javascript calling function as array element

Question

I came across this example in a Toptal youtube video that uses syntax that won't run in Chrome, unless I am missing something. This example comes up here (JavaScript closure inside loops – simple practical example) and the same syntax is used. Why is this not running for me/does the indicated line below contain valid syntax?

var x, y, funcs = [];

for(x=0; x<5; x++){

    (function(){
        var r = x;
        funcs.push(function(){
        console.log(r);
        });

    });
};  
for (var y=0; y<5; y++){
    funcs[y]();  //<< is this valid JS syntax/why is it not working for me?
};

You declare y as [] and then again as index, this might not be working as expected. Also retrieve a return value when iterating your array and try to check the result in a debugger. — bash.d
– bash.d, Commented Aug 16, 2015 at 14:28
The problem is not with the function call syntax; that's fine. The problem is that your first for loop isn't putting anything in the array. You've got an anonymous function that's never called, so the first loop does nothing. — Pointy
– Pointy, Commented Aug 16, 2015 at 14:31

connexo · Accepted Answer · 2015-08-16 14:38:50Z

3

You need to make this anonymous function inside your first for-loop self-executing by adding a pair of brackets () after the definition.

(function(){
    var r = x;
    funcs.push(function(){
    console.log(r);
    });

})();

Without this, you keep declaring a function inside your first loop which is never executed and thus, never pushes anything into your array.

After your first loop, do a console.log(funcs.length) and you will get a 0 since your array has no elements.

Your current code will give you uncaught type error: funcs[0] is not a function which is because funcs[0] is undefined.

edited Aug 16, 2015 at 14:38

answered Aug 16, 2015 at 14:28

connexo

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3 Comments

Pointy Over a year ago

Probably should pass x to the function as well to avoid the closure/loop problem.

connexo Over a year ago

That shouldn't be required since all anonymous functions in the array have declared their own var r.

Pointy Over a year ago

They do, that's true; I guess that should work fine!

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