Há três operações básicas ao se utilizar referências: atribuição por referência, passagem por referência, e retorno por referência. Esta seção fará uma introdução dessas operações, com atalhos para leituras posteriores.
Referências no PHP permitem criar duas variáveis que se referem ao mesmo conteúdo. Ou seja, quando se faz:
<?php
$a =& $b;
?>Nota:
$a e $b são completamente iguais aqui, mas não porque $a está apontando para $b ou vice-versa, mas sim que $a e $b apontam para o mesmo lugar.
Nota:
Se uma variável indefinida for atribuída, passada ou retornada por referência, ela será criada.
Exemplo #1 Usando referência com variáveis indefinidas
<?php
function foo(&$var) {}
foo($a); // $a é "criada" e definida como 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)
?>
A mesma sintaxe pode ser utilizada com funções que retornem referências:
<?php
$foo =& find_var($bar);
?>Usar essa mesma sintaxe com uma função que não retorna por referência irá gerar um erro, da mesma forma no caso do operador new. Ainda que objetos sejam internamente passados como ponteiros, isso não é o mesmo que uma referência, como explicado em Objetos e Referências.
Se for atribuída uma referência para uma variável declarada global
dentro da função, a referência será visível somente dentro da função.
Pode-se evitar isso usando o array $GLOBALS.
Exemplo #2 Referenciando variáveis globais de dentro de funções
<?php
$var1 = "Variável de exemplo";
$var2 = "";
function global_references($use_globals)
{
global $var1, $var2;
if (!$use_globals) {
$var2 =& $var1; // visível somente dentro da função
} else {
$GLOBALS["var2"] =& $var1; // visível também no contexto global
}
}
global_references(false);
echo "var2 é definida como '$var2'\n"; // var2 é definida como ''
global_references(true);
echo "var2 é definida como '$var2'\n"; // var2 é definida como 'Variável de exemplo'
?>global $var; como atalho para $var
=& $GLOBALS['var'];. Assim, atribuir outra referência
para $var somente modifica a variável de referência local.
Nota:
Se um valor for atribuído por referência a uma variável numa instrução foreach, a referência também é modificada.
Exemplo #3 Referências e o comando foreach
<?php
$ref = 0;
$row =& $ref;
foreach (array(1, 2, 3) as $row) {
// Faz alguma coisa
}
echo $ref; // 3 - último elemento do array iterado
?>
Ainda que não seja uma atribuição por referência explícita, expressões criadas
com o construtor
array() também podem
se comportar como tais com o prefixo & no elemento de array
a ser acrescentado. Exemplo:
<?php
$a = 1;
$b = array(2, 3);
$arr = array(&$a, &$b[0], &$b[1]);
$arr[0]++;
$arr[1]++;
$arr[2]++;
/* $a == 2, $b == array(3, 4); */
?>Note que referências dentro de arrays são potencialmente perigosas. Fazer uma atribuição normal (sem referência) com uma referência à direita não transforma a expressão a esquerda numa referência, mas referências dentro de arrays são preservadas nessas atribuições normais. Isso também se aplica a chamadas de função onde arrays são passados por valor. Exemplo:
<?php
/* Atribuição de variáveis escalares */
$a = 1;
$b =& $a;
$c = $b;
$c = 7; // $c não é referência; não modifica $a ou $b
/* Atribuição de variaveis do array */
$arr = array(1);
$a =& $arr[0]; // $a e $arr[0] estão no mesmo conjunto de referências
$arr2 = $arr; // Não atribui por referência!
$arr2[0]++;
/* $a == 2, $arr == array(2) */
/* O conteúdo de $arr é modificado embora ele não seja uma referência! */
?>A segunda coisa que referências fazem é passar variáveis por referência. Isso é feito com a criação de uma variável local em uma função e uma variável no escopo chamador que referenciam o mesmo conteúdo. Assim:
<?php
function foo(&$var)
{
$var++;
}
$a=5;
foo($a);
?>A terceira coisa que as referências fazem é retorno por referência.
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
?>It matters if you are playing with a reference or with a value
Here we are working with values so working on a reference updates original variable too;
$a = 1;
$c = 22;
$b = & $a;
echo "$a, $b"; //Output: 1, 1
$b++;
echo "$a, $b";//Output: 2, 2 both values are updated
$b = 10;
echo "$a, $b";//Output: 10, 10 both values are updated
$b =$c; //This assigns value 2 to $b which also updates $a
echo "$a, $b";//Output: 22, 22
But, if instead of $b=$c you do
$b = &$c; //Only value of $b is updated, $a still points to 10, $b serves now reference to variable $c
echo "$a, $b"//Output: 10, 22Watch out for this:
foreach ($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!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";
?>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)to reply to ' elrah [] polyptych [dot] com ', one thing to keep in mind is that array (or similar large data holders) are by default passed by reference. So the behaviour is not side effect. And for array copy and passing array inside function always done by 'pass by reference'...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( $baz );
?>points to post below me.
When you're doing the references with loops, you need to unset($var).
for example
<?php
foreach($var as &$value)
{
...
}
unset($value);
?>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)
?>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.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
?>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.
?>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.
?>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 $b2; //Echoes "Four"
?>If you set a variable before passing it to a function that takes a variable as a reference, it is much harder (if not impossible) to edit the variable within the function.
Example:
<?php
function foo(&$bar) {
$bar = "hello\n";
}
foo($unset);
echo($unset);
foo($set = "set\n");
echo($set);
?>
Output:
hello
set
It baffles me, but there you have it.In reply to Drewseph using foo($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.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 = 2When 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.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 $a1['a']; // will echo b (!)
?>
After reading the manual this looks like it is meant to happen. But it confused me for a few days!
(The solution I used was to turn the second foreach into a reference too)An interesting if offbeat use for references: Creating an array with an arbitrary number of dimensions.
For example, a function that takes the result set from a database and produces a multidimensional array keyed according to one (or more) columns, which might be useful if you want your result set to be accessible in a hierarchial manner, or even if you just want your results keyed by the values of each row's primary/unique key fields.
<?php
function array_key_by($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')));
?>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.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)!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 ,