考虑以下自定义Seq
工作:如何在Scala中创建具有有界类型参数的自定义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))
}
我想强加约束的类型参数B
。换句话说,我想这样的事情(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))
}
但我得到的指示线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
我曾试图解决这个问题将边界添加到CanBuildFrom和newBuilder的类型参数中,但随后出现其他错误消息。
如何创建一个自定义Seq
并绑定了一个类型参数?
你而扩展集合然后在包装类与上限要求封装它的任何理由? –
原因是它会更优雅。我会在这堂课上使用'map'和'filter',每次手动解包和重新包装都会很不方便。 – Bruno