<?php
$arr = array(1);
$a =& $arr[0];
$arr2 = $arr;
$arr2[0]++;
var_dump($arr);
This portion of code outputs 2.
Why?
We only touched first element of arr2 which isn't assigned by reference to $arr. Doesn't alias to the same array, so why is that?
When you assign one array to the next, a copy is made. But because the element at $arr[0] is a reference rather than a value, a copy of the reference is made, so that at the end, $arr[0] and $arr2[0] refer to the same thing.
This is more about references than arrays. Referenced values are not copied. This applies to objects as well. Consider:
$ageRef = 7;
$mike = new stdClass();
$mike->age = &$ageRef; // create a reference
$mike->fruit = 'apple';
$john = clone $mike; // clone, so $mike and $john are distinct objects
$john->age = 17; // the reference will survive the cloning! This will change $mike
$john->fruit = 'orange'; // only $john is affected, since it's a distinct object
echo $mike->age . " | " . $mike->fruit; // 17 | apple
See the first user note on this documentation page and also this one.
$arr[0] should also turn $arr[0] into a reference. I'm surprised that I haven't accidentally run across this by now.