PHPのお勉強！

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    リファレンスとは?

    PHP において、リファレンスとは同じ変数の内容を異なった名前で コールすることを意味します。これは C のポインタとは異なります。 リファレンスを使ってポインタの演算をすることはできませんし、 リファレンスは実メモリのアドレスでもありません。詳細は リファレンスが行わないこと を参照ください。 そうではなく、リファレンスは シンボルテーブル のエイリアスです。 PHP では、変数名と変数の内容は異なっており、 このため、同じ内容は異なった複数の名前を有する事が可能であることに 注意してください。最も良く似ているのは、Unix のファイル名とファイルの 関係です。この場合、変数名はディレクトリエントリ、変数の内容は ファイル自体に対応します。リファレンスは、Unix ファイルシステムの ハードリンクのようなものであると考えられます。

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    User Contributed Notes 3 notes

    up
    235
    273118949 at qq dot com
    7 years ago
    it just likes a person who has two different names.
    up
    48
    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
    16
    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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