如何在Scala中向Enumeration添加方法？

时间：2010-12-03 14:38:03

标签： java scala enums enumeration

在Java中你可以：

public enum Enum {
    ONE {
        public String method() {
            return "1";
        }
    },
    TWO {
        public String method() {
            return "2";
        }
    },
    THREE {
        public String method() {
            return "3";
        }
    };

    public abstract String method();
}

你如何在Scala中这样做？

编辑/有用的链接：

11 个答案:

答案 0 :(得分：34)

以下是通过扩展Enumeration.Val类向scala枚举添加属性的示例。

object Planet extends Enumeration { 
   protected case class Val(val mass: Double, val radius: Double) extends super.Val { 
     def surfaceGravity: Double = Planet.G * mass / (radius * radius) 
     def surfaceWeight(otherMass: Double): Double = otherMass * surfaceGravity 
   } 
   implicit def valueToPlanetVal(x: Value) = x.asInstanceOf[Val] 

   val G: Double = 6.67300E-11 
   val Mercury = Val(3.303e+23, 2.4397e6) 
   val Venus   = Val(4.869e+24, 6.0518e6) 
   val Earth   = Val(5.976e+24, 6.37814e6) 
   val Mars    = Val(6.421e+23, 3.3972e6) 
   val Jupiter = Val(1.9e+27, 7.1492e7) 
   val Saturn  = Val(5.688e+26, 6.0268e7) 
   val Uranus  = Val(8.686e+25, 2.5559e7) 
   val Neptune = Val(1.024e+26, 2.4746e7) 
} 

scala> Planet.values.filter(_.radius > 7.0e6) 
res16: Planet.ValueSet = Planet.ValueSet(Jupiter, Saturn, Uranus, Neptune)

答案 1 :(得分：30)

在Chris' solution的基础上，您可以通过隐式转换实现更好的语法：

object Suit extends Enumeration {
   val Clubs, Diamonds, Hearts, Spades = Value

   class SuitValue(suit: Value) {
      def isRed = !isBlack
      def isBlack = suit match {
         case Clubs | Spades => true
         case _              => false
      }
   }

   implicit def value2SuitValue(suit: Value) = new SuitValue(suit)
}

然后你可以打电话给Suit.Clubs.isRed。

答案 2 :(得分：12)

Scala enumerations与Java枚举不同。

目前，没有办法添加方法（以理智的方式）。有一些解决方法，但在所有情况下和的任何内容都不像语法垃圾。

我尝试了类似的东西（向枚举的类实例添加方法，同时能够在运行时创建新实例，并在object和new实例之间具有工作等效关系。但是被bug #4023阻止了（“getClasses / getDeclaredClasses似乎错过了某些（REPL）或所有（scalac）类（对象）声明的”）。

由我来看看这些相关问题：

老实说，我不会使用Enumeration。这是一个源自Scala 1.0（2004）的类，它有很奇怪的东西，并没有很多人（除了那些编写它的人）在没有教程的情况下理解如何使用它。

如果我绝对需要枚举，我会用Java编写该类。

答案 3 :(得分：11)

如果你不需要迭代枚举值或做一些其他的枚举，我建议使用ADT而不是Enumeration。

sealed abstract class Enum {
  def method: String = this match {
    case One => "1"
    case Two => "2"
    case Three => "3"
  }
}
case object One extends Enum
case object Two extends Enum
case object Three extends Enum

这种方法比Enumeration有一个优势，当你忘记match表达式中的一个或多个案例时，编译器会发出警告。

答案 4 :(得分：8)

使用Aaron's solution在Scala 2.10中使用implicit classes（一种更紧凑的形式）进行阐述：

object Suit extends Enumeration {
   val Clubs, Diamonds, Hearts, Spades = Value

   implicit class SuitValue(suit: Value) {
      def isRed = !isBlack
      def isBlack = suit match {
         case Clubs | Spades => true
         case _              => false
      }
   }
}

然后您可以像这样使用它：Suit.Clubs.isRed

答案 5 :(得分：7)

你可以这样做：

object Suit extends Enumeration {
  val Clubs, Diamonds, Hearts, Spades = Value

  def isRed(suit : Value) = !isBlack(suit)
  def isBlack(suit : Value) = suit match {
    case Clubs | Spades => true
    case _              => false
  }
}

