How can a variable be used if it's outside of curly braces in JS?

Question

I am not understanding this - how can a var that's defined within an if condition be used outside of that condition?

Example JS:

if (1===2) {
  var myVar = "I live in brackets";
}
$("#debug").append("myVar = " + myVar);
$("#debug").append("but I'm about to throw a 'not defined' exception right... now " + firstAppearanceVar);

Renders: myVar = I live in brackets

Isn't myVar's scope only inside of that if (1===2) condition?

In javascript, scope is by function. `if(){}` does not have its own scope — Sharlike, Feb 28 '13 at 19:17
MDN is a great resource for learning javascript: https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Values,_variables,_and_literals#Variable_scope — jholloman, Feb 28 '13 at 19:19

score 2 · Answer 1 · answered Feb 28 '13 at 19:16

2

Scopes only apply to functions, not other blocks.

answered Feb 28 '13 at 19:16

Niet the Dark Absol

320,036
81
464
592

so, definitely a difference between JS and C# in that matter, for sure. – Ian Davis Feb 28 '13 at 19:16
Considering the names are different, yeah, I'll say they're different! – Niet the Dark Absol Feb 28 '13 at 19:17
I just say that b/c the syntax is similar in both languages. – Ian Davis Feb 28 '13 at 19:22
@IanDavis, no, C# and js are completely different. [js has truthy values](http://stackoverflow.com/a/9579290/601179), unlike C#. a lot of differences. ohh and C# isn't a dynamic language... a bit different. – gdoron Feb 28 '13 at 19:24

score 2 · Answer 2 · answered Feb 28 '13 at 19:16

2

Javascript does not have block scope, it only has function scope.

In other words, variables declared by var are accessible within the scope of a function, everywhere, just there, not outside.

answered Feb 28 '13 at 19:16

jAndy

231,737
57
305
359

score 1 · Answer 3 · answered Feb 28 '13 at 19:17

1

Because of hoisting, every variable declaration pops to the top of the function scope.

alert(foo); // undefined (no error because of the hoisting.)
var foo = 2;
alert(bar); Error

answered Feb 28 '13 at 19:17

gdoron

147,333
58
291
367

but, I don't have any functions on the page. in that case, does JS just treat this all as under the umbrella of one wrapper function? – Ian Davis Feb 28 '13 at 19:18
1

@IanDavis, if it's not in a function it's inside the global object. – gdoron Feb 28 '13 at 19:20

score 0 · Answer 4 · answered Feb 28 '13 at 19:23

When a javascript variable is defined, the declaration is hoisted to the top of the function scope.

So this:

if (1===2) {
  var myVar = "I live in brackets";
}
$("#debug").append("myVar = " + myVar);
$("#debug").append("but I'm about to throw a 'not defined' exception right... now " + firstAppearanceVar);

is equivalent to this

var myVar;
if (1===2) {
  myVar = "I live in brackets";
}
$("#debug").append("myVar = " + myVar);
$("#debug").append("but I'm about to throw a 'not defined' exception right... now " + firstAppearanceVar);

Any variable defined in a function therefore, is accessible anywhere in that function, or inside any inner functions. They are not accessible outside the function.

so

(function(){
    if (1===2) {
      var myVar = "I live in brackets";
}}())
$("#debug").append("myVar = " + myVar); //reference exception

How can a variable be used if it's outside of curly braces in JS?

4 Answers4