[英]How can I pattern match on a range in Scala?
In Ruby I can write this: 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
How do I do this in Scala? 我如何在Scala中执行此操作?
Edit: preferably I'd like to do it more elegantly than using if
. 编辑:我最好比使用if
更优雅。
Inside pattern match it can be expressed with guards: 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"
}
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
Note that Contains instances should be named with initial caps. 请注意,Contains实例应以初始大写字母命名。 If you don't, you'll need to give the name in back-quotes (difficult here, unless there's an escape I don't know) 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)
Similar to @Yardena's answer, but using basic comparisons: 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"
}
Also works for floating point n 也适用于浮点数n
For Ranges of equal size, you can do it with old-school math: 对于相同大小的范围,您可以使用旧式数学来完成:
val a = 11
(a/10) match {
case 0 => println (a + " in 0-9")
case 1 => println (a + " in 10-19") }
11 in 10-19
Yes, I know: "Don't divide without neccessity!" 是的,我知道:“不要在没有必要的情况下分开!” But: Divide et impera! 但是:Divide et impera!
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