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在构造函数中创建引用可能会导致混淆的结果。本节以教程形式帮助避免问题。
下面来检查一下用拷贝运算符 = 创建的 $bar1
和用引用运算符 =& 创建的 $bar2
有没有区别...
显然没有区别,但实际上有一个非常重要的区别:$bar1
和 $globalref[0]
并没有被引用,它们不是同一个变量。这是因为“new”默认并不返回引用,而返回一个拷贝。
注:
在返回拷贝而不是引用中并没有性能上的损失(因为
PHP 4 及以上版本使用了引用计数)。相反更多情况下工作于拷贝而不是引用上更好,因为建立引用需要一些时间而建立拷贝实际上不花时间(除非它们都不是大的数组或对象,而其中之一跟着另一个变,那使用引用来同时修改它们会更聪明一些)。
要证明以上写的,看看下面的代码。
最后给出另一个例子,试着理解它。
aib
01-Jun-2006 02:20
Note that the "new" operator no longer returns a copy of the constructed object, but the object itself. This behavior may be as of PHP 5.x.
This means that
$foo = new class();
and
$foo =& new class();
are equivalent now, and you don't have to worry about the problems stated on this page.
Consider the following example:
<?php
class foo
{
var $num;
function foo()
{
global $globalref;
$globalref[] = &$this;
}
}
$bar = new foo();
$bar->num = 1;
$globalref[0]->num = 2;
echo "\$bar->num = ".$bar->num."\n";
echo "\$globalref[0]->num = ".$globalref[0]->num."\n";
?>
On PHP 5.0.1, it will print:
$bar->num = 2
$globalref[0]->num = 2
Whereas on PHP 4.4.0, it would print:
$bar->num = 1
$globalref[0]->num = 2
unless you assigned $bar to a new foo using a reference.
info at keenkelly dot com
11-Feb-2005 09:24
Regarding the use of assignment versus reference operators as they pertain to using classes which register shutdown functions.
Here's an example:
<?php
class testclass {
var $value;
function testclass() {
register_shutdown_function(array(&$this, 'tcshutdown'));
$this->value = 1;
print 'Construction value=[' . $this->value ."]\n";
}
function setvalue($value) {
$this->value = $value;
}
function tcshutdown() {
print 'Shutdown value=[' . $this->value ."]\n";
}
}
$testobj = new testclass();
$testobj->setvalue(2);
?>
Despite changing the value to 2 after the object is constructed, the above code will result in the following output:
Construction value=[1]
Shutdown value=[1]
The reason is new testclass() constructs the object and registers tcshutdown function from that copy of the object. Then assigning it to $testobj makes a copy of the object accessible as $testobj. Altering $testobj values confines those changes to the $testobj copy and doesn't affect the original ("new") copy of the object. When shutdown time comes around and the tcshutdown() function is called, it calls the original ("new") copy of tcshutdown() NOT $testobj's copy. Thus tcshutdown() sees only the original member value that was initialized by the constructor before the object was copied to $testobj.
The way to resolve this is to, again, convert from assignment to a reference with the following alteration:
<?php
$testobj =& new testclass();
?>
With that change, running the code again will result in what we expected:
Construction value=[1]
Shutdown value=[2]
This of course allows the tcshutdown() function to perform shutdown operations based on parameters which were altered subsequent to the construction of the object.
denis at yum
01-Sep-2004 06:22
Construction Methods
In case of PHP versions earlier 5.x (or both) you have to consider, that an instance of a class is not fully allocated while the construction method(s) aren't finished.
For example, this solution could fail:
class Child
{
var $c;
function Child(&$parent)
{
$this->c = &$parent;
}
}
class Test
{
var $a;
var $var;
function Test ()
{
$this->var = 123;
$this->a = &new Child($this);
}
}
$b = &new Test();
$b->a->c->var = 456;
echo $b->var;
[outputs: 123]
In order of really allocated instance references, append referenced objects after (and not during) the construction.
phpnotes at robertdhill dot com
03-Jul-2004 03:29
An earler note asked why the equal operator doesn't just make a reference, assuming that a direct reference would always be desired.
The answer is because a direct reference is *not* always desired.
When you instantiate a class, you are creating an object that has its own copy of all the class variables, access to the class methods, etc.
Now say you've created an instance of a class "MyClass" called "myObjectA" and you've done some processing to it, set some variables or whatever. Now say you want to create a duplicate of the object to do some changes, but you want to *keep the original.*
That is what the assignment (=) operator is for.
