Why is the output of this 4? Could someone be kind enough to explain?

Question

var r = (function (x, f = () => x) {
var x;
var y = x;
X = 2;

return (x + y + f() );

})
console.log (r)

Answer 1

You are probably missing something.
This code won't output 4 but will print the function r.

Answer 2

Important thing to note on this code is that the function

f = () => x

on second argument, operates on the x value that is in the arguments and not inside the function where x = 2

so f() will always return that value that is passed as the argument to x in the IIFE

Answer 3

var r = ( function(){
// code
})(1);

is a immediately invoked function expression IIFE ,

and in IIFE we can pass the argument to the function by the outer () like (1) , so 1st argument x is passed value 1

in next argument f = () => x we define a function based on argument x in es6 arrow notation , which simplifies to this

var f = function (){
return x ;
}

so doing f() later returns the passed arg value of x

Also note that final value at r depends on what is returned inside the IIFE,

return ( x + y + f() ) ; breaks down to

return ( 2 + 1 + 1); // 4

Answer 4

Ahhhh... I see. That was a really thoughtful response, thank you. This is the first I am hearing about IIFEs and arrow notations. I will have to study them in more depth but really appreciate the explanation. Helps a great deal.

Answer 5

as Swapnil said , but here are few modifications to produce 4, you must be missing something, and this one might be different

var r = (function (x, f = () => x) {
var x;
var y = x;
x = 2;

return (x + y + f() );

})(1);

console.log (r)

Answer 6

Yes Morpheus, you are very correct. I forgot the (1) at the end. But still not understanding how it outputs 4?