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
?>
참조는 무엇을 하는가
PHP 참조는 두 변수가 같은 내용을 참조할수 있게 한다. 다음과 같은 일을 한다:
<?php
$a =& $b;
?>
Note:
여기에서 $a와 $b는 완전히 동일합니다. $a는 $b를 가리키고, 그 반대일 수도 있습니다. $a와 $b는 같은 위치를 가리킵니다.
Note:
참조를 가진 배열이 복사되면, 그 값들은 참조인 상태로 남습니다. 이는 배열이 함수에 값으로 넘겨졌을 때에도 유효합니다.
Note:
정의되지 않은 변수를 참조를 통해 할당하거나, 넘기거나, 반환하면 변수가 생성됩니다.
Example #1 정의되지 않은 변수에 참조 사용하기
<?php
function foo(&$var) { }
foo($a); // $a가 "생성되고" null로 할당됩니다
$b = array();
foo($b['b']);
var_dump(array_key_exists('b', $b)); // bool(true)
$c = new StdClass;
foo($c->d);
var_dump(property_exists($c, 'd')); // boold(true)
?>
같은 문법을 참조를 반환하는 함수나, new 연산자(PHP 4.0.4부터)에도 사용합니다.
<?php
$bar =& new fooclass();
$foo =& find_var($bar);
?>
Note:
& 연산자를 사용하지않으면 생성된 객체의 복사본을 돌려받는다. 그 클래스에서 $this를 사용한다면 그것은 클래스의 현재 인스턴스에만 영향을 미친다. &가 없는 지정은 인스턴스(즉, 객체)를 복사할것이고 $this는 복사본에만 영향을 미칠것이다. 이와 같은 것은 퍼포먼스와 메모리의 효율적인 사용을 위해 오직 하나의 인스턴스만 취급하기를 원할 경우에 사용된다.
@new처럼 생성자에서 발생하는 모든 에러 메시지를 보이지 않게 하기 위해서 @연산자를 사용할수 있지만, &new구문을 사용할때는 이 연산자는 작동하지 않는다. 이것은 젠드 엔진의 한계로 이런 표현을 쓰면 해석 오류가 발생한다.
Warning
global $var;는 $var =&
$GLOBALS['var'];의 단축형임을 생각하십시오. 그러므로 다른
참조를 $var에 할당하는 것은 지역 변수의 참조를 변경할
뿐입니다.
함수 안에서 global로 선언한 변수에 참조를 할당하면, 그 참조는 그 함수 안에서만 보여집니다. 이를 피하려면 $GLOBALS 배열을 사용하십시오.
Example #2 함수 안에서 전역 변수 참조하기
<?php
$var1 = "Example variable";
$var2 = "";
function global_references($use_globals)
{
global $var1, $var2;
if ($use_globals) {
$var2 =& $var1; // 함수 안에서만 보여집니다
} else {
$GLOBALS["var2"] =& $var1; // 전역 문맥에서도 보여집니다
}
}
global_references(false);
echo "var2 is set to '$var2'\n"; // var2 is set to ''
global_references(true);
echo "var2 is set to '$var2'\n"; // var2 is set to 'Example variable'
?>
Note:
foreach 구문 안에서 참조 변수에 값을 할당하면, 참조도 변경됩니다.
Example #3 참조와 foreach 구문
<?php
$ref = 0;
$row =& $ref;
foreach (array(1, 2, 3) as $row) {
// do something
}
echo $ref; // 3 - 반복한 배열의 마지막 원소
?>
참조가 하는 두번째 일은 참조에 의해 변수를 전달하는 것이다. 이는 함수에서 지역 변수를 만들거나 같은 내용을 참조하는 호출을 사용하는 변수를 생성함으로써 가능해진다. 예를 들면:
<?php
function foo(&$var)
{
$var++;
}
$a=5;
foo($a);
?>
참조가 하는 세번째 일은 참조에 의해 돌려받기이다.
add a note
User Contributed Notes 23 notes
ladoo at gmx dot at ¶
19 years ago
Hlavac ¶
16 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!
elrah [] polyptych [dot] com ¶
13 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";
?>
amp at gmx dot info ¶
16 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)
nay at woodcraftsrus dot com ¶
12 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( $baz );
?>
Anonymous ¶
8 years ago
to reply to ' elrah [] polyptych [dot] com ', one thing to keep in mind is that array (or similar large data holders) are by default passed by reference. So the behaviour is not side effect. And for array copy and passing array inside function always done by 'pass by reference'...
charles at org oo dot com ¶
16 years ago
points to post below me.
When you're doing the references with loops, you need to unset($var).
for example
<?php
foreach($var as &$value)
{
...
}
unset($value);
?>
Amaroq ¶
14 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.
?>
Amaroq ¶
16 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"
?>
php.devel at homelinkcs dot com ¶
19 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.
dovbysh at gmail dot com ¶
16 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
?>
Oddant ¶
10 years 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 $point_to_test; // echo 'test' ! (still pointed somewhere in the memory)
?>
Drewseph ¶
15 years ago
If you set a variable before passing it to a function that takes a variable as a reference, it is much harder (if not impossible) to edit the variable within the function.
Example:
<?php
function foo(&$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.
dexant9t at gmail dot com ¶
2 years ago
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, 22
akinaslan at gmail dot com ¶
13 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 $reference->c; //Echoes 2.
?>
dnhuff at acm dot org ¶
15 years ago
In reply to Drewseph using foo($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.
firespade at gmail dot com ¶
17 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
Amaroq ¶
14 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.
php at hood dot id dot au ¶
17 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 $a1['a']; // will echo b (!)
?>
After reading the manual this looks like it is meant to happen. But it confused me for a few days!
(The solution I used was to turn the second foreach into a reference too)
strata_ranger at hotmail dot com ¶
14 years ago
An interesting if offbeat use for references: Creating an array with an arbitrary number of dimensions.
For example, a function that takes the result set from a database and produces a multidimensional array keyed according to one (or more) columns, which might be useful if you want your result set to be accessible in a hierarchial manner, or even if you just want your results keyed by the values of each row's primary/unique key fields.
<?php
function array_key_by($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')));
?>
joachim at lous dot org ¶
21 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.
butshuti at smartrwanda dot org ¶
11 years 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)!
admin at torntech dot com ¶
10 years 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 ,
