scala - Scala trait mixin 中的方法调用顺序

Question

我有一个程序结构如下:

abstract class IntQueue {
 def get(): Int
 def put(x: Int)
}
trait Doubling extends IntQueue{
 abstract override def put(x: Int) {
   println("In Doubling's put")
   super.put(2*x)
 }
}
trait Incrementing extends IntQueue {
 abstract override def put(x: Int) {
  println("In Incrementing's put")
  super.put(x + 1)
 }
}
class BasicIntQueue extends IntQueue {
 private val buf = new ArrayBuffer[Int]
  def get() = buf.remove(0)
  def put(x: Int) {
   println("In BasicIntQueue's put")
   buf += x
 }
}

当我做:

val incrThendoublingQueue = new BasicIntQueue with Doubling with 
                                            Incrementing
incrThendoublingQueue.put(10)
println(incrThendoublingQueue.get())

输出是:

In Incrementing's put

In Doubling's put

In BasicIntQueue's put

22

我对在这里订购有点困惑。我对这种情况的线性化顺序的理解是:

BasicIntQueue -> Incrementing -> Doubling -> IntQueue -> AnyRef -> Any

所以当我调用 put 时，不应该先调用 BasicIntQueue 的版本吗？

Answer 1

不。这种情况下的线性化是

{<anonymous-local>, Incrementing, Doubling, BasicIntQueue, IntQueue, AnyRef, Any}

你可以:

只需阅读规范并说服自己一定是这样

从规范中实现算法的玩具版本，看看它为各种类定义输出了什么(这有点启发，但主要是为了好玩)

阅读规范

section 5.1.2 of the Spec
准确地告诉您如何计算线性化。您似乎忘记了反转索引 1 ... n在 L(c_n) + ... + L(c_1) .

如果您应用正确的公式，您将获得涉及的特征和基类的以下线性化:

     IntQueue : {IntQueue, AnyRef, Any}
     Doubling : {Doubling, IntQueue, AnyRef, Any}
 Incrementing : {Incrementing, IntQueue, AnyRef, Any}
BasicIntQueue : {BasicIntQueue, IntQueue, AnyRef, Any}

如果您最终组合这些线性化以计算实例化为 incrThendoublingQueue 的匿名本地类的线性化。 :

<anonymous-local-class>, L(Incrementing) + L(Doubling) + L(BasicInt)

您获得了上面已经显示的线性化。因此，应按以下顺序调用这些方法:

递增

加倍

基本

这与实际输出一致。

重新实现线性化算法的乐趣

这实际上是规范的无依赖片段之一，您可以轻松地从头开始实现。这
可以复制带有替换的连接的定义
规范原样，它几乎是可运行的代码(除了带箭头的有趣加号有点难以输入，我希望它作为列表中的中缀运算符):

implicit class ConcatenationWithReplacementOps[A](list: List[A]) {
  def +^->(other: List[A]): List[A] = list match {
    case Nil => other
    case h :: t => 
      if (other contains h) (t +^-> other)
      else h :: (t +^-> other)
  }
}

对类声明建模 C extends C1 with ... with Cn也是
真的很简单:

case class ClassDecl(c: String, extendsTemplate: List[ClassDecl]) {
  def linearization: List[String] = c :: (
    extendsTemplate
      .reverse
      .map(_.linearization)
      .foldLeft(List.empty[String])(_ +^-> _)
  )
}

线性化公式在这里作为一种方法实现。注意 reverse .

规范中给出的示例:

val any = ClassDecl("Any", Nil)
val anyRef = ClassDecl("AnyRef", List(any))
val absIterator = ClassDecl("AbsIterator", List(anyRef))
val richIterator = ClassDecl("RichIterator", List(absIterator))
val stringIterator = ClassDecl("StringIterator", List(absIterator))
val iter = ClassDecl("Iter", List(stringIterator, richIterator))

println(iter.linearization.mkString("{", ", ", "}"))

产生与规范完全相同的输出:

{Iter, RichIterator, StringIterator, AbsIterator, AnyRef, Any}

现在，这是您示例的模型:

val intQueue = ClassDecl("IntQueue", List(anyRef))
val doubling = ClassDecl("Doubling", List(intQueue))
val incrementing = ClassDecl("Incrementing", List(intQueue))
val basicQueue = ClassDecl("BasicIntQueue", List(intQueue))

val incrThendoublingQueue = ClassDecl(
  "<anonymous-local>", 
  List(basicQueue, doubling, incrementing)
)

println(incrThendoublingQueue.linearization.mkString("{", ", ", "}"))    

它产生我上面已经显示的线性化顺序:

{<anonymous-local>, Incrementing, Doubling, BasicIntQueue, IntQueue, AnyRef, Any}

一切似乎都按预期工作，没有理由写信给 Scala 用户。