Reputation: 35
I am using a library which has a class that has a generic type that can be quite complicated. I need to write a method that takes a parameter with the generic type that a val of the library class has, and I would like to avoid having to write out the type in the method signature. I thought I might be able to create an implicit class which adds a type to the val that I could use in the method signature, kind of like:
// This comes from a library and can't be changed
case class LibraryClass[A](a: A)
//----------------------------------
object MyCode {
val thing = LibraryClass(3)
implicit class LibraryClassWithType[A](lc: LibraryClass[A]) {
type TheType = A
}
def doStuff(something: thing.TheType): Unit = {
println(something)
}
}
This does not compile (TheType is not a member of LibraryClass). But if I wrap it in the class myself, it works
val thingWithType = LibraryClassWithType(thing)
def doStuff(something: thingWithType.TheType): Unit = {
println(something)
}
Is there something I am missing that will make this work, or is this kind of implicit conversion not valid Scala?
Upvotes: 2
Views: 98
Reputation: 879
I haven't been able to do this sort of thing with implicits, but I have had to do something similar where I just instantiated these sorts of type holders:
case class LibraryClass[A](a: A)
object MyCode {
val thing = LibraryClass(3)
class HigherTypeHolder[A,F[A]](a: F[A]) {
type AT = A
}
val th = new HigherTypeHolder(thing)
def doStuff(something: th.AT): Unit = {
println(something)
}
}
Upvotes: 0
Reputation: 40500
You can do what (I think) you want like this:
implicit val thing = LibraryClass(3)
def doStuff[A](something: A)(implicit lc: LibraryClass[A])
What I don't understand is why this needs to be so complicated. Why not for example stick with your second approach, without implicits, that works, or why not just do
def doStuff[A](something: A)
to begin with?
Upvotes: 0