Odd thing with javascript "undefined"

Question

I have the following simple Javascript code.

var input = [
    'one',
    'two'
];

for(var i in input){
    if(typeof input[i+1] == undefined){
        console.log("i is the last index");
    }
}

I don't know if I did something wrong but the console.log() part never executes. Which means it never enters the if condition while clearly the index beyond the last index is undefined.

You can see it in this fiddle.

Please explain..

[Do not use `for in`-enumerations on arrays!!](http://stackoverflow.com/q/500504/1048572) — Bergi, May 23 '13 at 12:11
Yet another example of why people shouldn't use `typeof foo === "undefined"` to check for the `undefined` value. It causes far more bugs than it "solves". Just keep it simple and test for `input[i+1] == undefined`, and ignore the FUD people spread. They've usually not really thought things through. — , May 23 '13 at 12:23
...anyway, your test for the last index can fail since a defined index can have the `undefined` value. I really don't know why you're using `for-in` like this, but to test for the last index, you should be comparing the `i+1` to `input.length`. — , May 23 '13 at 12:30

score 6 · Accepted Answer · answered May 23 '13 at 11:59

6

if(typeof input[i+1] === 'undefined') { ... }

answered May 23 '13 at 11:59

Seer

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score 5 · Answer 2 · edited May 23 '13 at 12:07

5

This:

if(typeof input[i+1] == undefined){

Should be:

if(input[i+1] === undefined){

(no need to use typeof)

edited May 23 '13 at 12:07

Shadow The GPT Wizard

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answered May 23 '13 at 11:59

Rob Johnstone

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1

Ok, the downvotes mean I'm being dumb but can someone explain why? This works correctly (I've double-checked it in Chrome). Is it simply poor style? – Rob Johnstone May 23 '13 at 12:02
+1 You are right, i think the dumbest ones have down-voted this post – asim-ishaq May 23 '13 at 12:07
Sorry, was sure you wrote `if (typeof input[i+1] === undefined)` my mistake. Took the liberty too edit your post so I can undo my mistake. – Shadow The GPT Wizard May 23 '13 at 12:08
@ShadowWizard - Thanks. I probably should have made it clearer to start with – Rob Johnstone May 23 '13 at 12:09
I made the same mistake as ShadowWizard. Have removed my downvote – Dancrumb May 23 '13 at 12:17

score 3 · Answer 3 · answered May 23 '13 at 12:00

3

Undefined should be a string, "undefined", working fiddle: http://jsfiddle.net/asifrc/vRTsE/1/

answered May 23 '13 at 12:00

asifrc

5,781
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@asim-ishaq It works fine for me in Chrome.. from a console, from the fiddle I linked above, and even a quick page on my local server.. Does the fiddle not work in Chrome for you? I'd be curious to know why.. – asifrc May 23 '13 at 12:15

score 1 · Answer 4 · answered May 23 '13 at 12:05

That is because the typeof operator returns an string. You need to compare with a string "undefined" like so:

var input = [
    'one',
    'two'
];

for(var i in input){
    if(typeof input[i+1] == "undefined"){
        console.log("i is last");
    }
}

Dancrumb · Answer 5 · 2013-05-23T14:55:40.157

0

typeof returns a string. You're comparing it to the undefined value.

Use

if(typeof input[i+1] === "undefined"){

edited May 23 '13 at 14:55

answered May 23 '13 at 12:00

Dancrumb

26,597
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score 0 · Answer 6 · answered May 23 '13 at 12:01

0

javascript typeof operator always returns string, so you should compare against 'undefined' like this:

if(typeof input[i+1] === 'undefined')

Here is an updated fiddle - http://jsfiddle.net/Pharaon/V7EJZ/

answered May 23 '13 at 12:01

Sergey Rybalkin

3,004
22
26

score 0 · Answer 7 · answered May 23 '13 at 12:05

0

Turns out that the type of i is string, so to make the code work properly you need to cast it to integer:

for(var i in input){
    if(typeof input[parseInt(i, 10) + 1] === "undefined") {
        console.log(i + " is the last index");
    }
}

Updated fiddle.

answered May 23 '13 at 12:05

Shadow The GPT Wizard

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Odd thing with javascript "undefined"

7 Answers7