What is the idiomatic way to pattern match sequenc

问题:

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]

This is the same problem as MatchError when match receives an IndexedSeq but not a LinearSeq.

The question is, how to do it right? Adding .toList everywhere 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...

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 Seq extractor part of the standard library?

回答1:

Well, you can pattern-match any sequence:

case Seq(a, b, rest @ _ *) =>

For example:

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

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2:

One more solution for Vector in REPL:

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)