CODEWITHSUNDEEP
javascript
z.a.
z.a.

Reputation: 2777

javascript immediate function initialization

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;
                            }
                        }
                    };
                }());

Upvotes: 0

Views: 107

Answers (3)

jfriend00
jfriend00

Reputation: 707158

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.

Upvotes: 0

Jeremy J Starcher
Jeremy J Starcher

Reputation: 23863

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.

Upvotes: 1

Shomz
Shomz

Reputation: 37701

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

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

Upvotes: 0

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