"What does mean?"

Question

I'm trying to get the value returned from a function alert and I get the following in the popup:

[object Object]

The following is the JavaScript code:

<script type="text/javascript">
$(function ()
{
    var $main = $('#main'),
    $1 = $('#1'),
    $2 = $('#2');

    $2.hide(); // Hide div#2 when the page loads

    $main.click(function ()
    {
        $1.toggle();
        $2.toggle();
    });

    $('#senddvd').click(function ()
    {
       alert('hello');
       var a=whichIsVisible();
       alert(whichIsVisible());
    });

    function whichIsVisible()
    {
        if (!$1.is(':hidden')) return $1;
        if (!$2.is(':hidden')) return $2;
    }

});

</script>

whichIsVisible is the function I'm trying to check.

Answer 1

As others have pointed out, this is the default serialization of an object. But why is it [object Object] and not just [object]?

This is because there are different types of objects in Javascript!

• Function Object:
  stringify(function (){}) -> [object Function]
• Array object：
  stringify([]) -> [object Array]
• Regular Expression Object:
  stringify(/x/) -> [object RegExp]
• Date Object:
  stringify(new Date) -> [object Date]
• … there are more …
• AlsoObject object!
  stringify({}) -> [object Object]

This is because the constructor is called Object (capital "O"), and the term "object" (lowercase "o") refers to the structural nature of the object.

Usually, when talking about "objects" in Javascript, you actually mean "Object objects", not other types.

_{Where stringify should be like this:}

function stringify (x) {
    console.log(Object.prototype.toString.call(x));
}

Answer 2

The default result of converting an object to a string is "[object Object]".

Since you are dealing with jQuery objects, you may want to do the following

alert(whichIsVisible()[0].id);

To print the ID of the element.

As mentioned in the comments, you should use the tools included in browsers like Firefox or Chrome to inspect objects, instead of using alert, you can do console.log(whichIsVisible( )).

Note: ID should not start with a number.