x declared after fn function but returns its value. Why fn didn't return undefined?
function doWork(){
let fn = function(){ return x; }
let x = 2;
return fn();
}
console.log(doWork()); // 2
1 Answer
Inside of your doWork() function, first you set up a function and assign it to fn -- This function is not invoked yet. You then define x as 2. After this definition you invoke fn() by calling return fn().
Because JavaScript works from top-to-bottom, x is defined at the time you reference fn(), so fn() is able to return x correctly.
This can be seen in the following:
function doWork() {
let fn = function() {
return x;
}
let x = 2;
return fn();
}
console.log(doWork()); // 2
fn()is called afterxis assigned.letcreates variables in a scope. There’s no ordering within a scope, just a "temporal dead zone". If this didn't work, it would be super annoying.