Javascript - return value is not considered an integer

Question

I have a problem with the returning value of the test function. This is the code:

$(document).ready(function () {
    $("p").hide( test );

    function test() {
            return 5000;
    }
});

After a bit of research a found the parseInt() function. But it didn't work.

try with parseInt():

    $(document).ready(function () {
        $("p").hide( test );

        function test() {
            return parseInt(5000, 10);
        }
    });

try with parseInt():

    $(document).ready(function () {
        $("p").hide( parseInt(test, 10) );

        function test() {
            return 5000;
        }
    });

I know I could just type:

$("p").hide( 5000 );

But that function will get more complex after resolving this problem.

What you pass into into hide() is the function test, not the result of test(). jQuery interprets this as a completion callback - "call test after hiding the element". — user1233508
– user1233508, Commented Feb 16, 2013 at 16:53

Darren · Accepted Answer · 2013-02-16 16:52:09Z

4

You are not executing the function:

$(document).ready(function () {
    $("p").hide( test() );  // note the additional ()

 function test() {
  return 5000;
 }
});

answered Feb 16, 2013 at 16:52

Darren

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2 Comments

+1. But also tell him that without () was for assigning function (which will not work in this case) and with () it will execute.

:D I just can't believe I forgot those. Thank you very much.