One function changing another function's local variable?

Question

I had a problem with some JavaScript functions that had me scratching my head for about an hour until some well-placed alert()'s revealed something which surprised me. One function was changing another function's local variables, it seems. I wrote a simple test script:

function first() {
    msg = "1111";

    second();

    alert(msg);
    }


function second() {
    msg = "2222";
    }

When I call first() I'd expect to get an alert box saying "1111" but I get "2222" instead. How is it that second() is affecting a local variable belonging to first()? Am I missing something or is this a bug?

I'm using Firefox 12.0.

developer.mozilla.org/en/JavaScript/Guide/… (the whole guide is worth a read if you are new to JavaScript) — Felix Kling
– Felix Kling, Commented Jun 23, 2012 at 21:11
I used to be really confused by variable scope, and then I learned to shut up and love the closure scope. — Jared Farrish
– Jared Farrish, Commented Jun 23, 2012 at 21:13

matt3141 · Accepted Answer · 2012-06-23 21:11:14Z

5

The variable is only local when the var statement is used:

var msg = "1111";

Otherwise the value escapes into the global scope.

answered Jun 23, 2012 at 21:11

matt3141

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3 Comments

Jared Farrish Over a year ago

Not to be picky, but the "1111" and "2222" are also a little odd.

Iain Over a year ago

Haha, in several years on and off with JavaScript (I do it for hobby not for work) this had completely evaded me until now. I've put var in front of msg=... and it works as expected. Thanks!

matt3141 Over a year ago

Just think of it as without the var to declare the variable, it assumes the variable is present in the global scope.

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