引用做什么

目录搜索

文字

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。

Note:

$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。

Note:

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。

Note:

如果对一个未定义的变量进行引用赋值、引用参数传递或引用返回，则会自动创建该变量。

Example #1 对未定义的变量使用引用

<?php function foo (& $var ) { } foo ( $a ); // $a is "created" and assigned to null $b = array(); foo ( $b [ 'b' ]); var_dump ( array_key_exists ( 'b' , $b )); // bool(true) $c = new StdClass ; foo ( $c -> d ); var_dump ( property_exists ( $c , 'd' )); // bool(true) ?>

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。

Note:

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。

Warning

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。

Example #2 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。

Note:

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。

Example #3 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 ?>

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。

用户评论:

[#1] admin at torntech dot com [2013-08-28 19:58:39]

Something that has not been discussed so far is reference of a reference.
I needed a quick and dirty method of aliasing incorrect naming until a proper rewrite could be done.
Hope this saves someone else the time of testing since it was not covered in the Does/Are/Are Not pages.
Far from best practice, but it worked.

<?php
$a = 0;

$b =& $a;
$a =& $b;

$a = 5;
echo $a . ', ' . $b;
//ouputs: 5,5

echo ' | ';

$b = 6;
echo $a . ',' . $b;
//outputs: 6,6

echo ' | ';
unset( $a );
echo $a . ', ' . $b;

//outputs: , 6

class Product {

public $id;
private $productid;

public function __construct( $id = null ) {
$this->id =& $this->productid;
$this->productid =& $this->id;
$this->id = $id;
}

public function getProductId() {
return $this->productid;
}

}

echo ' | ';

$Product = new Product( 1 );
echo $Product->id . ', ' . $Product->getProductId();
//outputs 1, 1
$Product->id = 2;
echo ' | ';
echo $Product->id . ', ' . $Product->getProductId();
//outputs 2, 2
$Product->id = null;
echo ' | ';
echo $Product->id . ', ' . $Product->getProductId();
//outouts ,

[#2] Oddant [2013-06-01 08:07:32]

About the example on array references.
I think this should be written in the array chapter as well.
Indeed if you are new to programming language in some way, you should beware that arrays are pointers to a vector of Byte(s).

<?php $arr = array(1); ?>
$arr here contains a reference to which the array is located.
Writing :
<?php echo $arr[0]; ?>
dereferences the array to access its very first element.

Now something that you should also be aware of (even you are not new to programming languages) is that PHP use references to contains the different values of an array. And that makes sense because the type of the elements of a PHP array can be different.

Consider the following example :

<?php $arr = array(1, 'test'); $point_to_test =& $arr[1]; $new_ref = 'new'; $arr[1] =& $new_ref; echo $arr[1]; // echo 'new'; echo $point_to_test; // echo 'test' ! (still pointed somewhere in the memory) ?>

[#3] butshuti at smartrwanda dot org [2013-02-17 18:05:52]

This appears to be the hidden behavior: When a class function has the same name as the class, it seems to be implicitly called when an object of the class is created.
For instance, you may take a look at the naming of the function "reftest()": it is in the class "reftest". The behavior can be tested as follows:

<?php class reftest { public $a = 1; public $c = 1; public function reftest1() { $b =& $this->a; $b++; } public function reftest2() { $d =& $this->c; $d++; } public function reftest() { echo "REFTEST() called here!\n"; } } $reference = new reftest(); $reference->reftest1(); $reference->reftest2(); echo $reference->a."\n"; //Echoes 2, not 3 as previously noticed. echo $reference->c."\n"; //Echoes 2. ?>

The above outputs:

REFTEST() called here!
2
2

Notice that reftest() appears to be called (though no explicit call to it was made)!

[#4] nay at woodcraftsrus dot com [2011-07-08 09:35:46]

in PHP you don't really need pointer anymore if you want to share an object across your program