显然这不完美，但你可以这样做：

Suit.isBlack(Suit.Clubs)

答案 6 :(得分：4)

Scala的Enumeration不允许将属性和/或方法添加到枚举中的值。有了这个新的MyEnumeration你就可以。

abstract class MyEnumeration {
  // "Value" must be the name of the class defining your values type Value
  type Value

  // Contains your values in definition order
  private val vals = collection.mutable.LinkedHashMap[String, Value]()

  // A mixin for your values class to automatically collect the values
  protected trait ValuesCollector { self: Value =>
    private val ordinal = vals.size

    vals += (fieldNames(ordinal) -> self)

    def getName = fieldNames(ordinal)
    override def toString = getName
  }

  def apply(ordinal: Int) = vals(fieldNames(ordinal))
  def apply(fldName: String) = vals(fldName)

  def values = vals.values
  def namedValues: collection.Map[String, Value] = vals

  // Getting the field names through reflection.
  // Copied from scala.Enumeration
  private val fieldNames = getClass.getMethods filter (m =>
    m.getParameterTypes.isEmpty &&
    classOf[ValuesCollector].isAssignableFrom(m.getReturnType) &&
    m.getDeclaringClass != classOf[MyEnumeration]) map (_.getName)

}

在这里，您可以看到Scala中的Planet示例。

object Planet extends MyEnumeration {

  case class Value(val mass: Double, val radius: Double) extends ValuesCollector {
    // universal gravitational constant  (m3 kg-1 s-2)
    private val G = 6.67300E-11;

    def surfaceGravity = G * mass / (radius * radius)
    def surfaceWeight(otherMass: Double) = otherMass * surfaceGravity

  }

  val MERCURY = Value(3.303e+23, 2.4397e6)
  val VENUS = Value(4.869e+24, 6.0518e6)
  val EARTH = Value(5.976e+24, 6.37814e6)
  val MARS = Value(6.421e+23, 3.3972e6)
  val JUPITER = Value(1.9e+27, 7.1492e7)
  val SATURN = Value(5.688e+26, 6.0268e7)
  val URANUS = Value(8.686e+25, 2.5559e7)
  val NEPTUNE = Value(1.024e+26, 2.4746e7)
  val PLUTO = Value(1.27e+22, 1.137e6)

}

object PlanetTest {
  def main(args: Array[String]) {
    val earthWeight = 175
    val mass = earthWeight/Planet.EARTH.surfaceGravity
    for (p <- Planet.values) println("Your weight on %s is %f" format (p, p.surfaceWeight(mass)))
    /* Your weight on MERCURY is 66.107583
     * Your weight on VENUS is 158.374842
     * Your weight on EARTH is 175.000000
     * Your weight on MARS is 66.279007
     * Your weight on JUPITER is 442.847567
     * Your weight on SATURN is 186.552719
     * Your weight on URANUS is 158.397260
     * Your weight on NEPTUNE is 199.207413
     * Your weight on PLUTO is 11.703031
     */
  }

}

答案 7 :(得分：3)

object Unit extends Enumeration {
  abstract class UnitValue(var name: String) extends Val(name) {
    def m: Unit
  }
  val G = new UnitValue("g") {
    def m {
        println("M from G")
    }
  }
  val KG = new UnitValue("kg") {
    def m {
        println("M from KG")
    }
  }
}

答案 8 :(得分：1)

在查看scala.Enumeration的源代码后，我得到了这个：


object MyEnum extends Enumeration {
  val ONE = new Val { def method = "1" }
  val TWO = new Val { def method = "2" }
  val THREE = new Val { def method = "3" }
}

由于使用了匿名类，因此似乎很难摆脱'new'。 如果有人知道怎么做，请告诉我:)。

答案 9 :(得分：1)

如果你绝对需要每个枚举值的方法和需要能够迭代这些值，你可以这样做：

object BatchCategory extends Enumeration {
  class BatchCategory extends Val {
    val isOfficial, isTest, isUser = false
  }

  val OFFICIAL = new BatchCategory { override val isOfficial = true }
  val TEST =     new BatchCategory { override val isTest = true }
  val USER =     new BatchCategory { override val isUser = true }

