scala 模式匹配序列理解的惯用方法是什么？
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What is the idiomatic way to pattern match sequence comprehensions?
提问by letmaik
val x = for(i <- 1 to 3) yield i
x match {
case 1 :: rest => ... // compile error
}
constructor cannot be instantiated to expected type; found : collection.immutable.::[B] required: scala.collection.immutable.IndexedSeq[Int]
构造函数无法实例化为预期类型；发现：collection.immutable.::[B] 需要：scala.collection.immutable.IndexedSeq[Int]
This is the same problem as MatchError when match receives an IndexedSeq but not a LinearSeq.
当 match 接收到 IndexedSeq 而不是 LinearSeq 时，这与MatchError 的问题相同。
The question is, how to do it right? Adding .toListeverywhere doesn't seem right. And creating an own extractor which handles every Seq(as described in the answer of the other question) would lead to a mess if everybody would do it...
问题是，如何做才是正确的？.toList到处添加似乎不正确。Seq如果每个人都这样做，那么创建一个自己的提取器来处理每个（如另一个问题的答案中所述）会导致一团糟......
I guess the question is, why can't I influence what the return type of sequence comprehensions is, or: why isn't such a generalized Seqextractor part of the standard library?
我想问题是，为什么我不能影响序列推导式的返回类型是什么，或者：为什么Seq标准库中没有这种通用提取器的一部分？
回答by oxbow_lakes
Well, you can pattern-match any sequence:
好吧，您可以对任何序列进行模式匹配：
case Seq(a, b, rest @ _ *) =>
For example:
例如：
scala> def mtch(s: Seq[Int]) = s match {
| case Seq(a, b, rest @ _ *) => println("Found " + a + " and " + b)
| case _ => println("Bah")
| }
mtch: (s: Seq[Int])Unit
Then this will match any sequence with more than (or equal to) 2 elements
然后这将匹配具有多于（或等于）2 个元素的任何序列
scala> mtch(List(1, 2, 3, 4))
Found 1 and 2
scala> mtch(Seq(1, 2, 3))
Found 1 and 2
scala> mtch(Vector(1, 2))
Found 1 and 2
scala> mtch(Vector(1))
Bah
回答by Roman Kazanovskyi
One more solution for Vector in REPL:
REPL 中 Vector 的另一种解决方案：
Vector() match {
case a +: as => "head + tail"
case _ => "empty"
}
res0: String = "empty"
Vector(1,2) match {
case a +: as => s"$a + $as"
case _ => "empty" }
res1: String = 1 + Vector(2)
