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Syntax for Calling "super" in Java 8 Defender methods

09.07.2012
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This is a very interesting discussion. How to reference default methods from implemented interfaces throughout the class / interface hierarchy?

Situation:

interface K {
  int m() default { return 88; }
}

interface J extends K {
  int m() default { return K.super.m(); }
                        // ^^^^^^^^^^^^ How to express this?
}

Solution ideas:

  • K.super.m()
  • super.K.m()
  • ((K) super).m()
  • K::m()
  • K.default.m()
  • super<K>.m()
  • super(K).m()
  • super(K.class).m()
  • super[K].m()

Any other crazy ideas? See the discussion here:

http://mail.openjdk.java.net/pipermail/lambda-dev/2012-August/005616.html

 

 

Published at DZone with permission of Lukas Eder, author and DZone MVB. (source)

(Note: Opinions expressed in this article and its replies are the opinions of their respective authors and not those of DZone, Inc.)

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Comments

Qinxian Xiang replied on Sat, 2012/09/08 - 8:56pm

Why K.m not in sight? in the thread K.m simplest!

The key point is: K.m indicate direct subclass of K used. and JVM need maybe need new invokespecialinterface instruction!huf!

Konrad Malawski replied on Sun, 2012/09/09 - 2:15pm

I find Scala's solution to this pretty elegant (as with most things that java language design now struggles with due those default methods in interfaces (traits, duh!)):

trait M { def m = 88 }
trait K { def m = 44 }
trait J extends K with M {
  override def m = super.m // M#m, as it's the last mixin

  assert { super[K].m == 44 } 
  assert { m == 88 }
}

This could be easily adapted to Java syntax: super<K>.something

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