如何在 Scala 中使用有界类型参数创建自定义 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))
}

我想对类型参数施加限制 B。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

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

如何创建带有类型参数绑定的自定义 Seq？

Answer 1

我没有收到第 26 行的错误：

override def companion = MySeq

可能是其他原因造成的。

无论如何，问题是您不能拥有 GenericCompanion[MySeq]（SeqFactory 的超类型）。原因是 GenericCompanion[Coll] 意味着您可以为任何 A 构造一个 Coll[A]（参见 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 "generic collection" 因为它希望它的元素都是 MyA.

解决方案是 MySeq[B] <: GenericTraversableTemplate[B, Seq]，（extends Seq 免费提供）。然后您有两个选择 companion。它可以只是 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(...) 将产生一个 List（在运行时；在编译时它只是一个通用的 Seq）。在第二个中，它将取决于 as.s 是什么（同样，在运行时；编译时只会看到 Seq）。不过，您可以保留 extends SetLike。

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[_]])
}

Answer 2

这是一个完整的 MWE，它基于 HTNW 上面接受的答案，canBuildFrom 已实施：

class MyA

class MySeq[+A <: MyA](val s: Seq[A]) 
  extends Seq[A]
  with SeqLike[A, MySeq[A]]
{
  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 = MySeq.newBuilder
}

object MySeq {
  def newBuilder[A <: MyA] = new mutable.Builder[A, MySeq[A]] {
    private[this] val base = Seq.newBuilder[A]
    override def +=(elem: A) = { base += elem; this }
    override def clear() = base.clear()
    override def result() = new MySeq[A](base.result())
  }

  implicit def canBuildFrom[A <: MyA]: CanBuildFrom[MySeq[_], A, MySeq[A]] = new CanBuildFrom[MySeq[_], A, MySeq[A]] {
    def apply(from: Election[_]) = newBuilder
    def apply() = newBuilder
  }
}

如何在 Scala 中使用有界类型参数创建自定义 Seq？

How to create a custom Seq with bounded type parameter in Scala?

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