Alex Rivera | Logout

Extending a trait and types

Asked 2010-11-29T11:01:29.357
9

I would like to have a sealed trait which have a declared method that returns the actual class that extends the trait. Should I use an abstract type, a parameter type or is there any other nice way to solve this?

sealed trait Foo {
  type T
  def doit(other: T): T
}

or

sealed trait Foo[T] {
  def doit(other: T): T
}

Note that T must be a subtype of Foo in this example. If I do it like this the type information feels too repeated:

case class Bar(name: String) extends Foo[Bar] {
  def doit(other: Bar): Bar = ...
}
scala
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1 Answer

1

EDIT - Below is my original answer. Your comment indicates that you wish to return an arbitrary instance of a matching type but I don't really believe that this is in any way sensible. Suppose it were, via the T.type syntax:

trait T { def foo : T.type }

trait U extends T { def foo = new U } //must be a U

class W extends U

val w : W = (new W).foo //oh dear.

This is accomplishable via this.type:

scala> trait T {
 | def foo : this.type
 | }
defined trait T

scala> class W extends T {
 | def foo  = this
 | }
defined class W

scala> (new W).foo
res0: W = W@22652552

scala> res0.foo
res1: res0.type = W@22652552

And then also:

scala> ((new W) : T)
res4: T = W@45ea414e

scala> res4.foo.foo.foo
res5: res4.type = W@45ea414e
answered 2010-11-29T11:30:28.197

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