Default type parameters in scala

Question

i have this trait

trait Extractor extends Mapping[Node]

what's the simplest way to change it to take a type parameter with a default value

trait Extractor[T] extends Mapping[**T if given else Node**]

other similar questions involves some irrelevant details about upper and lower type bounds

@flavian yeah but suppose the types i'm intending to set for the Extractor are not related by any means to class Node , so upper and lower bounds are in applicable in this case , or is there something that fits ? — Hady Elsahar
– Hady Elsahar, Commented Nov 18, 2013 at 1:51

Miles Sabin · Accepted Answer · 2013-11-18 22:38:58Z

7

I think you want to overload Extractor to be of kind * and of kind * -> *. Overloading is usually best avoided, so instead I'd recommend,

scala> trait Mapping[T]
defined trait Mapping

scala> trait Node
defined trait Node

scala> trait Extractor[T] extends Mapping[T]
defined trait Extractor

scala> type NodeExtractor = Extractor[Node]
defined type alias NodeExtractor

scala> val ne = new NodeExtractor {}
ne: NodeExtractor = $anon$1@1ebdcc9a

If Extractor[Node] is the common case, then you could optimize for that and do something like this,

scala> trait ExtractorT[T] extends Mapping[T]
defined trait ExtractorT

scala> type Extractor = ExtractorT[Node]
defined type alias Extractor

scala> val ne = new Extractor {}
ne: Extractor = $anon$1@3d70fe39

answered Nov 18, 2013 at 22:38

Miles Sabin

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