Syntax for calling super

John Hendrikx hjohn at
Thu Aug 23 01:42:51 PDT 2012

  On 23-8-2012 3:35, Brenden Towey wrote:
> True.  In that case, I think I prefer your suggestion, super.K.m().
> K.super.m() seems to be invoking the super-class of K, not of J. It's a
> little confusing.  super.K.m() reads a bit more clearly. "my super
> class" . "oh, this interface K though" . "m()"
Another alternative might be to write it as:

or even

The general behaviour of super seems to be that it gives you a reference 
to the superclass.  If it is unclear which one it should 'return', then 
passing it a parameter seems fairly natural syntax.
> I can see a case for a cast, but super.K.m() is a convenient shorthand.
> Less typing (and less reading!) is sometimes the better option.
> Something like ((K) super).m() is a bit unnecessarily long.
> On 8/22/2012 6:28 PM, David Holmes wrote:
>> On 23/08/2012 11:09 AM, Brenden Towey wrote:
>>> On 8/22/2012 3:32 PM, Yuval Shavit wrote:
>>>> To my mind, the syntax "should" be K.m() (without the super). That
>>>> makes
>>> I'd tend to agree, except for the problems you mention.  What about
>>> "J.super.m()"?  "Don't call me call my parent class?"
>>> interface J extends K {
>>>       int m() default { return J.super.m(); }  // calls J's parent's m()
>>> }
>> The problem with that is that you need to be able to select from
>> multiple parents, so the parent's type has to be in there somewhere.
>> David

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