Question

因此，我尝试为Scala实现Builder pattern的版本，并且我的返回类型遇到了一些问题。这是我的问题：

abstract class Car() {
  protected var fuelConsumption  = 10.0

  def setFuelConsumption(con: Double): Car = {
    fuelConsumption = con
    this
  }
}

trait HasHorn extends Car {
  protected var hornSound = "Toot!"

  def setHornSound(sound: String): HasHorn = {
    hornSound = sound
    this
  }

}

class ModelT extends Car with HasHorn

// I can do this:
val aCar = new ModelT().setHornSound("Beep!").setFuelConsumption(5.0)
// But not this, because setFuelConsumption returns a Car
val bCar = new ModelT().setFuelConsumption(12.0).setHornSound("Beep!")

所以我的问题是：如何返回对象的运行时类型，以便可以像bCar这样的声明？

Answer 1

基本上，你只需要说Car.setFuelConsumption不会返回Car，而是返回它自己的类型，如下所示：

def setFuelConsumption(con: Double): this.type = {
    fuelConsumption = con
    this
}

Scala：返回构建器模式的运行时对象类型

1 个答案: