2010-07-01 50 views
30

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。

+2

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

回答

56

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" 
} 
13
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)

4

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!

8

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