Thanks for your reply, however, I've already seen that solution. What
I'm looking for is a more explicit solution.

I have an array and want to copy it into another array.

THx

--- In [email protected], "bill_sahlas" <[EMAIL PROTECTED]> wrote:
>
> 
> From the docs
> 
> 
> Cloning arrays
> The Array class has no built-in method for making copies of arrays. You
> can create a shallow copy of an array by calling either the concat() or
> slice() methods with no arguments. In a shallow copy, if the original
> array has elements that are objects, only the references to the objects
> are copied rather than the objects themselves. The copy points to the
> same objects as the original does. Any changes made to the objects are
> reflected in both arrays.
> 
> In a deep copy, any objects found in the original array are also copied
> so that the new array does not point to the same objects as does the
> original array. Deep copying requires more than one line of code, which
> usually calls for the creation of a function. Such a function could be
> created as a global utility function or as a method of an Array
> subclass.
> 
> The following example defines a function named clone() that does deep
> copying. The algorithm is borrowed from a common Java programming
> technique. The function creates a deep copy by serializing the array
> into an instance of the ByteArray class, and then reading the array back
> into a new array. This function accepts an object so that it can be used
> with both indexed arrays and associative arrays, as shown in the
> following code:
> import flash.utils.ByteArray;  function clone(source:Object):*{     var
> myBA:ByteArray = new ByteArray();     myBA.writeObject(source);    
> myBA.position = 0;     return(myBA.readObject()); }
> --- In [email protected], "Daniel" <danboh@> wrote:
> >
> > Good Day all!
> >
> > I need to clone or copy the content of an Array, not to keep reference
> > to the array neither the to objects inside, a complete new array but
> > with the same source as the other.
> >
> > Any ideas?
> >
> > Thanks!.
> >
>






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