For...each...in loop. Can I make assumptions about the order?

Question

I've always assumed that I can't make any assumptions about the order in which elements are indexed by a for...each...in loop.  Not even for a conventional indexed array.  (Emphasis!: I'm not talking about associative arrays here, I'm talking about normal indexed arrays.  Dense, no gaps.).
for each (var element:ElementClass in myArray) {}
May not traverse the array from first to last.  I've never assumed that it's guaranteed anyway.  Neither does the ECMA standard.  I assume the compiler/runtime is at liberty to index the array in any order, which may even be different each time. 
Whereas... for (var i:int=0;i<myArray.length;i++) {......var element:ElementClass=myArray[i]....}
....Is guaranteed to traverse the array from first to last.  So I always use this looping statement if the ordering is important.
MY QUESTION IS.... In ActionScript 3, Can I just assume that the for..each..in will always traverse a dense indexed array from first to last?
Actually, my specific case at the moment is where another programmer is passing me an array,assembled within a for...each..in loop.   But I need to assume the order of that array.  If this is always going to work in ActionScript3, then no problem.  Although I'll continue to worry about the ECMA standard for my own code, and wont make assumptions about ordering - but I won't force my anal-retentive programming style on the other guy.

Answer 1

I did a test on it
case1:
var arr:Array = new Array();
arr[0] = 'a';
arr[1] = 'b';
arr[2] = 'c';
arr[3] = 'd';
for each(var arrEle:string in arr)
     trace(arrEle);
always output a,b,c,d
case2:
var arr:Array = new Array();
arr[6] = 'd';
arr[0] = 'a';
arr[1] = 'b';
arr[3] = 'c';
for each(var arrEle:string in arr)
     trace(arrEle);
always output a,b,d,c
so I assume array is created in this way: alloc a memory block for array first, when attaching new element(numeric indexd by i) to array, check if i >= length, put new element in the end of array, else put element at i; so in case 2, array is alloced in this way
array:{'d'} //6>0
array:{'a', 'd'} //0<1
array:{'a', 'b', 'd'} //1<2
array:{'a', 'b', 'd', 'c'} //3>=3
disclaimer: totally assumption!!!