关于scala trait和子类重写如何结合的澄清

有人能确认我理解下面的测试吗？似乎当你在Scala中将一个特征添加到一个类中时，如果你引用super.symbol\u defined\u in\u trait，这个特征就会成为该类的超级特征。如果trait本身扩展了基类，那么trait的super将成为原始实现。我把事情搞砸了还是错过了什么大事

class Base {
  def doStuff = println("Base did stuff")
}

trait WrapBase extends Base {
  override def doStuff = {
    println("WrapBase did stuff"); 
    super.doStuff
  }
}

class Sub1 extends Base

class Sub2 extends Base with WrapBase

class Sub3 extends Base {
  override def doStuff = {
    println("Sub3 did stuff"); 
    super.doStuff
  }
}

class Sub4 extends Base with WrapBase {
  override def doStuff = {
    println("Sub4 did stuff"); 
    super.doStuff
  }
}

class Sub5 extends Base with WrapBase {
  override def doStuff = {
    println("Sub5 did stuff"); 
  }
}

此测试打印出：

scala> (new Sub1).doStuff
Base did stuff

scala> (new Sub2).doStuff
WrapBase did stuff
Base did stuff

scala> (new Sub3).doStuff
Sub3 did stuff
Base did stuff

scala> (new Sub4).doStuff
Sub4 did stuff
WrapBase did stuff
Base did stuff

scala> (new Sub5).doStuff
Sub5 did stuff

---

Scala中的遗传包含格有向无环图。可以实例化的最终类是像饼一样构造的：底部总是某个类，然后有一些特性。性状的顺序由属性决定——任何性状都不依赖于高于它的其它性状。或者，等效地，该特性只能在其所有依赖项之上混合

UPD：以下想法是完全错误的：

一个类的主体也是一种特征，可以混合在顶部，也可以混合在中间：

结果将是

实际上，我误解了早期成员定义的功能

此语法仅适用于初始化VAL

这个例子应该是：

trait Sub6Changed extends Base {
  override def doStuff = {
    println("Sub6Changed did stuff") 
    super.doStuff
  }       
} 

object Sub6 extends Base with Sub6Changed with WrapBase

---

另一方面，由于

可能的重复，尽管这涉及多个特征，这涉及混合特征和类别的遗传。虽然我的scala很弱，但也许这些都是内在的、真正光滑的东西。很好知道{}后面跟着with，代码有编译问题

scala> Sub6.doStuff
WrapBase did stuff
Sub6 did stuff
Base did stuff

trait Sub6Changed extends Base {
  override def doStuff = {
    println("Sub6Changed did stuff") 
    super.doStuff
  }       
} 

object Sub6 extends Base with Sub6Changed with WrapBase