Question

1) When an array is passed as an argument to a method or function, is it passed by reference, or by value?

2) When assigning an array to a variable, is the new variable a reference to the original array, or is it new copy?
What about doing this:

$a = array(1,2,3);
$b = $a;

Is $b a reference to $a?

Answer 1

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 😉