原创转载请注明出处:http://agilestyle.iteye.com/blog/2334476
Scala中创建一个枚举类型,通常需要定义一个object继承Enumeration抽象类
package org.fool.scala.enumeration object Level extends Enumeration { type Level = Value val Overflow, High, Medium, Low, Empty = Value } object EnumerationTest extends App { println(Level.Medium) println(Level(0)) println(Level(Level.maxId - 1)) for (n <- Range(0, Level.maxId)) println((n, Level(n))) // Vector((0,Overflow), (1,High), (2,Medium), (3,Low), (4,Empty)) println({ for (n <- Range(0, Level.maxId)) yield (n, Level(n)) }) // Vector(Overflow, High, Medium, Low, Empty) println({ for (level <- Level.values) yield level }.toIndexedSeq) def checkLevel(level: Level.Level) = level match { case Level.Overflow => ">>> Overflow!" case Level.Empty => "Alert: Empty" case other => s"Level $level OK" } println(checkLevel(Level.Low)) println(checkLevel(Level.Empty)) println(checkLevel(Level.Overflow)) }
Note:
创建object不会以创建class的方式创建新类型,如果想将其当做类型处理,那么必须使用type关键字,比如将Enumeration的Value起一个新的名字Level作为别名
Console Output