Scala abstract type representing type of subclass
I'm looking for a way to define a method that returns a type T where T = the type of the subclass.
I know I could possibly do this using abstract types, but dislike the overhead of having to redefine T for each subclass.
Some sample code:
object Helper {
def help[A <: MyClass](cls: A): Option[A] = { cls.foo() map { _.asInstanceOf[A] } }
}
class MyClass {
type T <: MyClass
def foo(): Option[T] = Some(this.asInstanceOf[T])
}
class ChildClass exten开发者_如何转开发ds MyClass {
type T = ChildClass
}
Possibly a new language feature has made this easier? Or can I use this.type in some way? It's important to me that I be able to define a helper class that can call into foo in this way.
If you are always returning this
, then you can indeed have as return type this.type
. Or have you tried it already?
this.type
is especially useful e.g. when you want to chain calls to the same object, or provide a static guarantee that you will be returning the same object (and not a copy). For instance, Buffer
s in Scala have the append operation :+
, which returns a Buffer[A]
, and +=
, which returns this.type
. The former duplicates the mutable sequence; the latter guarantees that you update the original object.
To follow up on Jean-Phillippe's answer, who wrote his exactly when I'm writing mine, here's the code:
trait SomeTrait {
def foo: this.type = this
}
class UsesTrait extends SomeTrait
object Main {
def main(args: Array[String]) {
println((new UsesTrait).foo) // prints UsesTrait@<hash value>
}
}
I found the following idiom useful:
class MyClass[T] {
self: T =>
def foo(): Option[T] = Some(this)
}
class ChildClass extends MyClass[ChildClass]
new ChildClass().foo()
//--> Option[ChildClass] = Some(ChildClass@2487b1)
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