How to call JS parent-methods without prototypical inheritance

Question

I'm working on a big project and decided against prototyping for inheritance (I wanted more control over var and this.* access).

Anyways, I have classes that look like this:

FirstScene.prototyp

FirstScene.prototype = Object.create(_SceneCanvas.prototype)
FirstScene.prototype.constructor = FirstScene

function FirstScene(){
     var arguments = [/*args*/]
     _SceneCanvas.apply(this, arguments)



     this.bar = function(){ /*do other stuff*/ }
}

If I was in foo() or bar() and wanted to call the parent method, how would I go about doing that? I tried some code that looked like _SceneCanvas.prototype.foo.call(params) to no avail.

Thanks!

The concept "parent method" doesn't really apply for "foo"; it's a local function in the constructor. — Pointy
– Pointy, Commented Aug 17, 2015 at 15:27
Also I don't understand what you mean by saying you decided against prototyping for inheritance, but then you go on to post code that uses prototype inheritance. — Pointy
– Pointy, Commented Aug 17, 2015 at 15:28
@Pointy my functions are not in the form of FirstScene.prototype.foo = function() Am I missing something? — 9TrSl9IAlaLfDmffagsw
– 9TrSl9IAlaLfDmffagsw, Commented Aug 17, 2015 at 15:36
Yes, that's just one way of putting properties on a prototype. You're still using prototype inheritance. — Pointy
– Pointy, Commented Aug 17, 2015 at 15:46

Zachary Kuhn · Accepted Answer · 2015-08-17 15:33:17Z

3

In JavaScript the common practice is to get a reference to the old function and call it from within the new function.

var bar = this.bar;
this.bar = function () {
  bar.call(this);
  // ...
};

Please check out this fiddle to see it in action.

edited Aug 17, 2015 at 15:33

answered Aug 17, 2015 at 15:29

Zachary Kuhn

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3 Comments

Pointy Over a year ago

... probably bar.call(this) to make sure there's a context.

9TrSl9IAlaLfDmffagsw Over a year ago

So why does this.bar() call the parent method instead of the immediate one? (Also, your solution works. Thanks!)

Zachary Kuhn Over a year ago

It is interpreted sequentially. When bar is assigned this.bar, this.bar is the parent's bar function. The next line then overwrites this.bar to be a new function. If you move var bar = this.bar; to after you overwrite it, you'll get the new function.

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