scala - Scala中具有构造函数参数的抽象类

我有以下课程模型：

sealed abstract class Tile(val coordinate: Int, val isOccupied: Boolean) {
  def isEmpty() : Boolean
  def getPiece() : Option[Piece]
}

case class EmptyTile(coordinate: Int) extends Tile(coordinate, false) {
  override def toString: String = "" +coordinate
  override def isEmpty() = true
  override def getPiece() = None
}

case class OccupiedTile(coordinate: Int, val piece: Piece) extends Tile(coordinate, true) {
  override def toString = piece.toString
  override def isEmpty = false
  override def getPiece = Some(piece)
}

我收到以下错误：

Error:(6, 22) overriding value coordinate in class Tile of type Int;
 value coordinate needs `override' modifier
case class EmptyTile(coordinate: Int) extends Tile(coordinate, false) {
                 ^

我究竟做错了什么？

编辑：请求查看Piece类，在这里添加：

import Alliance.Alliance
import PieceType.PieceType

abstract class Piece(val piecePosition: Int, val pieceType : PieceType, val alliance: Alliance) extends Movable {
}

object PieceType extends Enumeration {
  type PieceType = Value
  val PAWN, KNIGHT, BISHOP, ROOK, QUEEN, KING = Value
}

最佳答案

您的抽象类Tile声明一个val coordinate，使该值可公开访问。您的案例类EmptyTile也隐式地将coordinate声明为val（案例类“ magic”）。基本上，您的case类实际上是在尝试覆盖抽象类已经提供的值。

您可以在抽象类声明中删除val，也可以不创建EmptyTile和OccupiedTile案例类。

编辑：评论后建议的替代方案：

trait Tile {
  def coordinate: Int
  def isOccupied: Boolean
  def isEmpty() : Boolean = !isOccupied
  def getPiece() : Option[Piece]
}

case class EmptyTile(coordinate: Int) extends Tile {
  override def toString: String = "" +coordinate
  val isOccupied = false
  def getPiece() = None
}

case class OccupiedTile(coordinate: Int, val piece: Piece) extends Tile {
  override def toString = piece.toString
  val isOccupied = true
  def getPiece = Some(piece)
}

关于scala - Scala中具有构造函数参数的抽象类，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/24703786/