How to get around this scala compiler erasure warning?

Question

I am trying to compile this scala code and getting the following compiler warnings.

scala> val props: Map[String, _] = 
 |   x match {
 |     case t: Tuple2[String, _] => {
 |       val prop =
 |       t._2 match {
 |         case f: Function[_, _] => "hello"
 |         case s:Some[Function[_, _]] => "world"
 |         case _ => t._2
 |       }
 |     Map(t._1 -> prop)
 |   }
 |   case _ => null
 | }
<console>:10: warning: non-variable type argument String in type pattern (String, _) is unchecked since it is eliminated by erasure
       case t: Tuple2[String, _] => {
               ^
<console>:14: warning: non-variable type argument _ => _ in type pattern Some[_ => _] is unchecked since it is eliminated by erasure
           case s:Some[Function[_, _]] => "world"
                  ^

The answers given on How do I get around type erasure on Scala? Or, why can't I get the type parameter of my collections? seems to point to this same issue. But I couldn't infer a solution in this particular context.

x could be anything like an Any object or function – sobychacko Feb 14 '13 at 18:05 — sobychacko, Feb 14 '13 at 18:05

senia · Accepted Answer · 2013-02-14T07:09:09.240

4

Use case t @ (_: String, _) instead of case t: Tuple2[String, _] and case s @ Some(_: Function[_, _]) instead of case s:Some[Function[_, _]].

Scala cannot match on type parameters.

edited Feb 14 '13 at 07:09

answered Feb 14 '13 at 06:56

senia

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case t: WrappedArray[Tuple2[String, _]] => { -- also generates the same type erasure warning. Can you point me to why it is so? – sobychacko Feb 15 '13 at 04:03
@sobychacko It's so because of type erasure: in `jvm` in runtime `WrappedArray[(String, Int)]` and `WrappedArray[String]` are the same: `WrappedArray[Object]`. Information about type parameters is available only in compile time. See [this](http://www.scalafied.com/60/lightweight-type-erasure-matching) for some links and workaround. – senia Feb 15 '13 at 05:09

score 2 · Answer 2 · answered Feb 14 '13 at 07:29

2

I would rewrite it like this:

x match {
  case (name, method) => {
    val prop =
      method match {
        case f: Function[_, _] => "hello"
        case Some(f: Function[_, _]) => "world"
        case other => other
      }

    Map(name -> prop)
  }
  case _ => null
}

answered Feb 14 '13 at 07:29

EECOLOR

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Thanks for the response. I liked both of the approaches. – sobychacko Feb 14 '13 at 17:58

How to get around this scala compiler erasure warning?

2 Answers2