What References Do

There are three basic operations performed using references: assigning by reference, passing by reference, and returning by reference. This section will give an introduction to these operations, with links for further reading.

Assign By Reference

In the first of these, PHP references allow you to make two variables refer to the same content. Meaning, when you do:

<?php
$a 
=& $b;
?>
it means that $a and $b point to the same content.

Note:

$a and $b are completely equal here. $a is not pointing to $b or vice versa. $a and $b are pointing to the same place.

Note:

If you assign, pass, or return an undefined variable by reference, it will get created.

Example #1 Using references with undefined variables

<?php
function foo(&$var) { }

foo($a); // $a is "created" and assigned to null

$b = array();
foo($b['b']);
var_dump(array_key_exists('b'$b)); // bool(true)

$c = new StdClass;
foo($c->d);
var_dump(property_exists($c'd')); // bool(true)
?>

The same syntax can be used with functions that return references, and with the new operator (since PHP 4.0.4 and before PHP 5.0.0):

<?php
$foo 
=& find_var($bar);
?>
Since PHP 5, new returns a reference automatically, so using =& in this context is deprecated and produces an E_DEPRECATED message in PHP 5.3 and later, and an E_STRICT message in earlier versions. (Technically, the difference is that, in PHP 5, object variables, much like resources, are a mere pointer to the actual object data, so these object references are not "references" in the same sense used before (aliases). For more information, see Objects and references.)

Warning

If you assign a reference to a variable declared global inside a function, the reference will be visible only inside the function. You can avoid this by using the $GLOBALS array.

Example #2 Referencing global variables inside functions

<?php
$var1 
"Example variable";
$var2 "";

function 
global_references($use_globals)
{
    global 
$var1$var2;
    if (!
$use_globals) {
        
$var2 =& $var1// visible only inside the function
    
} else {
        
$GLOBALS["var2"] =& $var1// visible also in global context
    
}
}

global_references(false);
echo 
"var2 is set to '$var2'\n"// var2 is set to ''
global_references(true);
echo 
"var2 is set to '$var2'\n"// var2 is set to 'Example variable'
?>
Think about global $var; as a shortcut to $var =& $GLOBALS['var'];. Thus assigning another reference to $var only changes the local variable's reference.

Note:

If you assign a value to a variable with references in a foreach statement, the references are modified too.

Example #3 References and foreach statement

<?php
$ref 
0;
$row =& $ref;
foreach (array(
123) as $row) {
    
// do something
}
echo 
$ref// 3 - last element of the iterated array
?>

While not being strictly an assignment by reference, expressions created with the language construct array() can also behave as such by prefixing & to the array element to add. Example:

<?php
$a 
1;
$b = array(23);
$arr = array(&$a, &$b[0], &$b[1]);
$arr[0]++; $arr[1]++; $arr[2]++;
/* $a == 2, $b == array(3, 4); */
?>

Note, however, that references inside arrays are potentially dangerous. Doing a normal (not by reference) assignment with a reference on the right side does not turn the left side into a reference, but references inside arrays are preserved in these normal assignments. This also applies to function calls where the array is passed by value. Example:

<?php
/* Assignment of scalar variables */
$a 1;
$b =& $a;
$c $b;
$c 7//$c is not a reference; no change to $a or $b

/* Assignment of array variables */
$arr = array(1);
$a =& $arr[0]; //$a and $arr[0] are in the same reference set
$arr2 $arr//not an assignment-by-reference!
$arr2[0]++;
/* $a == 2, $arr == array(2) */
/* The contents of $arr are changed even though it's not a reference! */
?>
In other words, the reference behavior of arrays is defined in an element-by-element basis; the reference behavior of individual elements is dissociated from the reference status of the array container.

Pass By Reference

The second thing references do is to pass variables by reference. This is done by making a local variable in a function and a variable in the calling scope referencing the same content. Example:

<?php
function foo(&$var)
{
    
$var++;
}

$a=5;
foo($a);
?>
will make $a to be 6. This happens because in the function foo the variable $var refers to the same content as $a. For more information on this, read the passing by reference section.

