a little addition to the example of pixel at minikomp dot com here below
<?php
function &func(){
static $static = 0;
$static++;
return $static;
}
$var1 =& func();
echo "var1:", $var1; // 1
func();
func();
echo "var1:", $var1; // 3
$var2 = func(); // assignment without the &
echo "var2:", $var2; // 4
func();
func();
echo "var1:", $var1; // 6
echo "var2:", $var2; // still 4
?>
引用返回
引用返回用在当想用函数找到引用应该被绑定在哪一个变量上面时。 不要用返回引用来增加性能，引擎足够聪明来自己进行优化。 仅在有合理的技术原因时才返回引用！ 使用此语法返回引用：
<?php
class foo {
public $value = 42;
public function &getValue() {
return $this->value;
}
}
$obj = new foo;
$myValue = &$obj->getValue(); // $myValue 是对 $obj->value 的引用，即 42。
$obj->value = 2;
echo $myValue; // 打印 $obj->value 的新值，即 2。
?>注意: 和参数传递不同，这里必须在两个地方都用
&符号——指出返回的是一个引用，而不是通常的一个拷贝，同样也指出 $myValue 是作为引用的绑定，而不是通常的赋值。
注意: 如果试图这样从函数返回引用：
return ($this->value);，这将不会起作用， 因为在试图返回一个表达式的结果而不是一个引用的变量。 只能从函数返回引用变量——没别的方法。
要使用返回的引用，必须使用引用赋值：
<?php
function &collector() {
static $collection = array();
return $collection;
}
$collection = &collector();
$collection[] = 'foo';
?><?php
function &collector() {
static $collection = array();
return $collection;
}
array_push(collector(), 'foo');
?>注意: 注意
array_push(&collector(), 'foo');将 不 起作用，它会导致 Fatal 错误。
add a note
User Contributed Notes 19 notes
Spad-XIII ¶
16 years ago
stanlemon at mac dot com ¶
17 years ago
I haven't seen anyone note method chaining in PHP5. When an object is returned by a method in PHP5 it is returned by default as a reference, and the new Zend Engine 2 allows you to chain method calls from those returned objects. For example consider this code:
<?php
class Foo {
protected $bar;
public function __construct() {
$this->bar = new Bar();
print "Foo\n";
}
public function getBar() {
return $this->bar;
}
}
class Bar {
public function __construct() {
print "Bar\n";
}
public function helloWorld() {
print "Hello World\n";
}
}
function test() {
return new Foo();
}
test()->getBar()->helloWorld();
?>
Notice how we called test() which was not on an object, but returned an instance of Foo, followed by a method on Foo, getBar() which returned an instance of Bar and finally called one of its methods helloWorld(). Those familiar with other interpretive languages (Java to name one) will recognize this functionality. For whatever reason this change doesn't seem to be documented very well, so hopefully someone will find this helpful.
szymoncofalik at gmail dot com ¶
13 years ago
Sometimes, you would like to return NULL with a function returning reference, to indicate the end of chain of elements. However this generates E_NOTICE. Here is little tip, how to prevent that:
<?php
class Foo {
const $nullGuard = NULL;
// ... some declarations and definitions
public function &next() {
// ...
if (!$end) return $bar;
else return $this->nullGuard;
}
}
?>
by doing this you can do smth like this without notices:
<?php
$f = new Foo();
// ...
while (($item = $f->next()) != NULL) {
// ...
}
?>
you may also use global variable:
global $nullGuard;
return $nullGuard;
obscvresovl at NOSPAM dot hotmail dot com ¶
19 years ago
An example of returning references:
<?
$var = 1;
$num = NULL;
function &blah()
{
$var =& $GLOBALS["var"]; # the same as global $var;
$var++;
return $var;
}
$num = &blah();
echo $num; # 2
blah();
echo $num; # 3
?>
Note: if you take the & off from the function, the second echo will be 2, because without & the var $num contains its returning value and not its returning reference.
sandaimespaceman at gmail dot com ¶
16 years ago
The &b() function returns a reference of $a in the global scope.
