PHP: リファレンス渡し

リファレンス渡し

リファレンスにより関数に変数を渡すことが可能です。この場合、関数内でその引数を修正可能になります。構文は次のようになります。

<?php

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

$a=5;

foo($a);
// $a はここでは 6 です

?>

注意: 関数コールの際には、リファレンス記号がないことに注意してください。関数定義にのみリファレンス記号があります。リファレンスで正しく引数を渡すには、関数定義のみで十分です。

次のものはリファレンスで渡すことが可能です。

変数、すなわち、foo($a)
関数から返されるリファレンスは、次のようになります。

<?php function foo(&$var) { $var++; } function &bar() { $a = 5; return $a; } foo(bar()); ?>

リファレンスによる戻り値に関する説明も参照ください。

他の式は、結果が未定義となるため、リファレンスで渡すべきではありません。例えば、リファレンスで渡す次の例は、無効です。

<?php

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

function bar() // & がないことに注意
{
    $a = 5;
    return $a;
}

foo(bar()); // notice が発生

foo($a = 5); // 式、変数ではない
foo(5); // 致命的なエラーが発生する

class Foobar {}

foo(new Foobar()) // PHP 7.0.7 以降は E_NOTICE が発生
                  // Notice: Only variables should be passed by reference

?>

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User Contributed Notes 5 notes

down

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.

down

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
?>

down

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;
}

down

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.
?>

down

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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