Inner function cannot access outer functions variable

Question

I have created the following jsfiddle which highlights my problem. http://jsfiddle.net/UTG7U/

var ExampleObject = function() {
   var myArray = new Array();
   this.example = function() {
       alert(this.myArray);
   };
}

var exampleObj = new ExampleObject();
exampleObj.example();

I am new to JavaScript and trying to create an object, field and a method. I can't get my method to access my field variable.

It's recommended you use the array literal [] to create an array instead of new Array(); — David G
– David G, Commented Sep 2, 2012 at 19:44

Raymond Chen · Accepted Answer · 2012-09-02 20:10:19Z

7

You have confused two types of variables: Local variables and member variables. var myArray is a local variable. this.myArray is a member variable.

Solution using only local variables:

var ExampleObject = function() {
   var myArray = new Array(); // create a local variable
   this.example = function() {
       alert(myArray); // access it as a local variable
   };
}

var exampleObj = new ExampleObject();
exampleObj.example();

Solution using only member variables:

var ExampleObject = function() {
   this.myArray = new Array(); // create a member variable
   this.example = function() {
       alert(this.myArray); // access it as a member variable
   };
}

var exampleObj = new ExampleObject();
exampleObj.example();

answered Sep 2, 2012 at 20:10

Raymond Chen

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2 Comments

rjcarr Over a year ago

You answered this quite a while ago, but I don't think your second answer would work. In that case, 'this' is referring to the inner function, not the outer.

Raymond Chen Over a year ago

Try it. It works. That's because this refers to "the thing before the dot" which in this case is exampleObj.

Saket Patel · Accepted Answer · 2012-09-02 19:41:40Z

6

you were trying to access a local variable using this operator which is wrong, so here is the working example

var ExampleObject = function() {
   var myArray = new Array(1,2,3);
   this.example = function() {
       alert(myArray);
   };
}
var exampleObj = new ExampleObject();
exampleObj.example();

Link: http://jsfiddle.net/3QN37/

answered Sep 2, 2012 at 19:41

Saket Patel

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Comments

Rhyono · Accepted Answer · 2012-09-02 19:41:22Z

1

You don't need the this.myArray. Using myArray alone will suffice (and work).

answered Sep 2, 2012 at 19:41

Rhyono

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Comments

Oliver Moran · Accepted Answer · 2012-09-02 19:42:31Z

1

What this is changes with the scope of each function. However, myArray will be visible to inner function. Example:

var ExampleObject = function() {
   var myArray = new Array();
   this.example = function() {
       alert(myArray);
   };
}
var exampleObj = new ExampleObject();
exampleObj.example();

answered Sep 2, 2012 at 19:42

Oliver Moran

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Comments

athspk · Accepted Answer · 2012-09-02 19:45:24Z

1

alert(myArray); should work fine I think

edited Sep 2, 2012 at 19:45

athspk

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answered Sep 2, 2012 at 19:41

SNAG

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