Why is jQuery.fx.off returning 'undefined'?

Question

In jquery jQuery.fx.off will return bool. But when I am executing the following line I am getting undefined.

alert(jQuery.fx.off);

Please tell me why is it so.

Very sure. Please see the link http://api.jquery.com/jQuery.fx.off/ — user2833192, Sep 17 '14 at 18:38
I think you need to explicitly set it. The value is set with `$.fx.off = !$.fx.off`, but `undefined` is considered falsey, meaning that `!undefined` will return `true`. — , Sep 17 '14 at 18:41
if you see in source - all checking like `jQuery.fx.off ? ... : ...` so by default it simply not defined, because _false_ and _undefined_ have same behavior in this case — Grundy, Sep 17 '14 at 18:43

score 2 · Accepted Answer · edited May 23 '17 at 11:52

In recent versions, at least, jQuery doesn't assign an initial value to jQuery.fx.off, leaving it undefined by default.

jQuery.speed = function( speed, easing, fn ) {
    // ...

    opt.duration = jQuery.fx.off ? 0 : // ...
        //         ^^^^^^^^^^^^^

To get a boolean based on its truthiness, you can use the Boolean() function (without new) or double !!:

alert(jQuery.fx.off); // undefined

var fxOff = Boolean(jQuery.fx.off);
alert(fxOff);         // false

1 Answers1