For the second part of your question, see the array page of the manual, which states (quoting) :
Array assignment always involves value copying. Use the reference operator to copy an array by reference.
And the given example :
<?php
$arr1 = array(2, 3);
$arr2 = $arr1;
$arr2[] = 4; // $arr2 is changed,
// $arr1 is still array(2, 3)
$arr3 = &$arr1;
$arr3[] = 4; // now $arr1 and $arr3 are the same
?>
For the first part, the best way to be sure is to try ;-)
Consider this example of code :
function my_func($a) {
$a[] = 30;
}
$arr = array(10, 20);
my_func($arr);
var_dump($arr);
It'll give this output :
array
0 => int 10
1 => int 20
Which indicates the function has not modified the "outside" array that was passed as a parameter : it's passed as a copy, and not a reference.
If you want it passed by reference, you'll have to modify the function, this way :
function my_func(& $a) {
$a[] = 30;
}
And the output will become :
array
0 => int 10
1 => int 20
2 => int 30
As, this time, the array has been passed "by reference".
Don't hesitate to read the References Explained section of the manual : it should answer some of your questions ;-)