如何在Scala中的范围上进行模式匹配？

Question

在Ruby中我可以这样写:

case n
when 0...5  then "less than five"
when 5...10 then "less than ten"
else "a lot"
end

我如何在Scala中执行此操作？

编辑:我最好比使用它更优雅if.

Answer 1

内部模式匹配可以用守卫表达:

n match {
  case it if 0 until 5 contains it  => "less than five"
  case it if 5 until 10 contains it => "less than ten"
  case _ => "a lot"
}

Answer 2

class Contains(r: Range) { def unapply(i: Int): Boolean = r contains i }

val C1 = new Contains(3 to 10)
val C2 = new Contains(20 to 30)

scala> 5 match { case C1() => println("C1"); case C2() => println("C2"); case _ => println("none") }
C1

scala> 23 match { case C1() => println("C1"); case C2() => println("C2"); case _ => println("none") }
C2

scala> 45 match { case C1() => println("C1"); case C2() => println("C2"); case _ => println("none") }
none

请注意,Contains实例应以初始大写字母命名.如果你不这样做,你需要在后面引用这个名字(这里很难,除非有一个我不知道的逃脱)

Answer 3

与@ Yardena的答案类似,但使用基本比较:

n match {
    case i if (i >= 0 && i < 5) => "less than five"
    case i if (i >= 5 && i < 10) => "less than ten"
    case _ => "a lot"
}

也适用于浮点数n