Implementing inner traits in Scala like we do with inner interfaces in Java

Question

This code in Java compiles without errors:

interface T {
    interface Q {
    }
}

class C implements T.Q {
}

whereas this code in Scala does not:

trait T {
    trait Q {
    }
}

class C extends T.Q {
}

What is the correct translation (if it exists) of the Java code listing into Scala?

Theoretical explanations about language design are welcome.

Agnesse

In addition to what others have already answered about path-dependent types, it is also worth noting that Scala's notation for Java's T.Q is T#Q. – Omentum

The T#Q syntax Oleg referred to is called a type projection. And note that you can extend Q if you do it within a type derived from T, e.g. class A extends T { class B extends Q }. – Gentlewoman

Answer 1

The inner type Q is defined only for specific instance implementation of the T trait. Since scala has path-dependent types, each instance of T will have his own subtrait Q.

scala> trait T {
     |   trait Q
     | }
defined trait T

scala> class C extends T {
     |   def getQ: this.Q = new this.Q {}
     | }
defined class C

scala> val inC = (new C).getQ
inC: C#Q = C$$anon$1@3f53073a


scala> val c = new C
c: C = C@1a7e4ff0

scala> new c.Q {}
res4: c.Q = $anon$1@36bbb2f5

---

If you need an interface for a generic behavior for your clients to implement, and not dependent on a specific C instance, you should define it within an Object

scala> object T {
     |   trait Q {
     |     def implementMe: Unit
     |   }
     | }
defined module T

scala> val inT = new T.Q {
     |   def implementMe = println("implemented!")
     | }
inT: T.Q = $anon$1@20f2a08b

scala> inT.implementMe
implemented!

---

Why path-dependent types?

As for the design reasons, look here

Zarf

Answer 2

You cannot do that. When types are nested you're creating what is called a path-dependent type, meaning the type of each instance of the inner entity is tied to the specific instance within which it's constructed.

In other words, you interface Q has no independent existence that would allow you to refer to it apart from an instance of T.

Berkman