle(x : Int, y : Int, size : Int) extends Shape with Growable {
    def grow() : Circle = this.copy(size = size + 1)
  }
  case class Square(x : Int, y : Int, size : Int) extends Shape with Growable {
    def grow() : Square = this.copy(size = size + 1)
  }