generics - 在scala中混合类型参数和抽象类型

Question

我正在尝试使用 a preceding question 的答案实现一个小型图形库。这个想法是将图视为集合，其中顶点包装集合元素。

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示例时，我遇到了编译错误。这是示例:

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 + ")"
  }

}

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] 我只会得到第一个错误。

你有什么想法吗？

Answer 1

用 -explaintypes 编译它产量:

<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，我正要写我不明白如何 T <: T可能是假的——几乎就像 T被定义了两次，这当然是整个问题。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，上课 GraphKind用 T 参数化并输入 G必须是 Graph[T] .现在，上课 Graph也是参数化的，它的参数也叫 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，这与您编写的内容完全相同。我只是为类型参数使用不同的名称，以免与 T 混淆。即参数化 GraphKind .

所以，这里是逻辑:

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。我不明白为什么不存在继承时问题不会出现。使固定:

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和 Graph将绑定(bind)到 GraphKind 的一个实例，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")
                ^