Scope of variable i in Javascript

Question

<html>
<body>
<script type= "text/javascript">

var i=10;
function f(){
   alert(i);
   if(!i){
     var i= 20;
   }
   alert(i);
}
f();
</script>
</body>
</html>

This is my html file. The outputs I get are undefined for first alert and 20 for the second alert. Why is this so? What is happening here?

You're creating private variable `var i = 20;` but previously you're alerting i so get undefined.... — Bhojendra Rauniyar, Jan 17 '15 at 14:26

beartech1 · Accepted Answer · 2015-01-17T14:39:51.850

0

I believe it's because the i is hoisted, but the assignment of i is not. Therefore, when you alert i, it is referring to the locally scoped i which hasn't yet been assigned a value.

It ends up working like this:

var i;
alert(i);
if(!i){
    i = 20;
}

edited Jan 17 '15 at 14:39

answered Jan 17 '15 at 14:25

beartech1

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10

This is why in javascript you should declare function variables at the top because javascript doesn't have block scope. – beartech1 Jan 17 '15 at 14:26
oh so it is getting declared at top inside of function again? – Anirudh K Jan 17 '15 at 14:32
your code ends up working like this:(see edited answer) – beartech1 Jan 17 '15 at 14:37
figured it out thanks – Anirudh K Jan 17 '15 at 14:40

Scope of variable i in Javascript

1 Answers1