blocks|key|2017951|text|从PHP5开始，所有的对象都是通过引用传递和赋值的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2017952|在PHP4中，您仍然需要通过显式地使用&操作符来指定希望通过引用传递对象的位置。|offset|length|style|CODE|2017953|entityMap^0|0|J|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

As of PHP 5, all objects are passed and assigned by reference.

In PHP 4, you still need to specify where you want objects passed by reference, by explicitly using the <code>&amp;</code> operator.

blocks|key|2018257|text|这不同于仅仅传递一个引用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2018258|在http://www.php.net/manual/en/language.oop5.references.php上，列出的示例很不错。在第一个集合中，$a+=+NULL；不会影响$b，因为$a只是一个指针。在第二个集合中，$c+=+NULL；会导致$d也为NULL，因为$d是对$c的引用。|offset|length|2018259|<?php
class+A+{
++++public+$foo+=+1;
}++

$a+=+new+A;
$b+=+$a;
$a->foo+=+2;
$a+=+NULL;
echo+$b->foo."\n";+//+2

$c+=+new+A;
$d+=+&$c;
$c->foo+=+2;
$c+=+NULL;
echo+$d->foo."\n";+//+Notice:++Trying+to+get+property+of+non-object...
?>+|code-block|syntax|javascript|2018260|entityMap|0|LINK|mutability|MUTABLE|url|http://www.php.net/manual/en/language.oop5.references.php^0|0|1|1L|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

It seems to be a little more precise, the value of an object is passed by value, but the value of an object itself is a pointer. This is different from just passing a reference.

On <a href="http://www.php.net/manual/en/language.oop5.references.php" rel="noreferrer">http://www.php.net/manual/en/language.oop5.references.php</a> the example listed is nice. In the first set, $a = NULL; doesn't affect $b since $a was just a pointer. In the second set, $c = NULL; causes $d to be NULL as well since $d is a reference to $c.

<pre><code>&lt;?php
class A {
 public $foo = 1;
} 

$a = new A;
$b = $a;
$a-&gt;foo = 2;
$a = NULL;
echo $b-&gt;foo."\n"; // 2

$c = new A;
$d = &amp;$c;
$c-&gt;foo = 2;
$c = NULL;
echo $d-&gt;foo."\n"; // Notice: Trying to get property of non-object...
?&gt; 
</code></pre>

blocks|key|1804737|text|这只是一个通过引用传递“对象”很有用的例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1804738|class+RefTest1
{
++++public+$foo;

++++public+function+__construct(RefTest2+&$ref+=+null)
++++{
++++++++$this->foo+=&+$ref;
++++}
}

class+RefTest2
{
++++public+$foo;

++++public+function+__construct(RefTest1+&$ref+=+null)
++++{
++++++++$this->foo+=&+$ref;
++++}
}

class+RefTest3
{
++++public+$foo;

++++public+function+__construct(RefTest2+&$ref+=+null)
++++{
++++++++$this->foo+=&+$ref;
++++}
}

class+DoCrossRef
{
++++public+$refTest1;
++++public+$refTest2;

++++public+function+__construct()
++++{
++++++++$this->refTest1+=+new+RefTest1($this->refTest2);
++++++++$this->refTest2+=+new+RefTest2($this->refTest1);
++++}

++++public+function+changeReference()
++++{
++++++++$this->refTest1+=+new+RefTest3($this->refTest2);
++++}
}|code-block|syntax|javascript|1804739|最后，RefTest1保存对RefTest2的引用，反之亦然，如果在创建RefTest1时RefTest2对象还不存在。|offset|length|style|CODE|1804740|在调用DoCrossRef->changeReference()之后，RefTest2对象还保存了对新RefTest3对象的引用。|1804741|entityMap^0|0|0|3|8|E|8|10|8|19|8|0|3|T|Z|8|1F|8|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]|$I|16|J|17|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|P|$]]

Just an example where passing "objects" by reference is useful:

<pre><code>class RefTest1
{
 public $foo;

 public function __construct(RefTest2 &amp;$ref = null)
 {
 $this-&gt;foo =&amp; $ref;
 }
}

class RefTest2
{
 public $foo;

 public function __construct(RefTest1 &amp;$ref = null)
 {
 $this-&gt;foo =&amp; $ref;
 }
}

class RefTest3
{
 public $foo;

 public function __construct(RefTest2 &amp;$ref = null)
 {
 $this-&gt;foo =&amp; $ref;
 }
}

class DoCrossRef
{
 public $refTest1;
 public $refTest2;

 public function __construct()
 {
 $this-&gt;refTest1 = new RefTest1($this-&gt;refTest2);
 $this-&gt;refTest2 = new RefTest2($this-&gt;refTest1);
 }

 public function changeReference()
 {
 $this-&gt;refTest1 = new RefTest3($this-&gt;refTest2);
 }
}
</code></pre>

At the end <code>RefTest1</code> holds a reference to <code>RefTest2</code> and the other way around, also if the <code>RefTest2</code> object did not exist at the time <code>RefTest1</code> was created.

After calling <code>DoCrossRef-&gt;changeReference()</code>, the <code>RefTest2</code> objects also holds a reference to the new <code>RefTest3</code> object.

blocks|key|1804819|text|下面是另一个示例，它确认对象是通过引用传递的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1804820|<?php

//+discussion:+https://stackoverflow.com/questions/2715026/are-php5-objects-passed-by-reference#_=_
class+A+{
++++public+$foo+=+1;
}

class+B+{
++++function+inc($obj)+{
++++++++$obj->foo%2B%2B;
++++}
}

$objA+=+new+A();
$objB+=+new+B();
$objB->inc($objA);

echo+"["+.+$objA->foo+.+"]";+//+Outputs+[2]|code-block|syntax|javascript|1804821|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's another example confirming that objects are passed by reference.

<pre><code>&lt;?php

// discussion: https://stackoverflow.com/questions/2715026/are-php5-objects-passed-by-reference#_=_
class A {
 public $foo = 1;
}

class B {
 function inc($obj) {
 $obj-&gt;foo++;
 }
}

$objA = new A();
$objB = new B();
$objB-&gt;inc($objA);

echo "[" . $objA-&gt;foo . "]"; // Outputs [2]
</code></pre>

I can't seem to get any consistent info on this. Different sources appear to say different things and the venerable php.net itself (appears) not to explicitly state this - although, I must admit, I only had a quick look.

In cases where I am passing around 'heavy' objects, I need to pass by reference, but I don't want to keep typing:

<pre><code>function foo(TypeName&amp; $obj)
</code></pre>

if I can get away with simply

<pre><code>function foo(TypeName $obj)
</code></pre>

So what does the standard say?

Are PHP5 objects passed by reference?

我似乎得不到任何关于这方面的一致信息。不同的来源似乎说了不同的事情，古老的php.net本身(似乎)没有明确地说明这一点-尽管，我必须承认，我只是快速地看了一眼。在我传递“重”对象的情况下，我需要通过引用传递，但我不想一直输入：function foo(TypeName& $obj)如果我能用简单的function f...

问PHP5对象是通过引用传递的吗？
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