Scala注册mixin构造函数的结果是不可改变的

Question

我想以编程方式将mixins中发送的值绑定到一个实例,我想知道是否有一个更不可变的方法来执行此操作然后使用隐藏的可变对象.主要是我想将它用于注册表.我目前的方法在施工后并非严格不变,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“超级”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)