Question

鉴于以下y：

sealed trait

给定scala> sealed trait Parent defined trait Parent scala> case object Boy extends Parent defined object Boy scala> case object Girl extends Parent defined object Girl：

xs

我不明白这个无穷无尽的匹配警告。 scala> val xs: (Parent, (Seq[Int], Seq[Int])) = (Boy, (Nil, Nil)) xs: (Parent, (Seq[Int], Seq[Int])) = (Boy,(List(),List())) scala> xs match { | case (Boy, (Nil, Nil)) => 1 | case (Boy, (ys, a :: as)) => 2 | case (Boy, (ys, Nil)) => 3 | case (Girl, (Nil, Nil)) => 4 | case (Girl, (a :: as, ys)) => 5 | case (Girl, (Nil, ys)) => 6 | } <console>:15: warning: match may not be exhaustive. It would fail on the following inputs: (Boy, _), (Girl, _) xs match { ^ res1: Int = 1和(Boy, _)的含义是什么？

我不确定(Girl, _)除了我在右手边的比赛之外还有什么其他比赛。

Answer 1

我不确定原因，但似乎Scala在使用unapply（或apply）时遇到了问题，因为Seq是使模式匹配成为可能的神奇之处。显然，匹配中的cons与Seq存在问题。如果您更改xs以使用List，则可以使用：

val xs: (Parent, (List[Int], List[Int])) = (Boy, (Nil, Nil))

编辑：

可能您可以使用Seq，但只需对构造进行模式匹配的方式略有不同，如下所示：

  xs match {
    case (Boy, (Nil, Nil)) => 1
    case (Boy, (_, Seq(x, _))) => 2
    case (Boy, (_, Nil)) => 3
    case (Girl, (Nil, Nil)) => 4
    case (Girl, (Seq(x, _), _)) => 5
    case (Girl, (Nil, _)) => 6
  }

具有元组的非详尽模式匹配警告

1 个答案: