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Home/ Questions/Q 8385839
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T17:46:09+00:00 2026-06-09T17:46:09+00:00

I have a class which source I cannot modify: class Foo { def bar()

  • 0

I have a class which source I cannot modify:

class Foo {

  def bar() = println("bar")

}

And a trait I’d like to mix into it at runtime

trait Zee { this: Foo =>

  abstract override def bar() = {
    println("before bar")
    super.bar()
  }
}

This is throwing that bar is not a member of Object with ScalaObject

What am I doing wrong? Is it possible to achieve this without modifying Foo source?

The ultimate client code needs to look like this:

val foo = new Foo with Zee
foo.bar() // should print 'before bar' and then 'bar'
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T17:46:11+00:00Added an answer on June 9, 2026 at 5:46 pm

    Your Zee trait has no super traits (except implicit inheritance from ScalaObject) thus super does not contain definition for bar and there is nothing to override or call (super.bar).

    why don’t you write this without self-reference?

    class Foo {
    
      def bar() = println("bar")
    
    }
    
    
    trait Zee extends Foo {
    
      abstract override def bar() = {
        println("before bar")
        super.bar()
      }
    }
    
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