  // Needed to get BatchCategory from Enumeration.values
  implicit def valueToBatchCategory(v: Value): BatchCategory = v match {
    case bc: BatchCategory => bc
    case x => throw new IllegalArgumentException("Value is not a BatchCategory: " + x)
  }

  def valueOf(catStr: String): BatchCategory = {
    BatchCategory.values.
      find { v => val s = v.toString; s.take(1) == catStr || s == catStr }.
      getOrElse(throw new IllegalArgumentException("Unknown category '" + catStr + "' !  "))
  }

  def main(args: Array[String]) {
    BatchCategory.values.foreach(v => println(v + " isOfficial=" + v.isOfficial))
  }
}

打印

OFFICIAL isOfficial=true
TEST isOfficial=false
USER isOfficial=false

这是针对一些遗留代码完成的，除了Enumeration之外，这些代码无法转移到更健全的枚举策略。

答案 10 :(得分：0)

answer, which tells that Scala enums do not support args/methods-customized values似乎错了。其他答案（其中一些涉及implicit）证明它可以做，但它们创建了需要名称重复的印象：您的值已将名称声明为java对象字段，其次，该名称作为字符串提供给值构造函数而Enums的重点是创建一个可迭代的名称 - ＆gt;值映射，而scala可以没有冗余：

object Ops1 extends Enumeration {

    protected case class OpsVal(f: Int => Int) extends super.Val(/*nextId*/)

    val ZERO = new FuncVal (x => 0)
    val DOUBLE = new FuncVal (x => 2 * x )

    implicit def convert(v: Value) = v.asInstanceOf[OpsVal]

}

// implicit is not needed
Ops1.ZERO.f(1)                            //> res0: Int = 0

// implicit is needed
Ops1.values map (v => (v + "=>" + v.f(1)))//> res1: scala.collection.immutable.SortedSet[String] = TreeSet(DOUBLE=>2, ZERO=>0)

我认为上述内容比

更简洁 
object Ops2 extends Enumeration {

    protected abstract class OpsVal extends Val() {
      def f(a: Int): Int
    }

    val ZERO = new OpsVal { def f(x: Int) = 0 }
    val DOUBLE = new OpsVal { def f(x: Int) = 2 * x }

    implicit def convert(valu: Value) = valu.asInstanceOf[OpsVal]
}
Ops2.ZERO.f(1) // implicit is not needed  //> res2: Int = 0

// implicit is needed
Ops2_3.values map (v => (v, v(1)))        //> res7: scala.collection.immutable.SortedSet[(e.Ops2_3.Value, Int)] = TreeSet
                                              //| ((ZERO,0), (DOUBLE,2))

由于每个值只有一个方法，我们可以将它们转换为函数

object Ops2_3 extends Enumeration {

    protected case class FuncVal(f: Int => Int) extends Val {
        def apply(x: Int) = f(x) // no need to extend Function1 explicitly
    }

    val ZERO = new FuncVal (x => 0)
    val DOUBLE = new FuncVal (x => 2 * x )

    implicit def convert(v: Value) = v.asInstanceOf[FuncVal]

}
Ops2_3.ZERO(1) // implicit is not needed  //> res6: Int = 0

// implicit is needed
Ops2_3.values map (v => (v, v(1)))        //> res7: scala.collection.immutable.SortedSet[(e.Ops2_3.Value, Int)] = TreeSet
                                              //| ((ZERO,0), (DOUBLE,2))

所有值共享的函数可以像这样定义（可在arg parser中使用）

val args: Array[String] = "-silent -samples 100 -silent ".split(" +").toArray
                                              //> args  : Array[String] = Array(-silent, -samples, 100, -silent)
object Opts extends Enumeration {

    val nopar, silent, samples = new Val() {
        def apply() = args.contains(toString)
        def asInt(default: Int) = { val i = args.indexOf(toString) ;  if (i == -1) default else args(i+1).toInt}
        def asInt: Int = asInt(-1)
        override def toString = "-" + super.toString
    }
}

Opts.nopar()                              //> res0: Boolean = false
Opts.samples.asInt                        //> res1: Int = 100

其他用户主张密封特征+宏的情况，Iteration over a sealed trait in Scala?

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