Javascript function and variables scope

Question

Quick question, because one thing confused me today. Below is piece of JavaScript code:

var a = [1, 2, 3, 4];

function init() {

    console.log(a);

    if (false) {
        var a = [9, 8, 7];
    }
}

init();

This simple code logs "undefined" in JS console. I found that the reason for this is 5th line:

var a = [9, 8, 7];

Without word var inside that condition, it will log correct result [1, 2, 3, 4].

My question here is: why JavaScript interpreter takes into consideration piece of code which never will be executed?

Example to play with: http://jsfiddle.net/5DPCz/

user229044 · Accepted Answer · 2014-04-25 14:24:47Z

4

You cannot declare variables in the middle of a function in JavaScript; all variables have function scope. Effectively, variable declarations are "hoisted" to the top of the function. Your function will only ever have access to the local variable a regardless of where in the function it is declared.

This is why JSLint (and most other linting tools) will advise you to move all variable declarations to the top of the function; that's where they're effectively written anyways.

Regardless of where in your function you place your var a, your code effectively executes this way:

var a = [1, 2, 3, 4];

function init() {
    var a;

    console.log(a);

    if (false) {
        a = [9, 8, 7];
    }
}

init();

answered Apr 25, 2014 at 14:24

user229044♦

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4 Comments

sowiq Over a year ago

Even if definition of that variable is surrounded by condition which will never be satisfied?

user229044 Over a year ago

@KrzysztofSowa Yes. The initialization won't run, but the variable is still being introduced in the function, and it will always have function scope.

Frédéric GRATI Over a year ago

@KrzysztofSowa, to go further : github.com/getify/You-Dont-Know-JS/blob/master/…

sowiq Over a year ago

@meagar, Freuh - thanks for your answers! It now makes more sense.

Tim · Accepted Answer · 2014-04-25 14:27:35Z

0

Your example is equivalent to

var a = [1, 2, 3, 4];

function init() {
    var a;
    console.log(a);

    if (false) {
        a = [9, 8, 7];
    }
}

init();

Now you can see why it prints undefined. The a that is local to the function has not been assigned a value at that point yet.

If you remove the var you turn it into

var a = [1, 2, 3, 4];

function init() {
    console.log(a);

    if (false) {
        a = [9, 8, 7];
    }
}

init();

So a in the function now refers to a from the global scope.

answered Apr 25, 2014 at 14:27

Tim

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2 Comments

sowiq Over a year ago

I know, but my question is - why? :) If I would do: if(false){ alert(1); } it would never show alert, because that line of code wouldn't be executed. I know it might be silly question, but I'm trying to understand mechanism which takes into consideration code which shouldn't even be reached by interpreter.

Tim Over a year ago

As meagar mentioned, it has to do with hoisting. Before the function runs, the JS interpreter will analyze it and find all parts where a variable declaration happens. It will then move the declarations to the beginning of the function, but keep the assignments at their positions. That's why all var declarations in the function scope shadow the outer scope, even if they only seem to be declared later in the code, and even if that code is unreachable.

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