Pattern 将 Seq 与 Range 匹配

Question

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))

当我们尝试匹配 foo(problem) 中的范围时出现问题。

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

使用 val problem = (1 to 10).toSeq 将 Range 转换为 Seq 是无用的，因为 toSeq 方法只是 returns Range 本身：

override def toSeq = this.

可以使用解决方法：

val problem = (1 to 10).toList.toSeq,

但这并不是我见过的最漂亮的东西。

将 Range 匹配到 [head:tail] 模式的正确方法是什么？

Answer 1

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。

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，只在 Seq 提取器上进行模式匹配：

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"
}

Pattern 将 Seq 与 Range 匹配

Pattern match a Seq with Range

scala

pattern-matching