<?php class foo{ protected $name; function __construct($str){ $this->name = $str; } function __toString(){ return 'my name is "'. $this->name .'" and I live in "' . __CLASS__ . '".' . "\n"; } function setName($str){ $this->name = $str; } } class MasterOne{ protected $foo; function __construct($f){ $this->foo = $f; } function __toString(){ return 'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n"; } function setFooName($str){ $this->foo->setName( $str ); } } class MasterTwo{ protected $foo; function __construct($f){ $this->foo = $f; } function __toString(){ return 'Master: ' . __CLASS__ . ' | foo: ' . $this->foo . "\n"; } function setFooName($str){ $this->foo->setName( $str ); } } $bar = new foo('bar'); print("\n"); print("Only Created \$bar and printing \$bar\n"); print( $bar ); print("\n"); print("Now \$baz is referenced to \$bar and printing \$bar and \$baz\n"); $baz =& $bar; print( $bar ); print("\n"); print("Now Creating MasterOne and Two and passing \$bar to both constructors\n"); $m1 = new MasterOne( $bar ); $m2 = new MasterTwo( $bar ); print( $m1 ); print( $m2 ); print("\n"); print("Now changing value of \$bar and printing \$bar and \$baz\n"); $bar->setName('baz'); print( $bar ); print( $baz ); print("\n"); print("Now printing again MasterOne and Two\n"); print( $m1 ); print( $m2 ); print("\n"); print("Now changing MasterTwo's foo name and printing again MasterOne and Two\n"); $m2->setFooName( 'MasterTwo\'s Foo' ); print( $m1 ); print( $m2 ); print("Also printing \$bar and \$baz\n"); print( $bar ); print( $baz ); ?>

[#5] elrah [] polyptych [dot] com [2011-01-27 22:38:44]

It appears that references can have side-effects. Below are two examples. Both are simply copying one array to another. In the second example, a reference is made to a value in the first array before the copy. In the first example the value at index 0 points to two separate memory locations. In the second example, the value at index 0 points to the same memory location.

I won't say this is a bug, because I don't know what the designed behavior of PHP is, but I don't think ANY developers would expect this behavior, so look out.

An example of where this could cause problems is if you do an array copy in a script and expect on type of behavior, but then later add a reference to a value in the array earlier in the script, and then find that the array copy behavior has unexpectedly changed.

<?php // Example one $arr1 = array(1); echo "\nbefore:\n"; echo "\$arr1[0] == {$arr1[0]}\n"; $arr2 = $arr1; $arr2[0]++; echo "\nafter:\n"; echo "\$arr1[0] == {$arr1[0]}\n"; echo "\$arr2[0] == {$arr2[0]}\n"; // Example two $arr3 = array(1); $a =& $arr3[0]; echo "\nbefore:\n"; echo "\$a == $a\n"; echo "\$arr3[0] == {$arr3[0]}\n"; $arr4 = $arr3; $arr4[0]++; echo "\nafter:\n"; echo "\$a == $a\n"; echo "\$arr3[0] == {$arr3[0]}\n"; echo "\$arr4[0] == {$arr4[0]}\n"; ?>

[#6] akinaslan at gmail dot com [2011-01-08 17:31:15]

In this example class name is different from its first function and however there is no construction function. In the end as you guess "a" and "c" are equal. So if there is no construction function at same time class and its first function names are the same, "a" and "c" doesn't equal forever. In my opinion php doesn't seek any function for the construction as long as their names differ from each others.

<?php class reftest_new { public $a = 1; public $c = 1; public function reftest() { $b =& $this->a; $b++; } public function reftest2() { $d =& $this->c; $d++; } } $reference = new reftest_new(); $reference->reftest(); $reference->reftest2(); echo $reference->a; //Echoes 2. echo $reference->c; //Echoes 2. ?>

[#7] Amaroq [2010-01-16 03:14:05]

I think a correction to my last post is in order.

When there is a constructor, the strange behavior mentioned in my last post doesn't occur. My guess is that php was treating reftest() as a constructor (maybe because it was the first function?) and running it upon instantiation.

<?php class reftest { public $a = 1; public $c = 1; public function __construct() { return 0; } public function reftest() { $b =& $this->a; $b++; } public function reftest2() { $d =& $this->c; $d++; } } $reference = new reftest(); $reference->reftest(); $reference->reftest2(); echo $reference->a; //Echoes 2. echo $reference->c; //Echoes 2. ?>

[#8] Amaroq [2010-01-15 06:08:10]

When using references in a class, you can reference $this-> variables.

<?php class reftest { public $a = 1; public $c = 1; public function reftest() { $b =& $this->a; $b = 2; } public function reftest2() { $d =& $this->c; $d++; } } $reference = new reftest(); $reference->reftest(); $reference->reftest2(); echo $reference->a; //Echoes 2. echo $reference->c; //Echoes 2. ?>

However, this doesn't appear to be completely trustworthy. In some cases, it can act strangely.

<?php class reftest { public $a = 1; public $c = 1; public function reftest() { $b =& $this->a; $b++; } public function reftest2() { $d =& $this->c; $d++; } } $reference = new reftest(); $reference->reftest(); $reference->reftest2(); echo $reference->a; //Echoes 3. echo $reference->c; //Echoes 2. ?>

In this second code block, I've changed reftest() so that $b increments instead of just gets changed to 2. Somehow, it winds up equaling 3 instead of 2 as it should.

