CODEWITHSUNDEEP
javascript
user1336103
user1336103

Reputation: 3877

unable to call self invoking function

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

Upvotes: -1

Views: 186

Answers (2)

kinsho
kinsho

Reputation: 536

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.

Upvotes: 0

jsha
jsha

Reputation: 652

You are evaluating the function at line 0, and assigning the return value "20" to test. Since 20 is a number, not a function, you can't call it. Try instead:

var test = function(a,b){
  return a*b;
};
alert(test(4,5));
alert(test(10,5));

Upvotes: 2

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