Question

有人可以确认我理解下面的测试吗？似乎当您在Scala中向类中添加特征时，如果您引用super，该特征将成为该类的super.symbol_defined_in_trait。如果特征本身扩展了基类，则特征super成为原始实现。我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"); 
  }
}

此测试打印出来：

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

Answer 1

Scala中的继承包括晶格导向的非循环图。可以实例化的终极类构造成一个馅饼：底部总是一些类，然后有一些特征。特征的顺序由属性决定 - 任何特征都不依赖于其上的其他特征。或者，等效地，特征可以仅在其所有依赖性之上混合。

UPD：以下提示完全错误：

一个类的主体也是一个“特征”，可以在顶部而不是在中间混合：

object Sub6 extends Base with {
  override def doStuff = {
    println("Sub6 did stuff")
    super.doStuff
  }       
} with WrapBase

结果将是

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

其实我误解了“早期会员定义”功能（http://www.scala-lang.org/old/sites/default/files/sids/nielsen/Tue,%202009-06-02,%2013:16/early-defs.pdf）。

此语法仅适用于初始化val。

示例应该看起来：

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

object Sub6 extends Base with Sub6Changed with WrapBase

P.S。感谢Cloud tech

关于scala特性和子类重写如何组合的澄清

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