References Explained

Another example of something to watch out for when using references with arrays.  It seems that even an usused reference to an array cell modifies the *source* of the reference.  Strange behavior for an assignment statement (is this why I've seen it written as an =& operator?  - although this doesn't happen with regular variables).
<?php
    $array1 = array(1,2);
    $x = &$array1[1];   // Unused reference
    $array2 = $array1;  // reference now also applies to $array2 !
    $array2[1]=22;      // (changing [0] will not affect $array1)
    print_r($array1);
?>
Produces:
    Array
    (
    [0] => 1
    [1] => 22    // var_dump() will show the & here
    )

I fixed my bug by rewriting the code without references, but it can also be fixed with the unset() function:
<?php
    $array1 = array(1,2);
    $x = &$array1[1];
    $array2 = $array1;
    unset($x); // Array copy is now unaffected by above reference
    $array2[1]=22;
    print_r($array1);
?>
Produces:
    Array
    (
    [0] => 1
    [1] => 2
    )

in the example below, you would get the same result if you change the function to something like:

function test_ref(&$arr) {
   $time = time();
   $size = sizeof($arr);       // <--- this makes difference...
   for($n=0; $n<$size; $n++) {
       $x = 1;
   }

   echo "<br />The function using a reference took ".(time() - $time)." s";
}

If you want to know the reference count of a particular variable, then here's a function that makes use of debug_zval_dump() to do so:

<?php
function refcount($var)
{
    ob_start();
    debug_zval_dump($var);
    $dump = ob_get_clean();

    $matches = array();
    preg_match('/refcount\(([0-9]+)/', $dump, $matches);

    $count = $matches[1];

    //3 references are added, including when calling debug_zval_dump()
    return $count - 3;
}
?>

debug_zval_dump() is a confusing function, as explained in its documentation, as among other things, it adds a reference count when being called as there is a reference within the function. refcount() takes account of these extra references by subtracting them for the return value.

It's also even more confusing when dealing with variables that have been assigned by reference (=&), either on the right or left side of the assignment, so for that reason, the above function doesn't really work for those sorts of variables. I'd use it more on object instances.

However, even taking into account that passing a variable to a function adds one to the reference count; which should mean that calling refcount() adds one, and then calling debug_zval_dump() adds another, refcount() seems to have aquired another reference from somewhere; hence subtracting 3 instead of 2 in the return line. Not quite sure where that comes from.

I've only tested this on 5.3; due to the nature of debug_zval_dump(), the results may be completely different on other versions.

A little gotcha (be careful with references!):

<?php
$arr = array('a'=>'first', 'b'=>'second', 'c'=>'third');
foreach ($arr as &$a); // do nothing. maybe?
foreach ($arr as $a);  // do nothing. maybe?
print_r($arr);
?>
Output:

Array
(
    [a] => first
    [b] => second
    [c] => second
)

Add 'unset($a)' between the foreachs to obtain the 'correct' output:

Array
(
    [a] => first
    [b] => second
    [c] => third
)

It seems like PHP has problems with references, like that it can't work properly with circular references or free properly structure with more references. See http://bugs.php.net/?id=30053.

I have big problem with this and I hope someone from PHP add proper warning with explanation IN manual, if they can't fix it.

I found a very useful summary of how references work in PHP4 (and some of the common pitfalls) in this article: http://www.obdev.at/developers/articles/00002.html

It deals with some subtle situations and I recommend it to anyone having difficulty with their references.

Simply put, here's an example of what referencing IS:

<?php
    $foo  = 5;
    $bar = &$foo;
    $bar++;
    
    echo $foo;
?>

The above example will output the value 6, because $bar references the value of $foo, therefore, when changing $bar's value, you also change $foo's value too.

You should have in mind that php4 keep assigned variables "automagically" referenced until they are overwritten. So the variable copy is not executed on assignment, but on modification. Say you have this:

$var1 = 5;
$var2 = $var1; // In this point these two variables share the same memory location
$var1 = 3; // Here $var1 and $var2 have they own memory locations with values 3 and 5 respectively

Don't use references in function parameters to speed up aplications, because this is automatically done. I think that this should be in the manual, because it can lead to confusion.

More about this here:
http://www.zend.com/zend/art/ref-count.php

I ran into a bit of a problem recently, with an array copy resulting in a reference copy of one of the elements instead of a clone. Sample code:

<?php
$a=array(1 => "A");
$b=&