Scala : Why can't we do super. val？

Question

我正在从 JavaTpoint 练习这段代码，用于在 Scala 中学习继承。但是我无法从值初始化为零的 Vehicle 类访问成员 Bike 。我尝试通过父类(super class)型引用，但它仍然显示覆盖的值。为什么它不允许访问父类(super class)字段并指向覆盖的子类字段(速度)。这是代码和输出。
提前致谢。

class Vehicle {
  val speed = 0
  println("In vehicle constructor " +speed)
  def run() {
    println(s"vehicle is running at $speed")
  }
}

class Bike extends Vehicle {
  override val speed = 100
  override def run() {
    super.run()
    println(s"Bike is running at $speed km/hr")
  }
}

object MainObject3 {
  def main(args:Array[String]) {
    var b = new Bike()
    b.run()
    var v = new Vehicle()
    v.run()
    var ve:Vehicle=new Bike()
    println("SuperType reference" + ve.speed)
    ve.run()
  }
}

Answer 1

类似问题的答案在这里overriding-vals-in-scala ，或在这里 cannot-use-super-when-overriding-values说:Scala compiler does not allow to use super on a val
这是为什么？上面最后一个链接中的讨论指向:SI-899 .那里的第一条评论如下:it was changed so that traits could override vals to be more uniform with classes