Javascript: the Scope Pitfall II

Hello, and welcome to the second part of ["Javascript: The Scope Pitfall"]( Let's get started, shall we? What does the following code print?

~~~~~~~~~ var n = 0; function add() { var m = n + 1; print(m); } add(); ~~~~~~~~~

That was easy, right? The answer, of course is `1`, because `0 + 1 == 1`. Alright, a slight variation on the same example:

~~~~~~~~~ var n = 0; function add2() { var n = n + 1; print(n); } add2(); ~~~~~~~~~

What does this print? If your answer is `NaN` (not a number), you can stop reading. If you think the answer is `1` (as I did), read on. What is happening here? In my previous post I explained that [Javascript lifts variable declarations to the top of the function]( So, this code is equivalent to:

~~~~~~~~~ var n = 0; function add2() { var n; n = n + 1; print(n); } add2(); ~~~~~~~~~

It also turns out that when a previously undeclared variable (in this scope) is declared it is initialized to `undefined`. So we get:

~~~~~~~~~ var n = 0; function add2() { var n = undefined; n = n + 1; print(n); } add2(); ~~~~~~~~~

Any calculation involving `undefined` will return `NaN`, so naturally it will indeed print `NaN`. You can rewrite the code a little bit to make it print `1` again:

~~~~~~~~~ var n = 0; var oldN = n; function add3() { var n = oldN + 1; print(n); } add3(); ~~~~~~~~~

It took me a while to figure this one out, though, when I ran into it. Just thought I would share.