[#9] strata_ranger at hotmail dot com [2009-09-26 15:29:21]

An interesting if offbeat use for references: Creating an array with an arbitrary number of dimensions.

For example, a function that takes the result set from a database and produces a multidimensional array keyed according to one (or more) columns, which might be useful if you want your result set to be accessible in a hierarchial manner, or even if you just want your results keyed by the values of each row's primary/unique key fields.

<?php function array_key_by($data, $keys, $dupl = false) { // Sanity check if (!is_array($data)) return null; // Allow passing single key as a scalar if (is_string($keys) or is_integer($keys)) $keys = Array($keys); elseif (!is_array($keys)) return null; // Our output array $out = Array(); // Loop through each row of our input $data foreach($data as $cx => $row) if (is_array($row)) { // Loop through our $keys foreach($keys as $key) { $value = $row[$key]; if (!isset($last)) // First $key only { if (!isset($out[$value])) $out[$value] = Array(); $last =& $out; // Bind $last to $out } else // Second and subsequent $key.... { if (!isset($last[$value])) $last[$value] = Array(); } // Bind $last to one dimension 'deeper'. // First lap: was &$out, now &$out[...] // Second lap: was &$out[...], now &$out[...][...] // Third lap: was &$out[...][...], now &$out[...][...][...] // (etc.) $last =& $last[$value]; } if (isset($last)) { // At this point, copy the $row into our output array if ($dupl) $last[$cx] = $row; // Keep previous else $last = $row; // Overwrite previous } unset($last); // Break the reference } else return NULL; // Done return $out; } // A sample result set to test the function with $data = Array(Array('name' => 'row 1', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_a'), Array('name' => 'row 2', 'foo' => 'foo_a', 'bar' => 'bar_a', 'baz' => 'baz_b'), Array('name' => 'row 3', 'foo' => 'foo_a', 'bar' => 'bar_b', 'baz' => 'baz_c'), Array('name' => 'row 4', 'foo' => 'foo_b', 'bar' => 'bar_c', 'baz' => 'baz_d') ); // First, let's key it by one column (result: two-dimensional array) print_r(array_key_by($data, 'baz')); // Or, key it by two columns (result: 3-dimensional array) print_r(array_key_by($data, Array('baz', 'bar'))); // We could also key it by three columns (result: 4-dimensional array) print_r(array_key_by($data, Array('baz', 'bar', 'foo'))); ?>

[#10] dnhuff at acm dot org [2008-06-09 11:33:33]

In reply to Drewseph using foo($a = 'set'); where $a is a reference formal parameter.

$a = 'set' is an expression. Expressions cannot be passed by reference, don't you just hate that, I do. If you turn on error reporting for E_NOTICE, you will be told about it.

Resolution: $a = 'set'; foo($a); this does what you want.

[#11] Drewseph [2008-05-29 16:15:06]

If you set a variable before passing it to a function that takes a variable as a reference, it is much harder (if not impossible) to edit the variable within the function.

Example:
<?php function foo(&$bar) { $bar = "hello\n"; } foo($unset); echo($unset); foo($set = "set\n"); echo($set); ?>

Output:
hello
set

It baffles me, but there you have it.

[#12] Amaroq [2008-03-31 22:56:16]

The order in which you reference your variables matters.

<?php $a1 = "One"; $a2 = "Two"; $b1 = "Three"; $b2 = "Four"; $b1 =& $a1; $a2 =& $b2; echo $a1; //Echoes "One" echo $b1; //Echoes "One" echo $a2; //Echoes "Four" echo $b2; //Echoes "Four" ?>

[#13] charles at org oo dot com [2007-10-19 03:59:16]

points to post below me.
When you're doing the references with loops, you need to unset($var).

for example
<?php foreach($var as &$value) { ... } unset($value); ?>

[#14] Hlavac [2007-10-09 02:25:54]

Watch out for this:

foreach ($somearray as &$i) {
// update some $i...
}
...
foreach ($somearray as $i) {
// last element of $somearray is mysteriously overwritten!
}

Problem is $i contians reference to last element of $somearray after the first foreach, and the second foreach happily assigns to it!

[#15] dovbysh at gmail dot com [2007-07-06 00:50:57]

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 ?>

[#16] amp at gmx dot info [2007-06-08 10:59:57]

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)

[#17] firespade at gmail dot com [2007-04-03 07:11:06]

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

[#18] php at hood dot id dot au [2007-03-04 22:56:07]

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)

[#19] ladoo at gmx dot at [2005-04-17 02:05:46]

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 ?>

[#20] php.devel at homelinkcs dot com [2004-11-15 15:16:38]

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.

[#21] joachim at lous dot org [2003-04-10 15:46:22]

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.