php教程

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！要返回引用，使用此语法：

<?php
class foo {
    public $value = 42;

    public function &getValue() {
        return $this->value;
    }
}

$obj = new foo;
$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 42.
$obj->value = 2;
echo $myValue;                // prints the new value of $obj->value, i.e. 2.
?>
本例中 getValue 函数所返回的对象的属性将被赋值，而不是拷贝，就和没有用引用语法一样。

Note: 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。

Note: 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。如果代码试图返回一个动态表达式或 new 运算符的结果，自 PHP 4.4.0 和 PHP 5.1.0 起会发出一条 E_NOTICE 错误。

User Contributed Notes

fabian dot picone at gmail dot com 14-Aug-2018 09:00
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;
civilization28 at gmail dot com 17-Sep-2014 01:49
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                                           # is no longer in effect at all.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.

?>
Anonymous 05-Mar-2014 04:18
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.
}
?>
benjamin dot delespierre at gmail dot com 23-Feb-2011 05:36
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
?>
szymoncofalik at gmail dot com 28-Jan-2011 12:41
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;
spidgorny at gmail dot com 19-Mar-2010 06:18
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;
    }
?>
sandaimespaceman at gmail dot com 02-Nov-2008 08:52
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
Spad-XIII 31-Jan-2008 05:01
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

?>
pixel at minikomp dot com 30-Dec-2007 04:21
<?php

    function &func(){
        static $static = 0;
        $static++;
        return $static;
    }

    $var =& func();
    echo $var; // 1
    func();
    func();
    func();
    func();
    echo $var; // 5

?>
stanlemon at mac dot com 28-Sep-2007 10:51
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.
php at thunder-2000 dot com 02-Feb-2007 02:31
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.
rwruck 27-Feb-2006 09:22
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);
  }
?>
willem at designhulp dot nl 09-Oct-2005 04:54
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.
obscvresovl at NOSPAM dot hotmail dot com 24-Dec-2004 01:09
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.
hawcue at yahoo dot com 16-Mar-2004 07:58
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.
contact at infopol dot fr 12-Feb-2004 09:36
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).
zayfod at yahoo dot com 03-Dec-2003 09:23
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)

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