javascript - 在扩展 native 类的 ScalaJS 类中调用重载的 super 构造函数

Question

我有这个 JavaScript 类/构造函数:

function Grid(size, tileFactory, previousState, over, won) {
    this.size        = size;
    this.tileFactory = tileFactory;
    this.cells       = previousState ? this.fromState(previousState) : this.empty();
    this.over        = over ? over : false;
    this.won         = won ? won : false;
}

我已经使用这个 ScalaJS 门面映射了:

@js.native
class Grid[T <: Tile](val size: Int,
                      val tileFactory: TileFactory[T],
                      previousState: js.Array[js.Array[TileSerialized]],
                      val over: Boolean,
                      val won: Boolean) extends js.Object {

  val cells: js.Array[js.Array[T]] = js.native

  def this(size: Int, tileFactory: TileFactory[T]) = this(???, ???, ???, ???, ???)

  ...

}

我想扩展 Grid 类，我是这样做的:

@ScalaJSDefined
class ExtendedGrid(
                    override val size: Int,
                    override val tileFactory: TileFactory[Tile],
                    previousState: js.Array[js.Array[TileSerialized]],
                    override val over: Boolean,
                    override val won: Boolean) extends Grid(size, tileFactory, previousState, over, won) {

  ...

}

但现在我还需要为这个 ExtendedGrid 类实现重载的构造函数。

问题是，我该怎么做？

---

理想情况下，我想做类似的事情:

def this(size: Int, tileFactory: TileFactory[Tile]) = super(size: Int, tileFactory: TileFactory[Tile])

但据我了解，这在 Scala 中是不可能的。

为了尝试一下，我尝试简单地复制我在外观中定义的原始重载构造函数:

def this(size: Int, tileFactory: TileFactory[T]) = this(???, ???, ???, ???, ???)

确实编译但显然导致浏览器错误:

Uncaught scala.NotImplementedError: an implementation is missing

然后我尝试了:

def this(size: Int, tileFactory: TileFactory[Tile]) = this(size, tileFactory, this.empty(), false, false)

模仿原始 JavaScript 函数的行为但无济于事。它会产生此错误:

this can be used only in a class, object, or template

Answer 1

您尝试调用的构造函数确实没有重载。它更接近于具有可选值的默认参数。在JS中，默认参数基本都是undefined。因此，您可以对父构造函数进行不同的建模:

@js.native
class Grid[T <: Tile](val size: Int,
                      val tileFactory: TileFactory[T],
                      previousState: js.UndefOr[js.Array[js.Array[TileSerialized]]] = js.undefined,
                      _over: js.UndefOr[Boolean] = js.undefined,
                      _won: js.UndefOr[Boolean] = js.undefined) extends js.Object {
  val over: Boolean = js.native
  val won: Boolean = js.native
  val cells: js.Array[js.Array[T]] = js.native

  ...
}

然后您可以在定义类时模仿相同的结构:

@ScalaJSDefined
class ExtendedGrid(size: Int,
                   tileFactory: TileFactory[Tile],
                   previousState: js.UndefOr[js.Array[js.Array[TileSerialized]]] = js.undefined,
                   _over: js.UndefOr[Boolean] = js.undefined,
                   _won: js.UndefOr[Boolean] = js.undefined) extends Grid(size, tileFactory, previousState, _over, _won) {

  ...

}

顺便说一句，不要使用 override val，因为您将值传递给父构造函数，并且您从父类(super class)中获取了 val。