In PHP 5, when you pass an array by value, but the array contains references, then, you will get some subtly weird behavior.

$args = array();
$args['x'] =& $_REQUEST['x'];

echo "<p>$args[x]</p>";

f($args);

echo "<p>$args[x]</p>";

$x = $_REQUEST['x'];

echo "<p>$x</p>";

function f($a)
{
	$a['x'] = 100;
}

Calling it like this: test.php?x=5 Displays: 5 100 100

$args = array();
$args['x'] = $_REQUEST['x'];

echo "<p>$args[x]</p>";

f($args);

echo "<p>$args[x]</p>";

$x = $_REQUEST['x'];

echo "<p>$x</p>";

function f($a)
{
	$a['x'] = 100;
}

That displays: 5 5 5 This second example is the "expected" behavior. The array is passed to f() by value, so we expect change made within f() to remain local to f(). However, the array $args is an array of references. That is, it's an array, but it's element 'x' refers to $_REQUEST['x']. So, when $args is copied as an argument to f(), it's copies the reference. Then, within f(), $a['x'] refers to $_REQUEST['x']. So changes to $a['x'] affect $_REQUEST['x']. This can cause problems if you are creating objects based on $args, like this:

$args =& $_REQUEST['x'];
$obj1 = new Obj($args);
$obj2 = new Obj($args);

class Obj {
function Obj($args) {
  $this->args = $args;
}
}

In this situation, changes to $obj1->args['x'] will affect $obj2->args['x'], because they refer to the same variable!