在Ruby中我可以這樣寫:我如何在Scala範圍內進行模式匹配?
case n
when 0...5 then "less than five"
when 5...10 then "less than ten"
else "a lot"
end
如何在斯卡拉做到這一點?
編輯:最好我想比使用if
更優雅。
在Ruby中我可以這樣寫:我如何在Scala範圍內進行模式匹配?
case n
when 0...5 then "less than five"
when 5...10 then "less than ten"
else "a lot"
end
如何在斯卡拉做到這一點?
編輯:最好我想比使用if
更優雅。
內側的圖案匹配它可以與衛兵表示:
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"
}
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
請注意,包含實例應以首字母大寫命名。如果不這樣做,你需要在反引號給名稱(這裏很難,除非有一種逃避,我不知道)
對於相同大小的範圍,你可以用老派的數學做到這一點:
val a = 11
(a/10) match {
case 0 => println (a + " in 0-9")
case 1 => println (a + " in 10-19") }
11 in 10-19
是的,我知道:「不要沒有neccessity鴻溝」 但是:除以等於!
類似@亞丹娜的答案,但使用基本比較:
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"
}
也適用於浮動點n
請參閱相關的計算器問題:能否在範圍在斯卡拉相匹配?(HTTP: //stackoverflow.com/questions/1346127/can-a-range-be-matched-in-scala) – 2011-09-26 01:35:56