Return By Reference

The third thing references can do is return by reference.

add a note add a note

User Contributed Notes 21 notes

up
9
elrah [] polyptych [dot] com
3 years ago
It appears that references can have side-effects.  Below are two examples.  Both are simply copying one array to another.  In the second example, a reference is made to a value in the first array before the copy.  In the first example the value at index 0 points to two separate memory locations. In the second example, the value at index 0 points to the same memory location.

I won't say this is a bug, because I don't know what the designed behavior of PHP is, but I don't think ANY developers would expect this behavior, so look out.

An example of where this could cause problems is if you do an array copy in a script and expect on type of behavior, but then later add a reference to a value in the array earlier in the script, and then find that the array copy behavior has unexpectedly changed.

<?php
// Example one
$arr1 = array(1);
echo
"\nbefore:\n";
echo
"\$arr1[0] == {$arr1[0]}\n";
$arr2 = $arr1;
$arr2[0]++;
echo
"\nafter:\n";
echo
"\$arr1[0] == {$arr1[0]}\n";
echo
"\$arr2[0] == {$arr2[0]}\n";

// Example two
$arr3 = array(1);
$a =& $arr3[0];
echo
"\nbefore:\n";
echo
"\$a == $a\n";
echo
"\$arr3[0] == {$arr3[0]}\n";
$arr4 = $arr3;
$arr4[0]++;
echo
"\nafter:\n";
echo
"\$a == $a\n";
echo
"\$arr3[0] == {$arr3[0]}\n";
echo
"\$arr4[0] == {$arr4[0]}\n";
?>
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3
charles at org oo dot com
7 years ago
points to post below me.
When you're doing the references with loops, you need to unset($var).

for example
<?php
foreach($var as &$value)
{
...
}
unset(
$value);
?>
up
4
ladoo at gmx dot at
9 years ago
I ran into something when using an expanded version of the example of pbaltz at NO_SPAM dot cs dot NO_SPAM dot wisc dot edu below.
This could be somewhat confusing although it is perfectly clear if you have read the manual carfully. It makes the fact that references always point to the content of a variable perfectly clear (at least to me).

<?php
$a
= 1;
$c = 2;
$b =& $a; // $b points to 1
$a =& $c; // $a points now to 2, but $b still to 1;
echo $a, " ", $b;
// Output: 2 1
?>
up
3
Hlavac
7 years ago
Watch out for this:

foreach ($somearray as &$i) {
  // update some $i...
}
...
foreach ($somearray as $i) {
  // last element of $somearray is mysteriously overwritten!
}

Problem is $i contians reference to last element of $somearray after the first foreach, and the second foreach happily assigns to it!
up
2
Drewseph
6 years ago
If you set a variable before passing it to a function that takes a variable as a reference, it is much harder (if not impossible) to edit the variable within the function.

Example:
<?php
function foo(&$bar) {
   
$bar = "hello\n";
}

foo($unset);
echo(
$unset);
foo($set = "set\n");
echo(
$set);

?>

Output:
hello
set

It baffles me, but there you have it.
up
1
php at hood dot id dot au
7 years ago
I discovered something today using references in a foreach

<?php
$a1
= array('a'=>'a');
$a2 = array('a'=>'b');

foreach (
$a1 as $k=>&$v)
$v = 'x';

echo
$a1['a']; // will echo x

foreach ($a2 as $k=>$v)
{}

echo
$a1['a']; // will echo b (!)
?>

After reading the manual this looks like it is meant to happen. But it confused me for a few days!

(The solution I used was to turn the second foreach into a reference too)
up
1
Amaroq
4 years ago
When using references in a class, you can reference $this-> variables.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b = 2;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>

However, this doesn't appear to be completely trustworthy. In some cases, it can act strangely.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 3.
echo $reference->c; //Echoes 2.
?>

In this second code block, I've changed reftest() so that $b increments instead of just gets changed to 2. Somehow, it winds up equaling 3 instead of 2 as it should.
up
1
Amaroq
4 years ago
I think a correction to my last post is in order.

