That's not technically correct with regards to PHP. Your statement that any changes to $arr1 or $arr2 only impact the one in question, however, is accurate. If no changes are made they still refer to the same data in memory. It's copy-on-write semantics.

$arr1 = [1,2,3]; // $arr1 is a pointer to a zval array [1,2,3] and refcount:1

$arr2 = $arr1; // $arr1 and $arr2 are pointers to the same zval array but incremented refcount to 2.

$arr2[] = 4; // at this point, $arr2 creates a new zval array, copies over the data from the first, sets the recount to 1, and appends int(4). The [1,2,3] array has its refcount decremented to 1 as it's still referenced by $arr1.

[1] http://hengrui-li.blogspot.com/2011/08/php-copy-on-write-how...