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..
if(typeof input[i+1] === 'undefined') { ... }
This:
if(typeof input[i+1] == undefined){
Should be:
if(input[i+1] === undefined){
(no need to use typeof)
Undefined should be a string, "undefined", working fiddle: http://jsfiddle.net/asifrc/vRTsE/1/
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");
}
}
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/
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");
}
}
typeof returns a string. You're comparing it to the undefined value.
Use
if(typeof input[i+1] === "undefined"){
typeofreturns a string.undefinedis... undefined.for in-enumerations on arrays!!typeof foo === "undefined"to check for theundefinedvalue. It causes far more bugs than it "solves". Just keep it simple and test forinput[i+1] == undefined, and ignore the FUD people spread. They've usually not really thought things through.undefinedvalue. I really don't know why you're usingfor-inlike this, but to test for the last index, you should be comparing thei+1toinput.length.