How to declare traits as taking implicit “constructor parameters”?

Question

I\'m designing a class hierarchy, which consists of a base class along with several traits. The base class provides default implementations of several methods, and the trai

Answer 1

This isn't possible.

But you can use implicitly and Scala's type inference to make this as painless as possible.

trait MyTrait {

    protected[this] implicit def e: ClassName

}

and then

class MyClass extends MyTrait {

    protected[this] val e = implicitly // or def

}

Succinct, and doesn't even require writing the type in the extending class.