php教程

引用传递

可以将一个变量通过引用传递给函数，这样该函数就可以修改其参数的值。语法如下：

<?php
function foo(&$var)
{
    $var++;
}

$a=5;
foo($a);
// $a is 6 here
?>
注意在函数调用时没有引用符号——只有函数定义中有。光是函数定义就足够使参数通过引用来正确传递了。在最近版本的 PHP 中如果把 & 用在 foo(&$a); 中会得到一条警告说"Call-time pass-by-reference"已经过时了。

以下内容可以通过引用传递：

  • 变量，例如 foo($a)
  • New 语句，例如 foo(new foobar())

  • 从函数中返回的引用，例如：

    <?php
    function &bar()
    {
            $a = 5;
        return     $a;
    }
    foo(bar());
    ?>
    详细解释见引用返回。

任何其它表达式都不能通过引用传递，结果未定义。例如下面引用传递的例子是无效的：

<?php
function foo(&$var)
{
    $var++;
}
function bar() // Note the missing &
{
    $a = 5;
    return $a;
}
foo(bar()); // 自 PHP 5.0.5 起导致致命错误，自 PHP 5.1.1 起导致严格模式错误
            // 自 PHP 7.0 起导致 notice 信息
foo($a = 5) // 表达式，不是变量
foo(5) // 导致致命错误
?>
这些条件是 PHP 4.0.4 以及以后版本有的。

User Contributed Notes

ccb_bc at hotmail dot com 01-Jul-2019 05:58
<?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
?>
fladnag at zerezo dot com 18-Aug-2017 09:29
Beware of using references with anonymous function and "use" keyword :

If you have a PHP version between 5.3 and < 5.3.10, "use" keyword break the reference :

<?php
function withRef(&$arg) {
  echo 'withRef - BEGIN - '.$arg."\n"; // 1
  $func = function() use($arg) { /* do nothing, just declare using $arg */ };
  $arg = 2;
  echo 'withRef - END - '.$arg."\n"; // 2
}

$arg = 1;
echo 'withRef - BEFORE - '.$arg."\n"; // 1
withRef($arg);
// in PHP 5.3 < 5.3.10 : display 1
// in PHP 5.3 >= 5.3.10 : display 2
echo 'withRef - AFTER - '.$arg."\n";
?>

A workaround is to use a copy of the reference variable in "use" keyword :
<?php
  ...
  $arg2 = $arg;
  $func = function() use($arg2) { /* do nothing, just declare using $arg2 */ };
nickshanks at nickshanks dot com 26-May-2017 12:49
For anyone wondering, the copy-on-write behaviour just does the Right Thing™ when an array is passed to a function not by-ref which then passes it through to another function by-ref without writing to it. For example:

<?php

function do_sort(array $array) : array {
    usort($array, function ($a, $b) {
        return strnatcasecmp($a['name'], $b['name']);
    });

    return $array;
}

$data = [
    [
        'name' => 'one',
    ], [
        'name' => 'two',
    ], [
        'name' => 'three',
    ], [
        'name' => 'four',
    ],
];

var_dump($data);
do_sort($data); // does not affect value of $data
var_dump($data);
$data = do_sort($data);
var_dump($data);
pallsopp at gmail dot com 19-Dec-2016 08:32
The comment by tnestved at yahoo dot com is incorrect as it is based purely on perception and not architecture. The method passing the object should not care whether it is by ref or by val, and nor should the reader.

If we are talking about readability and perception, then the receiving method needs to show that the object coming in is a reference, not an object instance, otherwise the reader is perplexed why the object is not returned.

Good functional headers alleviate all issues in this case.
mike at eastghost dot com 19-May-2015 07:06
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;
}
no at spam dot please 05-May-2015 08:36
agreed : this change produces less readable code.

additionally, it breaks many existing perfectly working codes which are not portable anymore and in some cases will require complex modifications

another issue regards the fatal error that is produced : how the hell am i supposed to do if i want to allow the user to use a value that is not even in a variable, or the return or a function call, or use call_user_func... this produces many occasions for a code to even break at run time
tnestved at yahoo dot com 14-Oct-2014 01:54
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.
phpnet at holodyn dot com 26-Mar-2014 06:38
The notes indicate that a function variable reference will receive a deprecated warning in the 5.3 series, however when calling the function via call_user_func the operation aborts without fatal error.

This is not a "bug" since it is not likely worth resolving, however should be noted in this documentation.
diabolos @t gmail dot com 27-Jul-2012 02:46
<?php

/*

  This function internally swaps the contents between
  two simple variables using 'passing by reference'.

  Some programming languages have such a swap function
  built in, but PHP seems to lack such a function.  So,
  one was created to fill the need.  It only handles
  simple, single variables, not arrays, but it is
  still a very handy tool to have.

  No value is actually returned by this function, but
  the contents of the indicated variables will be
  exchanged (swapped) after the call.
*/

// ------------------------------------------
// Demo call of the swap(...) function below.