<?php
$a = 0;
function &b()
{
global $a;
return $a;
}
$c = &b();
$c++;
echo
"
\$a: $a
\$b: $c
"
?>
It outputs:
$a: 1 $b: 1
rwruck ¶
18 years ago
The note about using parentheses when returning references is only true if the variable you try to return does not already contain a reference.
<?php
// Will return a reference
function& getref1()
{
$ref =& $GLOBALS['somevar'];
return ($ref);
}
// Will return a value (and emit a notice)
function& getref2()
{
$ref = 42;
return ($ref);
}
// Will return a reference
function& getref3()
{
static $ref = 42;
return ($ref);
}
?>
civilization28 at gmail dot com ¶
10 years ago
Zayfod's example above is useful, but I feel that it needs more explanation. The point that should be made is that a parameter passed in by reference can be changed to reference something else, resulting in later changes to the local variable not affecting the passed in variable:
<?php
function & func_b ()
{
$some_var = 2;
return $some_var;
}
function func_a (& $param)
{
# $param is 1 here
$param = & func_b(); # Here the reference is changed and
# the "&" in "func_a (& $param)"
# is no longer in effect at all.
# $param is 2 here
$param++; # Has no effect on $var.
}
$var = 1;
func_a($var);
# $var is still 1 here!!! Because the reference was changed.
?>
hawcue at yahoo dot com ¶
20 years ago
Be careful when using tinary operation condition?value1:value2
See the following code:
$a=1;
function &foo()
{
global $a;
return isset($a)?$a:null;
}
$b=&foo();
echo $b; // shows 1
$b=2;
echo $a; // shows 1 (not 2! because $b got a copy of $a)
To let $b be a reference to $a, use "if..then.." in the function.
anisgazig at gmail dot com ¶
2 years ago
<?php
$a = 9;
function &myF(){
global $a;
return $a;
}
//before modified the value
$func =& myF();
echo "$a and $func";
echo "\n";
//after modified the value
$func++;
echo "$a and $func";
fabian dot picone at gmail dot com ¶
6 years ago
This note seems not to apply with PHP 7:
"Note: If you try to return a reference from a function with the syntax: return ($this->value); this will not work as you are attempting to return the result of an expression, and not a variable, by reference. You can only return variables by reference from a function - nothing else. Since PHP 5.1.0, an E_NOTICE error is issued if the code tries to return a dynamic expression or a result of the new operator."
Bug following code works without error output. Same result as i would not have braces around $this-value.
<?php
class foo {
public $value = 42;
public function &getValue() {
return ($this->value);
}
}
$obj = new foo;
$myValue = &$obj->getValue();
$obj->value = 2;
echo $myValue;
benjamin dot delespierre at gmail dot com ¶
13 years ago
Keep in mind that returning by reference doesn't work with __callStatic:
<?php
class Test {
private static $_inst;
public static function & __callStatic ($name, $args) {
if (!isset(static::$_inst)){
echo "create";
static::$_inst = (object)"test";
}
return static::$_inst;
}
var_dump($a = &Test::abc()); // prints 'create'
$a = null;
var_dump(Test::abc()); // doesn't prints and the instance still exists in Test::$_inst
?>
spidgorny at gmail dot com ¶
14 years ago
When returning reference to the object member which is instantiated inside the function, the object is destructed upon returning (which is a problem). It's easier to see the code:
<?php
class MemcacheArray {
public $data;
...
/**
* Super-clever one line cache reading AND WRITING!
* Usage $data = &MemcacheArray::getData(__METHOD__);
* Hopefully PHP will know that $this->data is still used
* and will call destructor after data changes.
* Ooops, it's not the case.
*
* @return unknown
*/
function &getData($file, $expire = 3600) {
$o = new MemcacheArray($file, $expire);
return $o->data;
}
?>
Here, destructor is called upon return() and the reference becomes a normal variable.
My solution is to store objects in a pool until the final exit(), but I don't like it. Any other ideas?