If you'd made a reference (&=), then any changes you make to the duplicate are also reflected in the original. This is undesireable.
So there are circumstances where the assignment (=) operator is required, and there are circumstances where the reference (&=) operator is required.
It just depends on what you're doing.
Hope this helps.
php dot net at game-point dot net
17-Jun-2004 03:02
As a bit of a newbie to objects, I found the final example very hard to work out at first. I'd like to try and explain more clearly:
It all centres around what the $this variable actually is. The answer is, it's an object. It's the object that the $this variable is currently inside.
So let's say we call the first instance of class A that is created by my script "Instance1". If I say,
<?php
class A {
function test() {
$foo = $this;
}
}
$myclass = new A();
?>
... then $this is the object Instance1. Note that it is *NOT* the same as the object $myclass, as $myclass is simply a *COPY* of the object Instance1. This means that if you create an object using the copy (=) operator, there is an instance of the object (in this case, Instance1) hanging around unreferenced.
Whereas if we use the reference (=&) operator to create an object, the first instance (Instance1) is directly referenced by the variable specified, and no copy is created. eg.
<?php
class A {
function test() {
$foo = $this;
}
}
$myclass =& new A();
?>
... this assigns the variable $myclass directly to Instance1, with no need for Instance1 to be copied. There is only one instance of the class created, and it is assigned to $myclass.
In brief, if we use <--> to indicate a reference between a variable and an instance of a class:
Statement | Results in...
$myclass = new A(); | Instance1 Instance2<-->$myclass (Instance2 is a copy)
$myclass =& new A(); | Instance1<-->$myclass
All seems simple when you see the diagram, eh? :-) Of course, you may be asking the question, "Why doesn't the '=' operator just make $myclass reference first object Instance1? Why would you ever need Instance1 to be copied, surely all you'd ever want or need is a direct reference to the newly created object Instance1?" Well, that's the same question I and others are asking. Hopefully this behaviour will change in a later version of PHP.
Hope that helps.
blueshadow822 at hotmail dot com
10-Feb-2004 07:20
Original : rick_php_net at roundeye dot net
Date : 01-Oct-2002 05:37
Modify : bitmore.co.kr
class StaticList {
var $list = 0;
function StaticList($name = '') {
global $master;
if (!isset($master)) {
print "creating new list $name<Br>\n";
//$master = $this;
$master =& $this;
$master->list = array();
}else {
print "creating new list $name... using global list<br>\n";
}
//$this->list =& $master->list;
//$this->list = $master->list;
return ($this);
}
function show() {
if (is_array($this->list)) {
if (count($this->list)) {
reset($this->list);
while (list($key, $value) = each($this->list)) {
print "list[$key] => $value<br>\n";
}
}else {
print "list empty.<br>\n";
}
}else {
print "list is not a list.<br>\n";
}
}
function append($val) {
if (!is_array($this->list)) {
print "forced to create list...<br>\n";
$this->list = array();
}else {
print "appending $val to list<br>\n";
$this->list[] = $val;
}
}
}
$a = new StaticList("a");
$a->append(1);
$a->append(2);
$b = new StaticList("b");
$b->append(3);
$b->append(4);
$a->append(5);
$b->append(6);
print "Contents of a:<br>\n";
$a->show();
print "Contents of a:<br>\n";
$b->show();
this Show Printing
creating new list a
appending 1 to list
appending 2 to list
creating new list b
appending 3 to list
appending 4 to list
appending 5 to list
appending 6 to list
Contents of a:
list[0] => 1
list[1] => 2
list[2] => 5
Contents of a:
list[0] => 3
list[1] => 4
list[2] => 6
kyong
30-Jan-2004 10:45
Here's a reminder to those of you are going to use
PHP 5.0.0b3. You don't need to add & before $this any more
to get that object reference.
jjh at gmx dot at
26-Apr-2003 04:52
This is another sample script to demonstrate the use of references on objects.
To test the script first replace all <b r> (note the space!) tags by normal
HTML-linke-brakes! Normal line brakes (what a shame!) cannot be shown inside a
php.net-note!!!
<?
// Class C contains instances of class A and class B.
//
// A should "know" B and B should "know" A.
