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Home/ Questions/Q 9161799
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T14:03:19+00:00 2026-06-17T14:03:19+00:00

i am trying to call self invoking function but it doesn’t seem to work,

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i am trying to call self invoking function but it doesn’t seem to work, as you can see code below i am able to alert (test) but not when it is called upon another function. Please advise – Thank you

var test = (function(a,b){
       return a*b;
           })(4,5);

function myFunc() {};

alert(test); // working
alert(test.call(myFunc, 10,5)); // not working
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  1. Editorial Team
    Editorial Team
    2026-06-17T14:03:20+00:00Added an answer on June 17, 2026 at 2:03 pm

    An immediately invoking function is one that executes right when the script is loaded. In your example, the function next to test is executed right away, and it returns a value of 20.

    I have a feeling what you really want is something like this:

    var test = (function()
    {
        var a = 4,
            b = 5;
    
        return function()
        {
            return a*b;
        }
    }());
    

    So in what I wrote above, test will NOT be set to 20. Instead, it’ll be set to a function that multiplies a against b and returns 20. Why? Cause when I immediately invoke the function, it’s not returning the actual value; it’s returning yet another function, and that function then returns the actual value that I’m trying to calculate.

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