scala - 如何在 Scala 中创建具有有界类型参数的自定义 Seq？

Question

考虑以下工作习惯 Seq :

class MySeq[B](val s: Seq[B]) 
extends Seq[B]
with GenericTraversableTemplate[B, MySeq]
with SeqLike[B, MySeq[B]] {
  override def companion = MySeq

  def iterator = s.iterator

  def apply(i: Int) = s(i)

  def length = s.length

  override def toString = s map { _.toString } mkString("\n")
}
object MySeq extends SeqFactory[MySeq] {
  implicit def canBuildFrom[B]: CanBuildFrom[Coll, B, MySeq[B]] =
       new GenericCanBuildFrom[B]
  def newBuilder[B] = new ListBuffer[B] mapResult (x => new MySeq(x.toSeq))
}

我想对类型参数 B 施加一个界限.换句话说，我想要这样的东西(不工作):

class MyA

class MySeq[+B <: MyA](val s: Seq[B]) 
extends Seq[B]
with GenericTraversableTemplate[B, MySeq]
with SeqLike[B, MySeq[B]] {
  override def companion = MySeq  // Type Mismatch Here

  def iterator = s.iterator

  def apply(i: Int) = s(i)

  def length = s.length

  override def toString = s map { _.toString } mkString("\n")
}
object MySeq extends SeqFactory[MySeq] {
  implicit def canBuildFrom[B]: CanBuildFrom[Coll, B, MySeq[B]] =
       new GenericCanBuildFrom[B]
  // Type Mismatch in the line below
  def newBuilder[B] = new ListBuffer[B] mapResult (x => new MySeq(x.toSeq))
}

但我在指示的行中收到以下类型不匹配错误:

inferred type arguments [B] do not conform to 
class MySeq's type parameter bounds [+B <: MyA] 
Main.scala  line 49

type mismatch;  
found   : countvotes.structures.MySeq.type  
required: scala.collection.generic.GenericCompanion[Seq]    
Main.scala  line 36

type mismatch;  
found   : MySeq[B(in class MySeq)]  
required: MySeq[B(in method newBuilder)]    
Main.scala  line 49

type mismatch;  
found   : scala.collection.immutable.Seq[B(in method newBuilder)]  
required: Seq[B(in class MySeq)]    
Main.scala  line 49

我试图通过向 CanBuildFrom 和 newBuilder 的类型参数添加边界来解决此问题，但随后我收到其他错误消息。

如何创建自定义 Seq绑定(bind)类型参数？

Answer 1

我没有收到第 26 行的错误:

override def companion = MySeq

也许是其他原因造成的。

无论如何，问题是您不能拥有 GenericCompanion[MySeq] ( SeqFactory 的父类(super class)型)。原因是 GenericCompanion[Coll]表示您可以构造一个 Coll[A]对于任何 A (见 newBuilder 的签名)。你也不能有 MySeq[A] <: GenericTraversableTemplate[A, MySeq] , 因为 genericBuilder是不可能的。这是有道理的； MySeq并不是真正的“通用集合”，因为它希望它的元素都是 MyA .

解决方案是拥有 MySeq[B] <: GenericTraversableTemplate[B, Seq] ,( extends Seq 免费提供)。然后你有两个选择 companion .它可以是 Seq 中的默认值, 也可以是 s.companion .在第一种情况下， ((as: MySeq[A]): Seq[A]).map(...)将产生 List (在运行时；在编译时它只是一个通用的 Seq )。第二，这将取决于 as.s是(同样，在运行时；编译时只会看到 Seq )。您可以保留 extends SetLike ， 尽管。

然后，您需要提供自定义 CanBuildFrom : MySeq.canBuildFrom[A <: MyA]: CanBuildFrom[MySeq[A], A, MySeq[A]] , 并定义 MySeq#newBuilder .

class MySeq[+B <: MyA](val s: Seq[B]) 
  extends Seq[B]
  with SeqLike[B, MySeq[B]]
{
  override def iterator = s.iterator
  override def apply(i: Int) = s(i)
  override def length = s.length

  override def toString = s.map(_.toString).mkString("\n")

  override def companion = s.companion
  protected[this] override def newBuilder: mutable.Builder[B, MySeq[B]] = new mutable.Builder[B, MySeq[B]] {
    private[this] val base = s.genericBuilder[B]
    override def +=(elem: B) = { base += elem; this }
    override def clear() = base.clear()
    override def result() = new MySeq[B](base.result())
  }
}

object MySeq {
  implicit def canBuildFrom[A <: MyA]: CanBuildFrom[MySeq[_], A, MySeq[A]] = ???
}

val list = List(new MyA, new MyA, new MyA, new MyA)
val vect = list.toVector
val mLst = new MySeq(list)
val mVec = new MySeq(vect)
{
  val res = mLst.filter(_.hashCode != list.head.hashCode)
  implicitly[res.type <:< MySeq[MyA]]
}
{
  val res = (mVec: Seq[MyA]).map(identity)
  assert(res.isInstanceOf[Vector[_]])
}
{
  val res = (mLst: Seq[MyA]).map(identity)
  assert(res.isInstanceOf[List[_]])
}