When there is a constructor, the strange behavior mentioned in my last post doesn't occur. My guess is that php was treating reftest() as a constructor (maybe because it was the first function?) and running it upon instantiation.

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
__construct()
    {
        return
0;
    }

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>
up
1
amp at gmx dot info
7 years ago
Something that might not be obvious on the first look:
If you want to cycle through an array with references, you must not use a simple value assigning foreach control structure. You have to use an extended key-value assigning foreach or a for control structure.

A simple value assigning foreach control structure produces a copy of an object or value. The following code

$v1=0;
$arrV=array(&$v1,&$v1);
foreach ($arrV as $v)
{
  $v1++;
  echo $v."\n";
}

yields

0
1

which means $v in foreach is not a reference to $v1 but a copy of the object the actual element in the array was referencing to.

The codes

$v1=0;
$arrV=array(&$v1,&$v1);
foreach ($arrV as $k=>$v)
{
    $v1++;
    echo $arrV[$k]."\n";
}

and

$v1=0;
$arrV=array(&$v1,&$v1);
$c=count($arrV);
for ($i=0; $i<$c;$i++)
{
    $v1++;
    echo $arrV[$i]."\n";
}

both yield

1
2

and therefor cycle through the original objects (both $v1), which is, in terms of our aim, what we have been looking for.

(tested with php 4.1.3)
up
0
php.devel at homelinkcs dot com
9 years ago
In reply to lars at riisgaardribe dot dk,

When a variable is copied, a reference is used internally until the copy is modified.  Therefore you shouldn't use references at all in your situation as it doesn't save any memory usage and increases the chance of logic bugs, as you discoved.
up
-2
admin at torntech dot com
1 year ago
Something that has not been discussed so far is reference of a reference.
I needed a quick and dirty method of aliasing incorrect naming until a proper rewrite could be done.
Hope this saves someone else the time of testing since it was not covered in the Does/Are/Are Not pages.
Far from best practice, but it worked.

<?php
$a
= 0;

$b =& $a;
$a =& $b;

$a = 5;
echo
$a . ', ' . $b;
//ouputs: 5,5

echo ' | ';

$b = 6;
echo
$a . ',' . $b;
//outputs: 6,6

echo ' | ';
unset(
$a );
echo
$a . ', ' . $b;

//outputs: , 6

class Product {

    public
$id;
    private
$productid;

    public function
__construct( $id = null ) {
       
$this->id =& $this->productid;
       
$this->productid =& $this->id;
       
$this->id = $id;
    }

    public function
getProductId() {
        return
$this->productid;
    }

}

echo
' | ';

$Product = new Product( 1 );
echo
$Product->id . ', ' . $Product->getProductId();
//outputs 1, 1
$Product->id = 2;
echo
' | ';
echo
$Product->id . ', ' . $Product->getProductId();
//outputs 2, 2
$Product->id = null;
echo
' | ';
echo
$Product->id . ', ' . $Product->getProductId();
//outouts ,
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-1
Oddant
1 year ago
About the example on array references.
I think this should be written in the array chapter as well.
Indeed if you are new to programming language in some way, you should beware that arrays are pointers to a vector of Byte(s).

<?php $arr = array(1); ?>
$arr here contains a reference to which the array is located.
Writing :
<?php echo $arr[0]; ?>
dereferences the array to access its very first element.

Now something that you should also be aware of  (even you are not new to programming languages) is that PHP use references to contains the different values of an array. And that makes sense because the type of the elements of a PHP array can be different.

Consider the following example :

<?php

$arr
= array(1, 'test');

$point_to_test =& $arr[1];

$new_ref = 'new';

$arr[1] =& $new_ref;

echo
$arr[1]; // echo 'new';
echo $point_to_test; // echo 'test' ! (still pointed somewhere in the memory)

?>
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-2
butshuti at smartrwanda dot org
1 year ago
This appears to be the hidden behavior: When a class function has the same name as the class, it seems to be implicitly called when an object of the class is created.
For instance, you may take a look at the naming of the function "reftest()": it is in the class "reftest". The behavior can be tested as follows:

