Trait with a method returning the actual type that extends the trait

Question
1

Often I have to write code like this:

  trait MyTrait[A <: MyParentTrait]:
    def myMethod: A = ... impl in trait ...

I have to have a type arg A with the only purpose that a call to myMethod returns the actual type extending MyTrait:

case class MyClassExt(...) extends MyParentTrait with MyTrait[MyClassExt]

Having the type arg A is complicated/hard to manage. Is there any way of having myMethod return MyClassExt (or whatever type extends MyTrait) without the type arg?

2

There’s some reading on this subject at Scala FAQ | Scala Documentation

1 Like
3

Among the things mentioned in that FAQ, I would recommend using a type member instead.

trait MyTrait:
    type This
    def myMethod: This = ... impl in trait ...

class MyClass extends MyTrait:
    type This = MyClass
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4

Just to add, the usage shown by Martin above is explained in much more detail in Programming in Scala, 5th edition, Chapter 20.6 “Abstract Types”.

Screenshot from 2024-02-25 13-07-41
Screenshot from 2024-02-25 13-07-41789Ă—970 109 KB

Reading the books pays off! :smiley:

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5

Thanks all, I ended up using the type This solution. I have more than 100 case classes (and more will be added in the future) and for a start it seemed the smallest change. Also not sure what 100+ implicits will do for compilation times if I ended up doing the type classes solution.

Anyway a few hours later I changed several traits that was using the trait X[A] pattern. It all works well and the code is simpler now. Only thing I have to declare type This on every case class, I suppose there is no shortcut to declare it on the parent trait like trait MyParentTrait: type This = this.class.type

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