Closure in Javascript (Simple)

Question

I'm very new to programming and have a quick question regarding a demonstration of closure given here: http://youtu.be/hQVTIJBZook?t=27m19s.

The code is:

var digit_name = function () {
    var names = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine'];
    return function (n) {
        return names[n];
    };
} ();

alert(digit_name(3));

My question is: Why is there an additional (); outside of the function digit_name? Are we calling the anonymous secondary function? I ran it without the (); and it came back with " function (n) {return names[n];} ". I would really appreciate a description of what happens to the parameter 3 when you pass it to the function digit_name, since that function doesn't have a specified parameter.

I apologize if I'm asking a basic question or using incorrect terminology. I've looked into a few related questions but alas, to no avail! Thank you in advance to those kind enough to provide a well-fashioned answer.

"()" are there to actually call the function to get digit_name = <result of calling the function> — mishik
– mishik, Commented Jun 9, 2013 at 4:25
Of course, don't forget to the read the StackOverflow explanation of Javascript Closures — zastrowm
– zastrowm, Commented Jun 9, 2013 at 4:40
possible duplicate of What do empty parentheses () after a function declaration do in javascript? — Felix Kling
– Felix Kling, Commented Jun 9, 2013 at 7:13

Joseph · Accepted Answer · 2013-06-09 04:37:00Z

2

In that code, it makes the outer function execute immediately, thus returning the inner function as digit_name. But since the returned function carries a reference to names, the scope of the outer function does not terminate like it normally should. Thus, the scope lives and it becomes a closure.

It's like doing:

function anon() {
  var names = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine'];
  return function (n) {
    return names[n];
  };
}

var digit_name = anon();

alert(digit_name(3));

So when you call digit_name(), you are actually calling the returned inner function which carries it's own names.

I suggest you read about closures at MDN. I also have an article discussing about closures in simple terms.

edited Jun 9, 2013 at 4:37

answered Jun 9, 2013 at 4:26

Joseph

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