Scala 不可变地注册 mixin 构造函数结果

Question

我想以编程方式将 mixin 中发送的值绑定(bind)到实例，我想知道是否有一种更不可变的方法可以使用隐藏的可变对象来执行此操作。我主要想将其用于注册表。我当前的方法在构建后并不是严格不可变的，有什么建议吗？

trait Numbers {
  lazy val values = holding
  private var holding = Set.empty[Int]
  protected def includes(i:Int) {
    holding += i
  }
}

trait Odd extends Numbers{
  includes(1)
  includes(3)
  includes(5)
  includes(7)
  includes(9)
}

trait Even extends Numbers {
  includes(2)
  includes(4)
  includes(6)
  includes(8)
}

这给出了我想要的结果

val n = new Odd with Even
println(n.values)

Set(5, 1, 6, 9, 2, 7, 3, 8, 4)

Answer 1

方法重写怎么样？然后您可以在特征线性化中引用“ super ”对象，

trait Numbers {
  def holding = Vector[Int]()
  lazy val values = holding
}

trait Odd extends Numbers {
  override def holding = super.holding ++ Vector(1,3,5)
}

trait Even extends Numbers {
  override def holding = super.holding ++ Vector(0,2,4)
}

(new Odd with Even).values // Vector(1, 3, 5, 0, 2, 4)
(new Even with Odd).values // Vector(0, 2, 4, 1, 3, 5)