Why doesn't returning a function defined outside the current function create a closure?

Question

In the function outer, I am returning a function that uses the same name as a variable declared/defined inside outer.

Why then is a closure not created? Why does the following code print undefined, and not Yolo!?

function inner(){
    console.log('theVar', theVar);
}

function outer(){
    var theVar = 'Yolo!';
    return inner;
}

console.log('Starting...');
outer()();

Because `theVar` is not in the scope of `inner()`. The fact that you're returning the function reference from `outer` is irrelevant to what `inner` has access to. — VLAZ, Sep 17 '18 at 18:12
Because you haven't defined `inner` *inside* `outer`. https://jsfiddle.net/khrismuc/upchb9dv/ — , Sep 17 '18 at 18:12
A closure is created where the function is **declared** (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/function) not in all places it is assigned — Denys Séguret, Sep 17 '18 at 18:12
Read the text of the `lexical-scope` tag that you yourself added. They're called *lexical* closures for a reason. — Jared Smith, Sep 17 '18 at 18:16
`function a() { var x = 5; } function b() { console.log(x); } a(); b()` - would you expect these to log the value `5` just because the variables have the same name? No, the `var x` declares a *local* variable — Bergi, Sep 17 '18 at 18:17

score 0 · Answer 1 · answered Sep 18 '18 at 01:54

0

var declarations inside a function scope are only accessible within that function, in your case outer()

answered Sep 18 '18 at 01:54

kamp

375
3
13

Why doesn't returning a function defined outside the current function create a closure?

1 Answers1