在scala中混合类型参数和抽象类型

Question

我试图使用前面问题的答案来实现一个小图库.我们的想法是将图形视为一种选择,其中顶点包裹集合元素.

我想使用抽象类型来表示Vertex和Edge类型(因为类型安全),我想使用类型参数来表示集合元素的类型(因为我想在实例化时轻松定义它们).

但是,在尝试我能想到的最基本的例子时,我遇到了编译错误.这是一个例子:

package graph

abstract class GraphKind[T] {

  type V <: Vertex[T]
  type G <: Graph[T]

  def newGraph(): G

  abstract class Graph[T] extends Collection[T]{
    self: G =>
    def vertices(): List[V]
    def add(t: T): Unit
    def size(): Int
    def elements(): Iterator[T]
  }

  trait Vertex[T] {
    self: V =>
      def graph(): G
      def value(): T
  }

}

这是基本的实现:

class SimpleGraphKind[T] extends GraphKind[T] {

  type G = GraphImpl[T]
  type V = VertexImpl[T]

  def newGraph() = new GraphImpl[T]

  class GraphImpl[T] extends Graph[T] {
    private var vertices_ = List[V]()
    def vertices = vertices_
    def add( t: T ) {  vertices_ ::= new VertexImpl[T](t,this) }
    def size() = vertices_.size
    def elements() = vertices.map( _.value ).elements
  }

  class VertexImpl[T](val value: T, val graph: GraphImpl[T]) extends Vertex[T] {
    override lazy val toString = "Vertex(" + value.toString + ")"
  }

}

在尝试编译时,我得到:

/prg/ScalaGraph/study/Graph.scala:10: error: illegal inheritance;
 self-type GraphKind.this.G does not conform to Collection[T]'s selftype Collection[T]
  abstract class Graph[T] extends Collection[T]{
                              ^
/prg/ScalaGraph/study/Graph.scala:33: error: illegal inheritance;
 self-type SimpleGraphKind.this.GraphImpl[T] does not conform to   SimpleGraphKind.this.Graph[T]'s selftype SimpleGraphKind.this.G
  class GraphImpl[T] extends Graph[T] {
                         ^
/prg/ScalaGraph/study/Graph.scala:36: error: type mismatch;
 found   : SimpleGraphKind.this.VertexImpl[T]
 required: SimpleGraphKind.this.V
    def add( t: T ) {  vertices_ ::= new VertexImpl[T](t,this) }
                                 ^
/prg/ScalaGraph/study/Graph.scala:38: error: type mismatch;
 found   : Iterator[T(in class SimpleGraphKind)]
 required: Iterator[T(in class GraphImpl)]
    def elements() = vertices.map( _.value ).elements
                                         ^
/prg/ScalaGraph/study/Graph.scala:41: error: illegal inheritance;
 self-type SimpleGraphKind.this.VertexImpl[T] does not conform to   SimpleGraphKind.this.Vertex[T]'s selftype SimpleGraphKind.this.V
  class VertexImpl[T](val value: T, val graph: GraphImpl[T]) extends Vertex[T] {
                                                                 ^
5 errors found

我完全不知道这些错误的含义......但是,如果我在实现中专门化类型T(class SimpleGraphKind extends GraphKind[Int]我只得到第一个错误.

你有什么想法吗？

Answer 1

用以下结果编译-explaintypes:

<console>:11: error: illegal inheritance;
 self-type GraphKind.this.G does not conform to Collection[T]'s selftype Collection[T]
         abstract class Graph[T] extends Collection[T]{
                                         ^
    GraphKind.this.G <: Collection[T]?
      Iterable[T] <: Iterable[T]?
        T <: T?
          T <: Nothing?
            <notype> <: Nothing?
            false
            Any <: Nothing?
              <notype> <: Nothing?
              false
            false
          false
          Any <: T?
            Any <: Nothing?
              <notype> <: Nothing?
              false
            false
          false
        false
      false
      GraphKind.this.Graph[T] <: Iterable[T]?
        Iterable[T] <: Iterable[T]?
          T <: T?
            T <: Nothing?
              <notype> <: Nothing?
              false
              Any <: Nothing?
                <notype> <: Nothing?
                false
              false
            false
            Any <: T?
              Any <: Nothing?
                <notype> <: Nothing?
                false
              false
            false
          false
        false
      false
    false

现在,我即将写作我不明白怎么T <: T可能是假的 - 它几乎就像T被定义了两次,当然,这是整个问题.这里:

abstract class GraphKind[T] { 

  type V <: Vertex[T] 
  type G <: Graph[T] 

  def newGraph(): G 

  abstract class Graph[T] extends Collection[T]{ 

好的,类GraphKind是参数化的T,类型G必须是a Graph[T].现在,class Graph也被参数化,并且它的参数也被调用T.为了防止混淆,让我们重写它:

  abstract class Graph[T2] extends Collection[T2]{
    self: G =>
    def vertices(): List[V]
    def add(t: T2): Unit
    def size(): Int
    def elements(): Iterator[T2]
  }

请注意,这与您所写的完全相同.我只是为类型参数使用了一个不同的名称,因此它不会与T参数化相混淆GraphKind.

所以,这是逻辑:

G <: Graph[T]
Graph[T2] <: Collection[T2]
Graph[T2] <: G  // self type

这意味着

Graph[T2] <: Graph[T]

而且,因为Graph扩展Collection:

Collection[T2] <: Collection[T]

但不能保证这是真的.我不明白为什么在没有继承时问题不会出现.固定:

abstract class GraphKind[T] {

  type V <: Vertex
  type G <: Graph

  def newGraph(): G

  abstract class Graph extends Collection[T]{
    self: G =>
    def vertices(): List[V]
    def add(t: T): Unit
    def size(): Int
    def elements(): Iterator[T]
  }

  trait Vertex {
    self: V =>
      def graph(): G
      def value(): T
  }

}

class SimpleGraphKind[T] extends GraphKind[T] {

  type G = GraphImpl
  type V = VertexImpl

  def newGraph() = new GraphImpl

  class GraphImpl extends Graph {
    private var vertices_ = List[V]()
    def vertices = vertices_
    def add( t: T ) {  vertices_ ::= new VertexImpl(t,this) }
    override def size() = vertices_.size
    override def elements() = vertices.map( _.value ).elements
  }

  class VertexImpl(val value: T, val graph: GraphImpl) extends Vertex {
    override lazy val toString = "Vertex(" + value.toString + ")"
  }
}

由于Vertex并将Graph绑定到一个实例GraphKind,然后T将固定为为该实例定义的任何内容.例如:

scala> new SimpleGraphKind[Int]
res0: SimpleGraphKind[Int] = SimpleGraphKind@1dd0fe7

scala> new res0.GraphImpl
res1: res0.GraphImpl = line10()

scala> res1.add(10)

scala> res1.add("abc")
<console>:9: error: type mismatch;
 found   : java.lang.String("abc")
 required: Int
       res1.add("abc")
                ^