How are local variables referenced in closures? [duplicate]

Answer 1

The loop ends when i becomes 3, but the "item" variable stored in the closure, and displayed by alert, is set to

var item = 'item' + list[i];

the text 'item' + the value at list[2]. The third list item is 3, so the text is item3

Answer 2

The list variable is stored in closure as you say.

Actually you can access the list variable, but you are trying to access list[3]. After all, the i variable is also stored as a closure and it's value is 3 when the console.log function is called.

Answer 3

I think the point you are missing is that list[i] is underfined because i is 3, and list is only defined for 0..2.

Answer 4

The for loop within buildList completes before you do the following:

for (var j = 0; j < fnlist.length; j++) {
  fnlist[j]();
}

... therefore, by that time (when you call each function), the variable item will be whatever was last assigned to it (i.e. "item3"), and i will be 3 (as a result of the last i++ operation), and list[3] is undefined.

It's all to do with the fact that the loop completes before you call the closure'd function. To prevent this, you could create a new closure, like so:

function buildList(list) {
  var result = [];
  for (var i = 0; i < list.length; i++) {
    var item = 'item' + list[i];
    result.push(
        (function(item, i){
            // Now we have our own "local" copies of `item` and `i`
            return function() {
                console.log(item + ' ' + list[i])
            };
        })(item, i)
    );
  }
  return result;
}

Answer 5

You're close...

Why does item have a value of 'item3'? Doesn't the for loop end when i becomes 3?

Yes.

If it ends shouldn't item still be 'item2'?

Nope. This example is a little tricky. During the last iteration of the loop, i is 2, but it references the 3rd element of the list array, which is 3. In other words, item == 'item' + list[2] == 'item3'

Or is the variable item created again when testList calls the functions?

No, you were almost right the first time. I think you just missed that item[2] has the value of 3.