Syntax for calling super

[Peter Levart]

On Monday, August 27, 2012 12:19:09 PM Dan Smith wrote:
> > What about if there are several outer classes that all directly implement
> > K  and you don't want to choose the innermost?
> 
> There is no feature (in the current JSR 335 spec) to refer to a
> superinterface method of an enclosing instance.  The rule for 'Foo.super'
> is that 'Foo' must refer to either a lexically enclosing class name or a
> direct superinterface of the immediately enclosing class.  Otherwise, it's
> an error.

Luckily there is a work-around when you put it all together :-)


public class Outer implements K {
    @Override
    public void m() {

        Runnable K_super_m = K.super::m;

        new K()
        {
            @Override
            public void m()
            {
                K_super_m.run();
            }
        }.m();
    }
}


Regards, Peter