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Home/ Questions/Q 737787
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T07:45:42+00:00 2026-05-14T07:45:42+00:00

When we have code like: function a(){ var x =0; this.add=function(){ alert(x++); } }

  • 0

When we have code like:

function a(){
  var x =0;
  this.add=function(){
    alert(x++);
  }
}

   var test = new a();
   test.add(); // alert 0
   test.add(); // alert 1
   test.add(); // alert 2

How does this work?
Doesn’t that the value of ‘x’ in a() should be ‘gone’ as soon as test = new a() is complete? The stack contains x should also be gone as well, right? Or, does javascript always keep all the stacks ever created in case they will be referenced in future? But that wouldn’t be nice, would it…?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T07:45:43+00:00Added an answer on May 14, 2026 at 7:45 am

    What you’re seeing is the effect of a closure. The function being defined within the other function gets access to all of the variables and such in scope where it is — even after the outer function returns. More here, but basically, the variables (and arguments) in the function all exist as properties on an object (called the “variable object”) related to that function call. Because the function you’ve bound to this.add is defined within that context, it has an enduring reference to that object, preventing the object from being garbage-collected, which means that that function can continue to access those properties (e.g., the variables and arguments to the function).

    You normally hear people saying that the function closes over the x variable, but it’s more complex (and interesting) than that. It’s the access to the variable object that endures. This has implications. For instance:

    function foo() {
        var bigarray;
        var x;
    
        bigarray = /* create a massive array consuming memory */;
        document.getElementById('foo').addEventListener('click', function() {
            ++x;
            alert(x);
        });
    }
    

    At first glance, we see that the click handler only ever uses x. So it only has a reference to x, right?

    Wrong, the reference is to the variable object, which contains x and bigarray. So bigarray‘s contents will stick around as well, even though the function doesn’t use them. This isn’t a problem (and it’s frequently useful), but it emphasizes the underlying mechanism. (And if you really don’t need bigarray‘s contents within the click handler, you might want to do bigarray = undefined; before returning from foo just so the contents are released.)

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