 $a = 123.456;
 $b = 'abcDEF';
 
 print "<pre>Define:\na = $a\nb = '$b'</pre>";
 swap($a,$b);
 print "<pre>After swap(a,b):\na = '$a'\nb = $b</pre>";

// -------------------------------

   function swap (&$arg1, &$arg2)
{

// Swap contents of indicated variables.
   $w=$arg1;   $arg1=$arg2;   $arg2=$w;
}

?>
fdelizy at unfreeze dot net 12-Aug-2006 09:32
Some have noticed that reference parameters can not be assigned a default value. It's actually wrong, they can be assigned a value as the other variables, but can't have a "default reference value", for instance this code won't compile :

<?php
function use_reference( $someParam, &$param =& $POST )
{
 ...
}
?>

But this one will work :

<?php
function use_reference( $someParam, &$param = null )
?>

So here is a workaround to have a default value for reference parameters :

<?php
$array1 = array ( 'test', 'test2' );

function AddTo( $key, $val, &$array = null)
{
    if ( $array == null )
    {
      $array =& $_POST;
    }

    $array[ $key ] = $val ;
}

AddTo( "indirect test", "test", $array1 );
AddTo( "indirect POST test", "test" );

echo "Array 1 " ;
print_r ( $array1);

echo "_POST ";
print_r( $_POST );

?>

And this scripts output is :

Array 1 Array
(
    [0] => test
    [1] => test2
    [indirect test] => test
)
_POST Array
(
    [indirect POST test] => test
)

Of course that means you can only assign default reference to globals or super globals variables.

Have fun
pillepop2003 at yahoo dot de 13-Feb-2005 08:08
PHP has a strange behavior when passing a part of an array by reference, that does not yet exist.

<?php
    function func(&$a)
    {
        // void();
    }
   
    $a['one'] =1;
    func($a['two']);
?>   

var_dump($a) returns

    array(2) {
        ["one"]=>
        int(1)
        ["two"]=>
        NULL
    }

...which seems to be not intentional!
obscvresovl at NOSPAM dot hotmail dot com 25-Dec-2004 10:51
Just a simple note...

<?php

$num = 1;

function blah(&$var)
{
    $var++;
}

blah($num);

echo $num; #2

?>

<?php

$num = 1;

function blah()
{
    $var =& $GLOBALS["num"];
    $var++;
}

blah();

echo $num; #2

?>

Both codes do the same thing! The second code "explains" how passage of parameters by reference works.
Sergio Santana: ssantana at tlaloc dot imta dot mx 10-Sep-2004 08:25
Sometimes we need functions for building or modifying arrays whose elements are to be references to other variables (arrays or objects for instance). In this example, I wrote two functions 'tst' and 'tst1' that perform this task. Note how the functions are written, and how they are used.

<?php
function tst(&$arr, $r) {
  // The argument '$arr' is declared to be passed by reference,
  // but '$r' is not;
  // however, in the function's body, we use a reference to
  // the '$r' argument
 
  array_push($arr, &$r);
  // Alternatively, this also could be $arr[] = &$r (in this case)
}
 
$arr0 = array();          // an empty array
$arr1 = array(1,2,3);   // the array to be referenced in $arr0

// Note how we call the function:
tst($arr0, &$arr1); // We are passing a reference to '$arr1' in the call !

print_r($arr0); // Contains just the reference to $arr1

array_push($arr0, 5); // we add another element to $arr0
array_push($arr1, 18); // we add another element to $arr1 as well

print_r($arr1); 
print_r($arr0); // Changes in $arr1 are reflected in $arr0

// -----------------------------------------
// A simpler way to do this:

function tst1(&$arr, &$r) {
  // Both arguments '$arr' and '$r" are declared to be passed by
  // reference,
  // again, in the function's body, we use a reference to
  // the '$r' argument
 
  array_push($arr, &$r);
  // Alternatively, this also could be $arr[] = &$r (in this case)
}

 
$arr0 = array();          // an empty array
$arr1 = array(1,2,3);   // the array to be referenced in $arr0

// Note how we call the function:
tst1($arr0, $arr1); // 'tst1' understands '$r' is a reference to '$arr1'

echo "-------- 2nd. alternative ------------ <br>\n";

print_r($arr0); // Contains just the reference to $arr1

array_push($arr0, 5); // we add another element to $arr0
array_push($arr1, 18);

print_r($arr1); 
print_r($arr0); // Changes in $arr1 are reflected in $arr0

// This outputs:
// X-Powered-By: PHP/4.1.2
// Content-type: text/html
//
// Array
// (
//     [0] => Array
//         (
//             [0] => 1
//             [1] => 2
//             [2] => 3
//         )
//
// )
// Array
// (
//     [0] => 1
//     [1] => 2
//     [2] => 3
//     [3] => 18
// )
// Array
// (
//     [0] => Array
//         (
//             [0] => 1
//             [1] => 2
//             [2] => 3
//             [3] => 18
//         )
//
//     [1] => 5
// )
// -------- 2nd. alternative ------------
// Array
// (
//     [0] => Array
//         (
//             [0] => 1
//             [1] => 2
//             [2] => 3
//         )
//
// )
// Array
// (
//     [0] => 1
//     [1] => 2
//     [2] => 3
//     [3] => 18
// )
// Array
// (
//     [0] => Array
//         (
//             [0] => 1
//             [1] => 2
//             [2] => 3
//             [3] => 18
//         )
//
//     [1] => 5
// )
?>

In both cases we get the same result.

I hope this is somehow useful

Sergio.

CopyRight © 2008-2022 verySource.Com All Rights reserved.　京ICP备17048824号-1　京公网安备：11010502034788