Scala type compilation error

Question

Try to understand how I can use type in scala:

object TypeSample extends App {

  type MyParams = Map[Int, String]

  def showParams(params: MyParams) = {
    params.foreach(x => x match { case (a, b) => println(a + " " + b) })
  }

  //val params = MyParams( 1 -> "one", 2 -> "two")
  val params = Map( 1 -> "one", 2 -> "two")

  showParams(params)

}

This line throws compilation exception: "Can not resolve symbol 'MyParams'"

//val params = MyParams( 1 -> "one", 2 -> "two")

Why? I can not use 'type' like this?

Also note that instead of `(x => x match { case (a, b) => ...` you can just write `{ case (a, b) => ...` — Alexey Romanov, Dec 16 '12 at 21:42

score 4 · Answer 1 · edited May 23 '17 at 12:21

4

Map( 1 -> "one", 2 -> "two") means Map.apply( 1 -> "one", 2 -> "two"). Map is a singleton object.

Try this:

val MyParams = Map.apply[Int, String] _

edited May 23 '17 at 12:21

Community

1
1

answered Dec 16 '12 at 20:33

senia

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score 4 · Accepted Answer · answered Dec 16 '12 at 20:34

4

Because MyParams is only an alias to the type Map[Int, String]. To make this work, you have to add a factory like

object MyParams {
  def apply(params: (Int, String)*) = Map(params: _*)
}

answered Dec 16 '12 at 20:34

drexin

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I guess, it is the option to make the compiler/scala even more smarter if one make an alias to some Class, and this class has companion object with apply method in it, then alias could use this.. since alias = synonym – ses Dec 16 '12 at 22:36

Scala type compilation error

2 Answers2