Remember that there is a difference between a variable and an object. A variable is just a named reference to an object, changing the variable just points it to another object (that is what = does, it changes where a variable points).

You can have two variables referencing the same object. Lets say we have “foo” and “bar” both pointing to the same array:

var foo = [1, 2, 3];
var bar = foo;

Now, if you change the array using the foo variable, that change can be seen by inspecting the bar variable too:

foo.pop();

trace(foo.length); // prints 2
trace(bar.length); // prints 2

However, if we do a similar excercise with ints it it doesn’t work the same way:

var foo = 5;
var bar = foo;

foo += 3;

trace(foo); // prints 8
trace(bar); // prints 5

This is because there is no way to change the object that was referenced by both foo and bar, if you want something else (like foo + 3) you will get a completely new object.

There are methods on Array that returns new objects, for example concat and slice, so not all operations are equal:

var foo = [1, 2, 3];
var bar = foo;

foo = foo.concat([4, 5]);

trace(foo.length); // prints 5
trace(bar.length); // print 3

What happened in this last example is that the foo variable got overwritten by a new array (the return value of the concat call), but bar still pointed at the old array.

I really appreciate that you took the time to post that, so many thanks for the explanation!

If you pass an array to a method and that method changes the array (by calling pop, for example), you can determine if the array was passed by reference or by value depending on whether or not the array was changed in both scopes, or just in the called method’s scope.

If you instead pass a string to a method it doesn’t matter whether that string is passed by value or by reference, because there is nothing you can do to prove that it was passed one way or the other. You cannot change the instance, so every copy of the instance will be equal.