javascript immediate function initialization

Question

In the code below, my return function inside the other function doesn't get run right away, but the first function is. My goal is to have the whole function running immediately.

var windowCheck = (function () {
                    var switcher = false;
                    return function () {
                        console.log(switcher);
                        if ($window.innerWidth > 480) {
                            if (!switcher) {
                                element.perfectScrollbar({
                                    wheelSpeed: scope.wheelSpeed || 50,
                                    wheelPropagation: $parse(attrs.wheelPropagation)() || false,
                                    minScrollbarLength: $parse(attrs.minScrollbarLength)() || false
                                });
                                console.log('Plugin On');
                                switcher = true;
                            }

                        }
                        else {
                            if (switcher) {
                                console.log('Plugin Off');
                                element.perfectScrollbar('destroy');
                                switcher = false;
                            }
                        }
                    };
                }());

Jeremy J Starcher · Accepted Answer · 2014-04-19 19:31:40Z

1

The function does indeed run the first time -- an anonymous function runs and it's result is --

Another function!

This 'second' function is then assigned to the value of windowCheck.

And windowCheck won't run until called.

answered Apr 19, 2014 at 19:31

Jeremy J Starcher

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2 Comments

z.a. Over a year ago

ok, how about when I add another set of parentheses after the returning function, it actually runs but then i can't call it again.

Jeremy J Starcher Over a year ago

Corrrect.. because when you add the second set of parens you are causeing it to invoke, and the return value of that function, undefined i this case, gets assigned to windowCheck. IIFE's are great for one-shot runs, but if you want to be able to run it a second time, you have to assign it to a variable and then call the function.

Shomz · Accepted Answer · 2014-04-19 19:31:30Z

0

Simply run it after defining it and avoid all the anonymous function mess:

var windowCheck = function() {
    ...
}
windowCheck();

answered Apr 19, 2014 at 19:31

Shomz

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1 Comment

jfriend00 Over a year ago

They are using the closure to house the switcher variable and keep it private. So, there is some point to the closure.

jfriend00 · Accepted Answer · 2014-04-19 19:35:55Z

0

The code as you have it now defines the windowCheck function and uses a closure to house a private switcher variable. That is all fine. If you want the function that gets defined to be executed, then you just have to add this to the end of what you already have:

windowCheck();

in order to actually execute it immediately.

answered Apr 19, 2014 at 19:35

jfriend00

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