<?php
class reftest
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest1()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
   
    public function
reftest()
    {
       echo
"REFTEST() called here!\n";
    }
}

$reference = new reftest();
/*You must notice the above will also implicitly call reference->reftest()*/

$reference->reftest1();
$reference->reftest2();

echo
$reference->a."\n"; //Echoes 2, not 3 as previously noticed.
echo $reference->c."\n"; //Echoes 2.
?>

The above outputs:

REFTEST() called here!
2
2

Notice that reftest() appears to be called (though no explicit call to it was made)!
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-2
nay at woodcraftsrus dot com
3 years ago
in PHP you don't really need pointer anymore if you want to share an  object across your program

<?php
class foo{
        protected
$name;
        function
__construct($str){
               
$this->name = $str;
        }
        function
__toString(){
                return 
'my name is "'. $this->name .'" and I live in "' . __CLASS__ . '".' . "\n";
        }
        function
setName($str){
               
$this->name = $str;
        }
}

class
MasterOne{
        protected
$foo;
        function
__construct($f){
               
$this->foo = $f;
        }
        function
__toString(){
                return
'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n";
        }
        function
setFooName($str){
               
$this->foo->setName( $str );
        }
}

class
MasterTwo{
        protected
$foo;
        function
__construct($f){
               
$this->foo = $f;
        }
        function
__toString(){
                return
'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n";
        }
        function
setFooName($str){
               
$this->foo->setName( $str );
        }
}

$bar = new foo('bar');

print(
"\n");
print(
"Only Created \$bar and printing \$bar\n");
print(
$bar );

print(
"\n");
print(
"Now \$baz is referenced to \$bar and printing \$bar and \$baz\n");
$baz =& $bar;
print(
$bar );

print(
"\n");
print(
"Now Creating MasterOne and Two and passing \$bar to both constructors\n");
$m1 = new MasterOne( $bar );
$m2 = new MasterTwo( $bar );
print(
$m1 );
print(
$m2 );

print(
"\n");
print(
"Now changing value of \$bar and printing \$bar and \$baz\n");
$bar->setName('baz');
print(
$bar );
print(
$baz );

print(
"\n");
print(
"Now printing again MasterOne and Two\n");
print(
$m1 );
print(
$m2 );

print(
"\n");
print(
"Now changing MasterTwo's foo name and printing again MasterOne and Two\n");
$m2->setFooName( 'MasterTwo\'s Foo' );
print(
$m1 );
print(
$m2 );

print(
"Also printing \$bar and \$baz\n");
print(
$bar );
print(
$baz );
?>
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-2
akinaslan at gmail dot com
3 years ago
In this example class name is different from its first function and however there is no construction function. In the end as you guess "a" and "c" are equal. So if there is no construction function at same time class and its first function names are the same, "a" and "c" doesn't equal forever. In my opinion php doesn't seek any function for the construction as long as their names differ from each others.

<?php
class reftest_new
{
    public
$a = 1;
    public
$c = 1;

    public function
reftest()
    {
       
$b =& $this->a;
       
$b++;
    }

    public function
reftest2()
    {
       
$d =& $this->c;
       
$d++;
    }
}

$reference = new reftest_new();

$reference->reftest();
$reference->reftest2();

echo
$reference->a; //Echoes 2.
echo $reference->c; //Echoes 2.
?>
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-2
Amaroq
6 years ago
The order in which you reference your variables matters.

<?php
$a1
= "One";
$a2 = "Two";
$b1 = "Three";
$b2 = "Four";

$b1 =& $a1;
$a2 =& $b2;

echo
$a1; //Echoes "One"
echo $b1; //Echoes "One"

echo $a2; //Echoes "Four"
echo $b2; //Echoes "Four"
?>
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-2
joachim at lous dot org
11 years ago
So to make a by-reference setter function, you need to specify reference semantics _both_ in the parameter list _and_ the assignment, like this:

class foo{
   var $bar;
   function setBar(&$newBar){
      $this->bar =& newBar;
   }
}

Forget any of the two '&'s, and $foo->bar will end up being a copy after the call to setBar.
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-2
dovbysh at gmail dot com
7 years ago
Solution to post "php at hood dot id dot au 04-Mar-2007 10:56":

<?php
$a1
= array('a'=>'a');
$a2 = array('a'=>'b');

foreach (
$a1 as $k=>&$v)
$v = 'x';

echo
$a1['a']; // will echo x

unset($GLOBALS['v']);

foreach (
$a2 as $k=>$v)
{}

echo
$a1['a']; // will echo x

?>
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-3
dnhuff at acm dot org
6 years ago
In reply to Drewseph using foo($a = 'set'); where $a is a reference formal parameter.

