PHP Conference Nagoya 2025

Gönderim Nedir?

PHP'de gönderimler aynı değişken içeriğine farklı isimlerle erişmek demektir. C'deki göstericilere benzemezler; örneğin, üzerlerinde gösterici aritmetiği kullanamazsınız, çünkü gerçekte bellek adresleri değillerdir. Ayrıntılar için Gönderimler Ne Değildir? bölümüne bakınız. Daha çok simge tablosu takma adlarıdır. PHP'de şuna dikkat edin: Değişken içeriği ve değişken ismi farklı şeylerdir. Bu bakımdan, aynı içeriğin farklı isimleri olabilir. En yakın benzerlik, Unix dosya isimleri ve dosyaları ile kurulabilir; değişken isimleri dizinlerse, değişken içeriği de dizinin içindeki dosyadır. Gönderimler ise Unix dosya sistemindeki sabit bağlar olarak düşünülebilir.

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User Contributed Notes 3 notes

up
234
273118949 at qq dot com
7 years ago
it just likes a person who has two different names.
up
46
Anonymous
6 years ago
Unlike in C, PHP references are not treated as pre-dereferenced pointers, but as complete aliases.

The data that they are aliasing ("referencing") will not become available for garbage collection until all references to it have been removed.

"Regular" variables are themselves considered references, and are not treated differently from variables assigned using =& for the purposes of garbage collection.

The following examples are provided for clarification.

1) When treated as a variable containing a value, references behave as expected. However, they are in fact objects that *reference* the original data.

<?php
var = "foo";
$ref1 =& $var; // new object that references $var
$ref2 =& $ref1; // references $var directly, not $ref1!!!!!

echo $ref; // >foo

unset($ref);

echo
$ref1; // >Notice: Undefined variable: ref1
echo $ref2; // >foo
echo $var; // >foo
?>

2) When accessed via reference, the original data will not be removed until *all* references to it have been removed. This includes both references and "regular" variables assigned without the & operator, and there are no distinctions made between the two for the purpose of garbage collection.

<?php
$var
= "foo";
$ref =& $var;

unset(
$var);

echo
$var; // >Notice: Undefined variable: var
echo $ref; // >foo
?>

3) To remove the original data without removing all references to it, simply set it to null.

<?php
$var
= "foo";
$ref =& $var;

$ref = NULL;

echo
$var; // Value is NULL, so nothing prints
echo $ref; // Value is NULL, so nothing prints
?>

4) Placing data in an array also counts as adding one more reference to it, for the purposes of garbage collection.

For more info, see http://php.net/manual/en/features.gc.refcounting-basics.php
up
17
aldo dot caruso at argencasas dot com
5 years ago
The following three code snippets show the effect of using references in scalar variables, arrays and objects under different circumstances.

In any case the result is the expected one if you stick to the concept that a reference is an alias to a variable. After assigning by reference ( no matter if $a =& $b or $b =& $a ) both variable names refer to the same variable.

References with scalars

<?php
/*
References are aliases for the same variable
*/

$a = 1;
$b =& $a;

$b = 2;
echo
"$a,$b\n"; // 2,2

$a = 3;
echo
"$a,$b\n"; // 3,3

// Variables can be bound before being assigned
$c =& $d;
$c = 4;
echo
"$c,$d\n"; // 4,4
?>

References with arrays

<?php
/*
Array elements referencing scalar variables
*/

$a = 1;
$b = 2;
$c = array(&$a, &$b);

$a = 3;
$b = 4;
echo
"c: $c[0],$c[1]\n"; // 3,4

$c[0] = 5;
$c[1] = 6;
echo
"a,b: $a,$b\n"; // 5,6

/*
Reference between arrays
*/

$d = array(1,2);
$e =& $d;

$d[0] = 3;
$d[1] = 4;
echo
"e: $e[0],$e[1]\n"; // 3,4

$e[0] = 5;
$e[1] = 6;
echo
"d: $d[0],$d[1]\n"; // 5,6

$e = 7;
echo
"d: $d\n"; // 7 ( $d is no more an array, but an integer )

/*
Iterating an array of references using foreach construct
*/

$a = 1;
$b = 2;
$f = array(&$a,&$b);

foreach(
$f as $x) // If $x is assigned by value it doesn't change referred variables.
$x = 3;
echo
"a,b: $a,$b\n"; // 1,2

foreach($f as &$x) // If $x is assigned by reference it changes referred variables.
$x = 3;
echo
"a,b: $a,$b\n"; // 3,3

// Be aware that, after the loop, $x still references $f[1] and so $b
$x = 4;
echo
"a,b: $a,$b\n"; // 3,4 ( $b affected )

// To avoid previous side effects it is advisable to unset x, unlinking it from $f[1] and $b
unset($x);
$x = 5;
echo
"a,b: $a,$b\n"; // 3,4 ( $b not affected )
?>

References with objects

<?php
/*
Object property referencing a scalar variable
*/

$a = 1;
$b = new stdClass();
$b->x =& $a;

$a = 2;
echo
"b->x: $b->x\n"; // 2

$b->x = 3;
echo
"a: $a\n"; // 3

/* Reference between objects */
$c = new stdClass();
$c->x = 1;
$d =& $c;

$d->x = 2;
echo
"c->x: $c->x\n"; // 2

$d = new stdClass();
$d->y = 3;
echo
"c->y: $c->y\n"; // 3
echo "c->x: $c->x\n"; // Undefined property: stdClass::$x
?>
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