Передача по ссылке

    Переменную по ссылке разрешается передавать в функцию, чтобы функция изменила переменную. Синтаксис выглядит вот так:

    <?php

    function foo(&$var)
    {
    $var++;
    }

    $a = 5;
    foo($a);
    // В этом месте переменная $a указывает на значение 6

    ?>

    Замечание: В вызове функции отсутствует знак ссылки — знак указывают только в определении функции. Определения функции достаточно, чтобы правильно передать аргумент по ссылке.

    По ссылке разрешается передавать:

    • Переменные: foo($a)

    • Ссылки, которые возвращаются функциями:

      <?php

      function foo(&$var)
      {
      $var++;
      }

      function &      bar()
      {
      $a = 5;
      return       $a;
      }

      foo(bar());

      ?>
      Подробнее об этом рассказывает раздел о возврате по ссылке.

    Другие выражения не передают по ссылке, поскольку результат будет неопределённым. Следующие примеры передачи по ссылке неправильные:

    <?php

    function foo(&$var)
    {
    return     $var++;
    }

    // Обратите внимание, перед названием функции отсутствует знак &
    function bar()
    {
    $a = 5;
    return     $a;
    }

    foo(bar()); // Выдаёт предупреждение

    foo($a = 5); // Выражение, а не переменная
    foo(5); // Константа, а не переменная

    class Foobar {}

    foo(new Foobar()); // Вызывает уведомление с PHP 7.0.7
    // Notice: Only variables should be passed by reference
    ?>

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    Примечания пользователей 5 notes

    up
    448
    tnestved at yahoo dot com
    10 years ago
    By removing the ability to include the reference sign on function calls where pass-by-reference is incurred (I.e., function definition uses &), the readability of the code suffers, as one has to look at the function definition to know if the variable being passed is by-ref or not (I.e., potential to be modified). If both function calls and function definitions require the reference sign (I.e., &), readability is improved, and it also lessens the potential of an inadvertent error in the code itself. Going full on fatal error in 5.4.0 now forces everyone to have less readable code. That is, does a function merely use the variable, or potentially modify it...now we have to find the function definition and physically look at it to know, whereas before we would know the intent immediately.
    up
    29
    ccb_bc at hotmail dot com
    5 years ago
    <?php
    // PHP >= 5.6

    // Here we use the 'use' operator to create a variable within the scope of the function. Although it may seem that the newly created variable has something to do with '$x' that is outside the function, we are actually creating a '$x' variable within the function that has nothing to do with the '$x' variable outside the function. We are talking about the same names but different content locations in memory.
    $x = 10;
    (function() use (    $x){
    $x = $x*$x;
    var_dump($x); // 100
    })();
    var_dump($x); // 10

    // Now the magic happens with using the reference (&). Now we are actually accessing the contents of the '$y' variable that is outside the scope of the function. All the actions that we perform with the variable '$y' within the function will be reflected outside the scope of this same function. Remembering this would be an impure function in the functional paradigm, since we are changing the value of a variable by reference.
    $y = 10;
    (function() use (&    $y){
    $y = $y*$y;
    var_dump($y); // 100
    })();
    var_dump($y); // 100
    ?>
    up
    37
    mike at eastghost dot com
    9 years ago
    beware unset() destroys references

    $x = 'x';
    change( $x );
    echo $x; // outputs "x" not "q23" ---- remove the unset() and output is "q23" not "x"

    function change( & $x )
    {
    unset( $x );
    $x = 'q23';
    return true;
    }
    up
    2
    Anonymous
    1 year ago
    Parameters passed by references can have default values.
    You can find out if a variable was actually passed by using func_num_args():

    <?php

    function refault( & $ref = 'Do I have to be calculated?'){
    echo     'NUM ARGS: '. func_num_args()."\n";
    echo     "ORI VALUE: {$ref}\n";
    if(     func_num_args() > 0 ) $ref = 'Yes, expensive to calculate result: ' . sleep(1);
    else     $ref = 'No.';
    echo     "NEW VALUE: {$ref}\n";
    }

    $result = 'Do I have to be calculated?';
    refault( $result );
    echo     "RESULT: {$result}\n";
    // NUM ARGS: 1
    // ORI VALUE: Do I have to be calculated?
    // NEW VALUE: Yes, expensive to calculate result: 0
    // RESULT: Yes, expensive to calculate result: 0

    refault();
    // NUM ARGS: 0
    // ORI VALUE: Do I have to be calculated?
    // NEW VALUE: No.
    ?>
    up
    2
    Jason Steelman
    5 years ago
    Within a class, passing array elements by reference which don't exist are added to the array as null. Compared to a normal function, this changes the behavior of the function from throwing an error to creating a new (null) entry in the referenced array with a new key.

    <?php

    class foo {
    public     $arr = ['a' => 'apple', 'b' => 'banana'];
    public function     normalFunction($key) {
    return     $this->arr[$key];
    }
    public function &    referenceReturningFunction($key) {
    return     $this->arr[$key];
    }
    }

    $bar = new foo();
    $var = $bar->normalFunction('beer'); //Notice Error. Undefined index beer
    $var = &$bar->referenceReturningFunction('beer'); // No error. The value of $bar is now null
    var_dump($bar->arr);
    /**
    [
    "a" => "apple",
    "b" => "banana",
    "beer" => null,
    ],
    */

    ?>
    This is in no way a "bug" - the framework is performing as designed, but it took careful thought to figure out what was going on. PHP7.3
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