$a = 'set' is an expression. Expressions cannot be passed by reference, don't you just hate that, I do. If you turn on error reporting for E_NOTICE, you will be told about it.

Resolution: $a = 'set'; foo($a); this does what you want.
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-3
firespade at gmail dot com
7 years ago
Here's a good little example of referencing. It was the best way for me to understand, hopefully it can help others.

$b = 2;
$a =& $b;
$c = $a;
echo $c;

// Then... $c = 2
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-5
strata_ranger at hotmail dot com
5 years ago
An interesting if offbeat use for references:  Creating an array with an arbitrary number of dimensions.

For example, a function that takes the result set from a database and produces a multidimensional array keyed according to one (or more) columns, which might be useful if you want your result set to be accessible in a hierarchial manner, or even if you just want your results keyed by the values of each row's primary/unique key fields.

<?php
function array_key_by($data, $keys, $dupl = false)
/*
* $data  - Multidimensional array to be keyed
* $keys  - List containing the index/key(s) to use.
* $dupl  - How to handle rows containing the same values.  TRUE stores it as an Array, FALSE overwrites the previous row.
*         
* Returns a multidimensional array indexed by $keys, or NULL if error.
* The number of dimensions is equal to the number of $keys provided (+1 if $dupl=TRUE).
*/  
{
   
// Sanity check
   
if (!is_array($data)) return null;
   
   
// Allow passing single key as a scalar
   
if (is_string($keys) or is_integer($keys)) $keys = Array($keys);
    elseif (!
is_array($keys)) return null;

   
// Our output array
   
$out = Array();
   
   
// Loop through each row of our input $data
   
foreach($data as $cx => $row) if (is_array($row))
    {
     
     
// Loop through our $keys
     
foreach($keys as $key)
      {
       
$value = $row[$key];

        if (!isset(
$last)) // First $key only
       
{
          if (!isset(
$out[$value])) $out[$value] = Array();
         
$last =& $out; // Bind $last to $out
       
}
        else
// Second and subsequent $key....
       
{
          if (!isset(
$last[$value])) $last[$value] = Array();
        }

       
// Bind $last to one dimension 'deeper'.
        // First lap: was &$out, now &$out[...]
        // Second lap: was &$out[...], now &$out[...][...]
        // Third lap:  was &$out[...][...], now &$out[...][...][...]
        // (etc.)
       
$last =& $last[$value];
      }
     
      if (isset(
$last))
      {
       
// At this point, copy the $row into our output array
       
if ($dupl) $last[$cx] = $row; // Keep previous
       
else       $last = $row; // Overwrite previous
     
}
      unset(
$last); // Break the reference
   
}
    else return
NULL;
   
   
// Done
   
return $out;
}

// A sample result set to test the function with
$data = Array(Array('name' => 'row 1', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_a'),
              Array(
'name' => 'row 2', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_b'),
              Array(
'name' => 'row 3', 'foo' => 'foo_a', 'bar' => 'bar_b', 'baz' => 'baz_c'),
              Array(
'name' => 'row 4', 'foo' => 'foo_b', 'bar' => 'bar_c', 'baz' => 'baz_d')
              );

// First, let's key it by one column (result: two-dimensional array)
print_r(array_key_by($data, 'baz'));

// Or, key it by two columns (result: 3-dimensional array)
print_r(array_key_by($data, Array('baz', 'bar')));

// We could also key it by three columns (result: 4-dimensional array)
print_r(array_key_by($data, Array('baz', 'bar', 'foo')));

?>
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