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Home/ Questions/Q 952737
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T23:56:34+00:00 2026-05-15T23:56:34+00:00

Given an Array Literal inside a JavaScript Object, accessing its own object’s properties does

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Given an Array Literal inside a JavaScript Object, accessing its own object’s properties does not seem to work:

 var closure =  {

         myPic : document.getElementById('pic1'),
         picArray: [this.myPic]
 }    

 alert(closure.picArray[0]); // alerts [undefined]

Whereas declaring an Array Item by accessing an other JavaScript Object seem to work

 ​var closure1 = {
 ​    
 ​     myPic : document.getElementById('pic1')
 ​}
 ​    
 ​var closure2 =  {
 ​  
 ​        picArray: [closure1.myPic]
 ​}    
 ​    
 ​alert(closure2.picArray[0]); // alerts [object HTMLDivElement]

Example:
http://jsfiddle.net/5pmDG/

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T23:56:35+00:00Added an answer on May 15, 2026 at 11:56 pm

    The this value will not work like that, it refers to a value determined by the actual execution context, not to your object literal.

    If you declare a function member of your object for example, you could get the desired result:

    var closure =  {
      myPic: document.getElementById('pic1'),
      getPicArray: function () {
        return [this.myPic];
      }
    };
    //...
    closure.getPicArray();
    

    Since the this value, inside the getPicArray function, will refer to your closure object.

    See this answer to another question, where I explain the behavior of the this keyword.

    Edit: In response to your comment, in the example that I’ve provided, the getPicArray method will generate a new Array object each time it is invoked, and since you are wanting to store the array and make changes to it, I would recommend you something like this, construct your object in two steps:

    var closure =  {
      myPic: document.getElementById('pic1')
    };
    closure.picArray = [closure.myPic];
    

    Then you can modify the closure.picArray member without problems.

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