Mixin calling its own thing instead of super's thing

[bcowgill]

I was looking at how the mixin code works. In the test case paraphrased below, If I wanted to implement a mixin where I called the Sub-class's thing before the implemented thing, how do you do that? Calling this.thing directly will just give an infinite loop.

test('should implement a wrapper method for mixins', 5, function () {
...
inst.implement({
thing: function (n) {
console.log('Implement#thing()');
ok(true, 'called implementer');
//this.supr(); would be called by subclass' thing, if we could
this.thing(); // WARNING infinite loop. want to call the Sub's thing, not the base thing
// this.myself(); would be a good expression for that
this.boooshr();
},
boooshr: function () {
console.log('Implement#boooshr()');
ok(true, 'called boooshr');
}
}).thing();