引用做什么

PHP 的引用允许用两个变量来指向同一个内容。意思是，当这样做时：


<?php
$a =& $b;
?>

这意味着 $a 和 $b 指向了同一个变量。

注意: $a 和 $b 在这里是完全相同的，这并不是 $a 指向了 $b 或者相反，而是 $a 和 $b 指向了同一个地方。

注意: 如果具有引用的数组被拷贝，其值不会解除引用。对于数组传值给函数也是如此。

同样的语法可以用在函数中，它返回引用，以及用在 new 运算符中（PHP 4.0.4 以及以后版本）：


<?php
$bar =& new fooclass();
$foo =& find_var($bar);
?>

自 PHP 5 起，new 自动返回引用，因此在此使用 =& 已经过时了并且会产生 E_STRICT 级别的消息。

注意: 不用 & 运算符导致对象生成了一个拷贝。如果在类中用 $this，它将作用于该类当前的实例。没有用 & 的赋值将拷贝这个实例（例如对象）并且 $this 将作用于这个拷贝上，这并不总是想要的结果。由于性能和内存消耗的问题，通常只想工作在一个实例上面。
尽管可以用 @ 运算符来抑制构造函数中的任何错误信息，例如用 @new，但用 &new 语句时这不起效果。这是 Zend 引擎的一个限制并且会导致一个解析错误。

警告

如果在一个函数内部给一个声明为 global 的变量赋于一个引用，该引用只在函数内部可见。可以通过使用 $GLOBALS 数组避免这一点。

例 21-1. 在函数内引用全局变量


<?php
$var1 = "Example variable";
$var2 = "";

function global_references($use_globals)
{
    global $var1, $var2;
    if (!$use_globals) {
        $var2 =& $var1; // visible only inside the function
    } else {
        $GLOBALS["var2"] =& $var1; // visible also in global context
    }
}

global_references(false);
echo "var2 is set to '$var2'\n"; // var2 is set to ''
global_references(true);
echo "var2 is set to '$var2'\n"; // var2 is set to 'Example variable'
?>

把 global $var; 当成是 $var =& $GLOBALS['var']; 的简写。从而将其它引用赋给 $var 只改变了本地变量的引用。

注意: 如果在 foreach 语句中给一个具有引用的变量赋值，被引用的对象也被改变。
例 21-2. 引用与 foreach 语句
<?php $ref = 0; $row =& $ref; foreach (array(1, 2, 3) as $row) { // do something } echo $ref; // 3 - last element of the iterated array ?>

警告

复杂数组最好拷贝而不是引用。下面的例子不会如期望中那样工作。

例 21-3. 复杂数组的引用


<?php
$top = array(
    'A' => array(),
    'B' => array(
        'B_b' => array(),
    ),
);

$top['A']['parent'] = &$top;
$top['B']['parent'] = &$top;
$top['B']['B_b']['data'] = 'test';
print_r($top['A']['parent']['B']['B_b']); // array()
?>

引用做的第二件事是用引用传递变量。这是通过在函数内建立一个本地变量并且该变量在呼叫范围内引用了同一个内容来实现的。例如：


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

$a=5;
foo($a);
?>

将使 $a 变成 6。这是因为在 foo 函数中变量 $var 指向了和 $a 指向的同一个内容。更多详细解释见引用传递。

引用做的第三件事是引用返回。

add a note User Contributed Notes

dovbysh at gmail dot com
06-Jul-2007 07:50


Solution to post "php at hood dot id dot au 04-Mar-2007 10:56":



<?php

$a1 = array('a'=>'a');

$a2 = array('a'=>'b');



foreach ($a1 as $k=>&$v)

$v = 'x';



echo $a1['a']; // will echo x



unset($GLOBALS['v']);



foreach ($a2 as $k=>$v)

{}



echo $a1['a']; // will echo x



?>

amp at gmx dot info
08-Jun-2007 05:59


Something that might not be obvious on the first look:

If you want to cycle through an array with references, you must not use a simple value assigning foreach control structure. You have to use an extended key-value assigning foreach or a for control structure.



A simple value assigning foreach control structure produces a copy of an object or value. The following code



$v1=0;

$arrV=array(&$v1,&$v1);

foreach ($arrV as $v)

{

  $v1++;

  echo $v."\n";

}



yields



0

1



which means $v in foreach is not a reference to $v1 but a copy of the object the actual element in the array was referencing to.



The codes



$v1=0;

$arrV=array(&$v1,&$v1);

foreach ($arrV as $k=>$v)

{

    $v1++;

    echo $arrV[$k]."\n";

}



and



$v1=0;

$arrV=array(&$v1,&$v1);

$c=count($arrV);

for ($i=0; $i<$c;$i++)

{

    $v1++;

    echo $arrV[$i]."\n";

}



both yield



1

2



and therefor cycle through the original objects (both $v1), which is, in terms of our aim, what we have been looking for.



(tested with php 4.1.3)

firespade at gmail dot com
03-Apr-2007 02:11


Here's a good little example of referencing. It was the best way for me to understand, hopefully it can help others.



$b = 2;

$a =& $b;

$c = $a;

echo $c;



// Then... $c = 2

php at hood dot id dot au
05-Mar-2007 06:56


I discovered something today using references in a foreach



<?php

$a1 = array('a'=>'a');

$a2 = array('a'=>'b');



foreach ($a1 as $k=>&$v)

$v = 'x';



echo $a1['a']; // will echo x



foreach ($a2 as $k=>$v)

{}



echo $a1['a']; // will echo b (!)

?>



After reading the manual this looks like it is meant to happen. But it confused me for a few days!



(The solution I used was to turn the second foreach into a reference too)

ladoo at gmx dot at
17-Apr-2005 09:05


I ran into something when using an expanded version of the example of pbaltz at NO_SPAM dot cs dot NO_SPAM dot wisc dot edu below.

This could be somewhat confusing although it is perfectly clear if you have read the manual carfully. It makes the fact that references always point to the content of a variable perfectly clear (at least to me).



<?php

$a = 1;

$c = 2;

$b =& $a; // $b points to 1

$a =& $c; // $a points now to 2, but $b still to 1;

echo $a, " ", $b;

// Output: 2 1

?>

php.devel at homelinkcs dot com
15-Nov-2004 11:16


In reply to lars at riisgaardribe dot dk, 



When a variable is copied, a reference is used internally until the copy is modified.  Therefore you shouldn't use references at all in your situation as it doesn't save any memory usage and increases the chance of logic bugs, as you discoved.

joachim at lous dot org
10-Apr-2003 10:46


So to make a by-reference setter function, you need to specify reference semantics _both_ in the parameter list _and_ the assignment, like this:



class foo{

   var $bar;

   function setBar(&$newBar){

      $this->bar =& newBar;

   }

}



Forget any of the two '&'s, and $foo->bar will end up being a copy after the call to setBar.

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