Question: I don't know if it is suppoused to do this, but I guess it is...

I thought at first that it may be something wrong with my whole script and I managed to make new file on localhost and test just `fadeOut();` function.

Apparently it returned function twice again so I went to jsfiddle to check what will happen there. And same thing happened. In this case `console.log();` returned twice.

* *What I did and what I am trying to do ?*

Well I want to return specified function, or in fiddle sample, specified `console.log();` only once. However I am fadingOut multiple elements (two, to be exact).

Is there any way to do that, instead of duplicating each element to fadeOut at the same time ?

**Sample that will return console.log(); twice.**

```
setTimeout(function () {

    $( ".one, .two" ).fadeOut(300, function () {
        console.log("Return Function!");
    });

}, 2000);

```

**Sample that will return console.log(); once.**

```
setTimeout(function () {

    $( ".one" ).fadeOut(300);

    $( ".two" ).fadeOut(300, function () {
        console.log("Return Function!");
    });

}, 2000);

```

**Fiddle Preview:**
[Fiddle Redirect](http://jsfiddle.net/dvLden/ApUYq/1/)

Answer: 
Perhaps a simple flag would suffice? Especially if you wanted to log to the console as soon as the first one was done (for whatever reason):

```
var already_returned = false;
setTimeout(function () {
    $( ".one, .two" ).fadeOut(300, function () {
        if (!already_returned) {
            already_returned = true;
            console.log("Return Function!");
        }
    });
}, 2000);

```
Need to see your HTML, but appears to me that you may have more than one element assigned with the class "two" - that would explain why in the second example you would outcome 2 log lines...

In the first example, you may (and probably is) selecting 2 elements (one with the class "one" and another with the class "two").