>PHP: Retorno por Referencia

add a note User Contributed Notes
Retorno por Referencia

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)

add a note

MANUAL DE PHP

windsurf mercedes camper

Retorno por Referencia