模式匹配具有范围的 Seq [英] Pattern match a Seq with Range

问题描述

考虑一段代码:

def foo(xs: Seq[Int]) = xs match {
  case Nil => "empty list"
  case head :: Nil => "one element list"
  case head :: tail => s"head is $head and tail is $tail"
}

val x1 = Seq(1,2,3)
println(foo(x1))

val x2 = Seq()
println(foo(x2))

val x3 = Seq(1)
println(foo(x3))

val problem = 1 to 10
println(foo(problem))

出现问题，当我们尝试匹配 foo(problem) 中的 Range 时.

A problem occurs, when we try to match a Range in foo(problem).

scala.MatchError: Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)(类 scala.collection.immutable.Range$Inclusive).

Converting Range to Seq with val problem = (1 to 10).toSeq 没用，因为 toSeq 方法只返回 Range 本身:

Converting Range to Seq with val problem = (1 to 10).toSeq is useless, as the toSeq method just returns the Range itself:

覆盖 def toSeq = this.

可以使用一种解决方法:

A workaround can be used:

val 问题 = (1 to 10).toList.toSeq,

但这并不是我见过的最漂亮的东西.

but it's not exactly the prettiest thing I've seen.

将 Range 匹配到 [head:tail] 模式的正确方法是什么?

What is the proper way to match a Range to [head:tail] pattern?

推荐答案

您可以使用 +: 运算符.就像 ::，除了它不仅适用于 List，还适用于任何 Seq.

You can use the +: operator. It's like ::, except instead of only being for List, it works on any Seq.

def foo(xs: Seq[Int]) = xs match {
  case Seq() => "empty list"
  case head +: Seq() => "one element list"
  case head +: tail => s"head is $head and tail is $tail"
}

或者，更好的是，只需在 Seq 提取器上进行模式匹配:

Or, even better, just pattern match on the Seq extractor:

def foo(xs: Seq[Int]) = xs match {
  case Seq() => "empty list"
  case Seq(head) => "one element list"
  case Seq(head, tail @ _*) => s"head is $head and tail is $tail"
}

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