// In other words: Object A should have a Reference on
// object B and vice versa.
class A {
var $a;
var $a_b; // this links to object B
function A() {
$this->a="set by constructor of A";
}
function getToKnowB(&$b) {
$this->a_b =& $b;
}
}
class B {
var $b;
var $b_a; // this links to object A
function B() {
$this->b="set by constructor of B";
}
function getToKnowA(&$a) {
$this->b_a =& $a;
}
}
class C {
var $myA;
var $myB;
function C() {
// Create Objects A and B
$this->myA = new A;
$this->myB = new B();
// "Introduce" A and B to each other
$this->myA->getToKnowB($this->myB);
$this->myB->getToKnowA($this->myA);
// Prove that A and B are using References (not copies!)
echo "Prove that myA->a equals myB->b_a->a<br>";
$this->myA->a = "one";
echo "this->myA->a is set to ".$this->myA->a."<br>";
$this->myB->b_a->a = "two";
echo "this->myB->b_a->a is set to ".$this->myB->b_a->a."<br>";
echo "this->myA->a: ".$this->myA->a."<br>";
echo "this->myB->b_a->a: ".$this->myB->b_a->a."<br>";
echo "<br>";
echo "Prove that myB->b equals myA->a_b->b<br>";
$this->myB->b = "one";
echo "this->myB->b is set to ".$this->myB->b."<br>";
$this->myA->a_b->b = "two";
echo "this->myA->a_b->b is set to ".$this->myA->a_b->b."<br>";
echo "this->myB->b: ".$this->myB->b."<br>";
echo "this->myA->a_b->b: ".$this->myA->a_b->b."<br>";
}
}
$myC = new C;
/*
evaluates to:
Prove that myA->a equals myB->b_a->a
this->myA->a is set to one
this->myB->b_a->a is set to two
this->myA->a: two
this->myB->b_a->a: two
Prove that myB->b equals myA->a_b->b
this->myB->b is set to one
this->myA->a_b->b is set to two
this->myB->b: two
this->myA->a_b->b: two
*/
?>
rick_php_net at roundeye dot net
01-Oct-2002 07:37
I was attempting to find a way to share data structures between objects and ideally wanted to be able to have a single shared object that would be returned as a reference from the constructor -- without having the caller use a different syntax.
I couldn't get the effect I wanted, but was able to share the internal data structures (which is where the bulk of the storage was in my application). This is an example class I constructed (requires php >= 4.0.4) which illustrates the idea:
class StaticList
{
var $list = 0;
function StaticList($name = '')
{
global $master;
if (!isset($master))
{
print "creating new list $name<Br>\n";
$master = $this;
$master->list = array();
}
else
{
print "creating new list $name... using global list<br>\n";
}
$this->list =& $master->list;
return ($this);
}
function show()
{
if (is_array($this->list))
{
if (count($this->list))
{
reset($this->list);
while (list($key, $value) = each($this->list))
{
print "list[$key] => $value<br>\n";
}
}
else
{
print "list empty.<br>\n";
}
}
else
{
print "list is not a list.<br>\n";
}
}
function append($val)
{
if (!is_array($this->list))
{
print "forced to create list...<br>\n";
$this->list = array();
}
else
{
print "appending $val to list<br>\n";
$this->list[] = $val;
}
}
}
$a = new StaticList("a");
$a->append(1);
$a->append(2);
$b = new StaticList("b");
$b->append(3);
$b->append(4);
$a->append(5);
$b->append(6);
print "Contents of a:<br>\n";
$a->show();
print "Contents of a:<br>\n";
$b->show();
rpav at NOSPAMusers dot sf dot net
14-Aug-2002 06:13
Ah HA! I just found an easy workaround with a bit of playing. This works:
--- cut ---
$OBJECTS = array();
class Object {
function &Object() {
global $OBJECTS;
$OBJECTS[] = &$this;
}
function _init() {
}
}
class Derived {
function &Derived() {
Object::Object();
}
}
function blah() {
global $FOO;
$x =& new Derived;
$GLOBALS['FOO'] = &$x;
$x->foo = 42;
display("x", $x);
}
function display($text, &$o) {
echo("($text) =>" . serialize($o));
}
blah();
display("GLOBALS[FOO]", $GLOBALS['FOO']);
display("OBJECTS", $OBJECTS);
--- cut ---
As you can see, simply adding & before the constructor to make it return a reference does the trick. The bit about "returning a copy by default" lead me to try this trick, and it seems to work. Have fun!
-rjp
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