JS variable scoping [duplicate]

Question

 var x = 5;

 function a() {

     function b() {
         console.log(x);
     }

     b();
     var x = 6;
 }

 a();

In console.log(x), output is undefined. But if I change to,

 var x = 5;

 function a() {

     function b() {
         console.log(x);
     }
     b();
     x = 6; // remove `var`
 }

 a();

then output is 5.

Why this behaviour? Please explain me.

Guffa · Accepted Answer · 2015-06-11 23:12:03Z

2

In the first example the local variable x is hoisted to the top of the scope, i.e. the function where it is declared. It works the same as if it was declared at the top:

var x = 5;

 function a() {
     var x;

     function b() {
         console.log(x);
     }

     b();
     x = 6;
 }

 a();

As the local variable isn't assigned a value before the function b is called, it's contains undefined.

answered Jun 11, 2015 at 23:12

Guffa

703k112 gold badges760 silver badges1k bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

RobG Over a year ago

Per ECMA-262 at step 8 c ii, variables created through a variable declaration are initially assigned a value of undefined, so return undefined until assigned some other value.

Collectives™ on Stack Overflow

JS variable scoping [duplicate]

1 Answer 1

1 Comment

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Linked

Related