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Home/ Questions/Q 117235
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Editorial Team
Asked: May 11, 20262026-05-11T03:18:15+00:00 2026-05-11T03:18:15+00:00

Is there a performance hit when iterating over object attributes vs. iterating an array?

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Is there a performance hit when iterating over object attributes vs. iterating an array?

Example, using objects:

var x:Object = {one: 1, two: 2, three: 3}; for (var s:String in x) {  trace(x[s]); } 

Vs using an array

var a:Array = [1, 2, 3]; var len:Number = a.length; for (var i:Number = 0; i < len; ++i) {  trace(a[i]); } 

So – which is faster and most importantly by what factor?

IIRC, in some JavaScript implementation iterating over objects attributes is slower up to 20x but I haven’t been able to find such measurement for ActionScript2.

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  1. 2026-05-11T03:18:15+00:00Added an answer on May 11, 2026 at 3:18 am

    I just tried a very similar test, but iterating just once over 200k elements, with opposite results:

    Task build-arr: 2221ms Task iter-arr: 516ms  Task build-obj: 1410ms Task iter-obj: 953ms 

    I suspect Luke’s test is dominated by loop overhead, which seems bigger in the array case. Also, note that the array took significantly longer to populate in the first place, so ymmv if your task is insert-heavy.

    Also, in my test, storing arr.length in a local variable gave a measurable performance increase of about 15%.

    Update:

    By popular demand, I am posting the code I used.

    var iter:Number = 200000; var time:Number = 0; var obj:Object = {}; var arr:Array = [];  time = getTimer(); for (var i:Number = 0; i < iter; ++i) {   arr[i] = i; } trace('Task build-arr: ' + (getTimer() - time) + 'ms');  time = getTimer(); for (var i:Number = 0; i < iter; ++i) {   arr[i] = arr[i]; } trace('Task iter-arr: ' + (getTimer() - time) + 'ms');  time = getTimer(); for (var i:Number = 0; i < iter; ++i) {   obj[String(i)] = i; } trace('Task build-obj: ' + (getTimer() - time) + 'ms');  time = getTimer(); for (var i:String in obj) {   obj[i] = obj[i]; } trace('Task iter-obj: ' + (getTimer() - time) + 'ms'); 
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