jQuery error - Uncaught TypeError: object is not a function

Question

I tried debugging but no luck as of yet! This Works:

function slider(direction){
this.direction = direction;
}

slider.prototype.move = function(){
console.log('moving');
};

var slider1 = new slider('backward');
console.log(slider1.direction);

var slider2 = new slider('forward');
console.log(slider2.direction);

This doesn't work:

var slider = new slider('backward');
console.log(slider.direction);

var slider2 = new slider('forward');
console.log(slider2.direction);

I know this would have something to do with the semicolon and I'd searched some other articles on this but couldn't find an answer to my problem

Is the slider function defined in the second block of code? — lonesomeday
– lonesomeday, Commented Oct 12, 2012 at 22:21
As lonesomeday suggests, this is likely a scope issue. If I was to guess, I'd bet that the first bit of code is hidden in a closure like $(document).ready(function(){/*closure*/}); — John Green
– John Green, Commented Oct 12, 2012 at 22:22
Yes. I just didn't want to copy the whole thing again. I am not sure why this error is happening.. tried debugging it in Chrome but doesn't give me any feedback? — Andrew
– Andrew, Commented Oct 12, 2012 at 22:23

codingbiz · Accepted Answer · 2012-10-12 22:23:29Z

2

It's hard to know the problem from your code but I think this line could cause it

var slider = new slider('backward');

You declared slider as function and as a variable at the same time. That is if my guess is right. This is because it is the only difference between the first code and the second

answered Oct 12, 2012 at 22:23

codingbiz

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2 Comments

lonesomeday Over a year ago

Ah yes, that's the problem. That's effectively saying new undefined('backward'). Good spot.

Andrew Over a year ago

Hmm I see, but having this along works: var slider = new slider('backward'); console.log(slider.direction);

Joe · Accepted Answer · 2012-10-12 22:24:34Z

1

When you run

var slider = new slider('backward');
console.log(slider.direction);

you now have 2 variables called slider - one is a function and one is an instance of the object that function generates. When you ask the second slider to generate:

var slider2 = new slider('forward');
console.log(slider2.direction);

you're actually asking it to use the instance of the object as a variable... this feels really clunky to explain because they have the same names.

What you're doing in the second example, which causes it to fail, is the same as trying to do this in the first example:

var slider2 = new slider1('forward');
console.log(slider2.direction);

because the function name is the same as a variable containing an instance of it.

If this hasn't been explained massively clearly (which wouldn't surprise me) let me know and I'll try to make it make more sense.

answered Oct 12, 2012 at 22:24

Joe

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1 Comment

Andrew Over a year ago

Yes sir. I got it now! Thank you all for your help!

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