Help to understand which method is called

[Jasper-M]

If none of your superclasses override equals it should be the same as this eq that.asInstanceOf[AnyRef].

[szeiger]

See https://github.com/scala/scala/pull/7348 for the new mutable HashMap and https://github.com/scala/scala/pull/6975 for cached hashcodes in the new immutable HashMap.

[tarsa]

@jimka

what would be the semantics of this?

  override def equals(obj:Any):Boolean = {
    super.equals(obj)
  }

That effectively changes nothing, except that your stacktraces that cross that method implementation will be longer.

[jimka]

@tarsa

That effectively changes nothing, except that your stacktraces that cross that method implementation will be longer.

Well one thing it does do is prevents scala from writing another equals method for the case class.

[jimka]

@tarsa

In Scala world the type of reference doesn’t matter on method selection, i.e. it matters at compile time to decide whether code compiles, but runtime selection of method implementation is done using actual object class, not reference type.

Is this statement still true in light of the other discussion? in Dynamic dispatch on object, but static …

[tarsa]

@jimka

Well one thing it does do is prevents scala from writing another equals method for the case class.

Yes, that’s right. I’ve forgot about things automatically generated by Scala compiler.

[tarsa]

@jimka

Is this statement still true in light of the other discussion? in Dynamic dispatch on object, but static …

I was somewhat imprecise when writing that statement. It’s true if you don’t combine overloading with upcasting (unless I forgot something) and can be false otherwise. That other discussion clears thing up.