How does an IIFE's being called immediately prevent it from polluting global scope?

Question

In a Udacity lesson on immediately invoked function expressions (regarding the provided code snippet) it says:

The function that is being returned clo

Answer 1

How does an IIFE's being called immediately prevent it from polluting global scope?

It doesn't.

Being a function stops it polluting the global scope.

Variables declared inside a function exist only within that function

Answer 2

In modern JavaScript, you have let and block scope (and I'm pretty sure const has block scope, too), so you could just do this:

let myFunction;
{
    let hi = 'Hi!';
    myFunction = function () {
        console.log(hi);
    };
}

This creates myFunction without leaking hi into the surrounding scope.

In traditional JavaScript, where you only have var and function scope, you could do this:

var myFunction;
function a_private_scope() {
    var hi = 'Hi!';
    myFunction = function () {
        console.log(hi);
    };
}
a_private_scope();

a_private_scope limits the scope of hi, but (it being a function declaration) it needs to be called explicitly, and we still leak a name to the surrounding scope (this time it's a_private_scope, the name of the function-serving-as-a-scope).

By using a function expression and immediately calling it, we avoid this second name pollution:

var myFunction;
(function () {
    var hi = 'Hi!';
    myFunction = function () {
        console.log(hi);
    };
})();

Now the only thing defined in the outer scope is myFunction. The anonymous function that serves as a scope for hi has no name it could pollute the surrounding scope with.

Finally we can clean it up a bit by using return values, so we don't have to mention myFunction twice:

var myFunction = function () {
    var hi = 'Hi!';
    return function () {
        console.log(hi);
    };
}();

(This also saves us a pair of ( ) because the function keyword doesn't appear at the beginning of a statement anymore.)