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Сброс переменных-ссылок Передача по ссылке

Last updated: Sat, 27 Jan 2007

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# Возвращение по ссылке

Возвращение по ссылке используется в тех случаях, когда вы хотите использовать функцию для выбора переменной, с которой должна быть связана данная ссылка. При возвращении по ссылке используйте такой синтаксис:

`<?php
function &find_var($param)
{
/* ... код ... */
return $found_var;
}
$foo =& find_var($bar);
$foo->x = 2;
?>`

В этом примере устанавливается свойство объекта, возвращённого функцией `find_var`, а не его копии, как было бы без использования ссылок.

> Замечание: В отличие от передачи параметров по ссылке, `&` здесь нужно использовать в обоих местах - для указания на то, что вы возвращаете ссылку, а не копию, как обычно, и для указания того, что происходит связывание по ссылке, а не обычное присвоение.

add a note User Contributed Notes
Возвращение по ссылке
cesoid at yahoo dot com
23-Feb-2007 12:27
`The note by zayfod below is very misleading. The following is not true:
"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."
As proof, you can change zayfod's second example slightly to show that $var does not, as he claims, become a reference to $some_var.
<?php
class some_class
{
function & func_b ()
{
$this->some_var = 2;
return $this->some_var;
}
function func_a (& $param)
{
# $param is 1 here
$param = $this->func_b();
# $param is 2 here
}
}
$var = 1;
$object = new some_class();
$object->func_a($var);
print $var; // this prints two
$var = 5;
print $object->some_var;
// $object->some_var is still 2, because $var is not bound to it
// (as the original post by zayfod claims) and therefore has no
// effect on it
?>
In actuality it is impossible to bind a variable passed by reference to a variable returned by reference, doing so only removes the link between the internal parameter variable and binds it to something new. This is why zayfod's first example shows $var having a value of 1 at the end. What is actually happening in zayfod's first example is the following:
1) $var is passed by reference into func_a, which binds $param to $var
2) func_b is evaluated, returning a reference to $some_var
3) $param is bounded to $some_var therefore losing it's binding to $var
4) func_a returns, with $var unchanged
In the second example step 1 and 2 are the same, but then this happens:
3) the value 2 is copied into $param, which still references $var, and therefore $var now also evaluates to 2
4) func_a returns, $var is 2, but is not bound to $some_var (and wouldn't be even in my example where some_var still exists afterward)
The mistake seems to be based on misuse of references. When you return a reference, it means that you want a variable that is tied to another, so that when you change one, the other one changes. In zayfod's example, returning by reference is pointless, because $some_var ceases to exist after the function is completed, it won't stay tied to anything.
There are two ways to change those functions so that in the end you have $var actually bound to $this->some_var:
1) Pass $param into func_b by reference and reference $this->some_var to func_b's parameter like so:
func_b (&$b_param)
{
$this->some_var =& $b_param; // be careful not to flip this
$this->some_var = 2;
}
2) change func_a to return the result of func_b by reference, and reference that to $var
And remember this simple rule: If a variable is already referenced to another variable and you reference it to a third (put it on the left side of =&) it loses its original reference. This is what happens to $param in zayfod's example. (Also don't forget that passing by reference actually references the internal parameter variable to the the passed variable.)`
php at thunder-2000 dot com
02-Feb-2007 02:31
`If you want to get a part of an array to manipulate, you can use this function
function &getArrayField(&$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.`
neozenkai at yahoo dot com
15-Jul-2006 01:38
`While trying to create a function to return Database connection objects, it took me a while to get this right:
<?php
class TestClass
{
var $thisVar = 0;
function TestClass($value)
{
$this->thisVar = $value;
}
function &getTestClass($value)
{
static $classes;
if (!isset($classes[$value]))
{
$classes[$value] = new TestClass($value);
}
return $classes[$value];
}
}
echo "<pre>";
echo "Getting class1 with a value of 432\n";
$class1 =& TestClass::getTestClass(432);
echo "Value is: " . $class1->thisVar . "\n";
echo "Getting class2 with a value of 342\n";
$class2 =& TestClass::getTestClass(342);
echo "Value is: " . $class2->thisVar . "\n";
echo "Getting class3 with the same value of 432\n";
$class3 =& TestClass::getTestClass(432);
echo "Value is: " . $class3->thisVar . "\n";
echo "Changing the value of class1 to 3425, which should also change class3\n";
$class1->thisVar = 3425;
echo "Now checking value of class3: " . $class3->thisVar . "\n";
?>
Which outputs:
Getting class1 with a value of 432
Value is: 432
Getting class2 with a value of 342
Value is: 342
Getting class3 with the same value of 432
Value is: 432
Changing the value of class1 to 3425, which should also change class3
Now checking value of class3: 3425
Note that PHP syntax is different from C/C++ in that you must use the & operator in BOTH places, as stated by the manual. It took me a while to figure this out.`
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);
}
?>`
warhog at warhog dot net
13-Dec-2005 12:04
`firstly a note on the post below: technically correct that -> "to get a reference to an exsisting class and it's properties" should be "...to an existing object..."
in PHP5 it's senseless to return objects by reference.. let's say you have, as in the post below, a class which should return a reference to its own instance (maybe it's been created using the singleton pattern..), than it's no problem to simply return that variable holding the instance.
In PHP5 variables holding objects are always references to a specific object, so when you return (=copy) a variable "having" an object you in fact return a reference to that object.
Because of that behaviour, which is very common for many programming languages, especially those, which are to be compiled (due to memory problems and so on), you have to use the clone-function which was introduced in PHP5.
Here an example to underline what i mean:
<?php
class sample_singleton
{
protected static $instance = NULL;
private function __construct()
{ }
public static function create()
{ self::$instance = new sample_singleton(); }
public static function getInstance()
{ return self::$instance; }
public function yea()
{ echo "wuow"; }
}
sample_singleton::create();
// $singleton will be a reference to sample_singleton::$instance
$singleton = sample_singleton::getInstance();
$singleton->yea();
?>
Note some more (maybe) strange behaviour: although $instance is private you can have a reference pointing on it. Maybe that does not seem strange to you in any way, but maybe you wondered as i did : )
The code posted here was just a sample to illustrate my posting, for using the singleton pattern please look it up in the manual. I just used this cause it was the simplest example which went in my mind right know.`
willem at designhulp dot nl
09-Oct-2005 04:54
`There is an important difference between php5 and php4 with references.
Lets say you have a class with a method called 'get_instance' to get a reference to an exsisting class and it's properties.
<?php
class mysql {
function get_instance(){
// check if object exsists
if(empty($_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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Сброс переменных-ссылок Передача по ссылке

Last updated: Sat, 27 Jan 2007
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