<?php
protected static $instances = array();
function &getData($file, $expire = 3600) {
$o = new MemcacheArray($file, $expire);
self::$instances[$file] = $o; // keep object from destructing too early
return $o->data;
}
?>
zayfod at yahoo dot com ¶
21 years ago
There is a small exception to the note on this page of the documentation. You do not have to use & to indicate that reference binding should be done when you assign to a value passed by reference the result of a function which returns by reference.
Consider the following two exaples:
<?php
function & func_b ()
{
$some_var = 2;
return $some_var;
}
function func_a (& $param)
{
# $param is 1 here
$param = & func_b();
# $param is 2 here
}
$var = 1;
func_a($var);
# $var is still 1 here!!!
?>
The second example works as intended:
<?php
function & func_b ()
{
$some_var = 2;
return $some_var;
}
function func_a (& $param)
{
# $param is 1 here
$param = func_b();
# $param is 2 here
}
$var = 1;
func_a($var);
# $var is 2 here as intended
?>
(Experienced with PHP 4.3.0)
php at thunder-2000 dot com ¶
17 years ago
If you want to get a part of an array to manipulate, you can use this function
function &getArrayField(&$array,$path) {
if (!empty($path)) {
if (empty($array[$path[0]])) return NULL;
else return getArrayField($array[$path[0]], array_slice($path, 1));
} else {
return $array;
}
}
Use it like this:
$partArray =& getArrayField($GLOBALS,array("config","modul1"));
You can manipulate $partArray and the changes are also made with $GLOBALS.
Anonymous ¶
10 years ago
I learned a painful lesson working with a class method that would pass by reference.
In short, if you have a method in a class that is initialed with ampersand during declaration, do not use another ampersand when using the method as in &$this->method();
For example
<?php
class A {
public function &hello(){
static $a='';
return $a;
}
public function bello(){
$b=&$this->hello(); // incorrect. Do not use ampersand.
$b=$this->hello(); // $b is a reference to the static variable.
}
?>
contact at infopol dot fr ¶
20 years ago
A note about returning references embedded in non-reference arrays :
<?
$foo;
function bar () {
global $foo;
$return = array();
$return[] =& $foo;
return $return;
}
$foo = 1;
$foobar = bar();
$foobar[0] = 2;
echo $foo;
?>
results in "2" because the reference is copied (pretty neat).
willem at designhulp dot nl ¶
19 years ago
There is an important difference between php5 and php4 with references.
Lets say you have a class with a method called 'get_instance' to get a reference to an exsisting class and it's properties.
<?php
class mysql {
function get_instance(){
// check if object exsists
if(empty($_ENV['instances']['mysql'])){
// no object yet, create an object
$_ENV['instances']['mysql'] = new mysql;
}
// return reference to object
$ref = &$_ENV['instances']['mysql'];
return $ref;
}
}
?>
Now to get the exsisting object you can use
mysql::get_instance();
Though this works in php4 and in php5, but in php4 all data will be lost as if it is a new object while in php5 all properties in the object remain.
jpenna ¶
4 years ago
You can set the value of the variable returned by reference, be it a `static` function variable or a `private` property of an object (which is quite dangerous o.o).
Static function variable:
<?php
function &func(){
static $static = 0;
return $static;
}
$var1 =& func();
echo "var1:", $var1, "\n"; // 0
func();
$var1 = 90;
echo "var1:", $var1, "\n"; // 90
echo "static:", func(), "\n"; // 90
?>
Private property
<?php
class foo {
private $value = 1;
public function &getValue() {
return $this->value;
}
public function setValue($val) {
$this->value = $val;
}
}
$obj = new foo;
$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 1.
echo $obj->getValue(); // 1
echo $myValue; // 1
$obj->setValue(5);
echo $obj->getValue(); // 5
echo $myValue; // 5
$myValue = 1000;
echo $obj->getValue(); // 1000
echo $myValue; // 1000
?>
pixel at minikomp dot com ¶
16 years ago
<?php
function &func(){
static $static = 0;
$static++;
return $static;
}
$var =& func();
echo $var; // 1
func();
func();
func();
func();
echo $var; // 5
?>
