PHP引用如何在数组引擎下工作？

我正在阅读有关 PHP变量引用的文章： http://derickrethans.nl/talks/phparch-php-variables-article.pdf
并且想要检查我的理解是否正确,关于何时创建新的变量容器.

对于非数组,只要您指定一个未指向设置了is_ref的容器的变量,就会创建变量容器.

Ex 1 (let {..} be a container):
$a = 1;     // "a" => {integer,1,is_ref = 0,ref_count = 1}

$b = $a;    // "a","b" => {integer,ref_count = 2}

$b = 2;     // "a" => {integer,ref_count = 1}
            // "b" => {integer,2,ref_count = 1}

Ex 2:
$a = 1;     // "a" => {integer,ref_count = 1}

$b = &$a;    // "a",is_ref = 1,ref_count = 2}

$b = 2;     // "a",ref_count = 2}

它如何适用于数组？它看起来不一样适用.例如,

$a = array(1,3);  
$b = $a;
$c = &$b[2];
$c = 4;
print_r($a); // prints (1,3) instead of (1,4)
print_r($b); // prints (1,4)

我的期望：

$a和$b指向同一个容器.在这个容器中,我们有3个numeric_keys“0”,“1”,“2”分别指向整数1,2和3的容器.

当我们执行$c =& $b [2]时,我们更新包含整数3的容器：

> is_ref = 0变为is_ref = 1
> ref_count = 1变为ref_count = 2.

当我们执行$c = 4时,我们更新包含整数3的容器：

>因为设置了is_ref,整数3变为整数4

但是,我的期望出了点问题,因为$a [2]！= 4最后.我想弄明白为什么.我最好的猜测是,当我们尝试引用数组的元素或对象的属性时,PHP引擎首先检查数组/对象本身以查看是否is_ref = 1.如果是,则一切都按照我的期望工作.如果is_ref = 0,则会发生其他事情,这就是我所看到的.有人可以填写我的“其他东西”是什么吗？

编辑
看起来这就是实际发生的事情.这段代码应该澄清一切！

$a = array(1,3);
$b = $a;
$c = &$b[2];      // $b points to a new container where $b[0],$b[1] still point to same container as $a[0],$a[1],but $b[2] points to a new container also pointed to by $c
$d = $b;        // $d points to $b's container,this means changing $c will also change $d[2]      
$d[0] = 5;      // The container pointed to by $d[0] is the same as the one pointed to by $a[0] and $b[0]. Since this container has is_ref = 0,$d[0] will Now point to a new container

// At this point $a = (1,3),$b = (1,$c = 3,$d = (5,3)

$d[2] = 25;     // The container pointed to by $d[2] is the same as the one pointed to by $b[2] and $c. Since this container has is_ref = 1,Changing $d[2] will affect both $b[2] and $c.

// At this point $a = (1,25),$c = 25,25)

$e = $d[2];     // Since $d[2]'s container has is_ref = 1,$e will point to its own container

$c = 4;         // Same idea as $d[2] = 25; except $e won't get affected

// At this point $a = (1,4),$c = 4,$e = 25

// only way to have $d[2] be different from $b[2] is to make the container's is_ref = 0
unset($b[2]);
unset($c);
$b[2] = $d[2];
$d[2] = 55;

// At this point $a = (1,$e = 25

你创造了什么$a它是一个简单的变量.但是当你创建$b时,默认情况下,PHP复制了变量.所以$b现在与$a完全分开,就像你在第一个例子中那样.

然后将$c设置为等于$b [2]的引用.所以他们都指向相同的内存地址.更新一个,它更新另一个.问题是你认为$a也应该更新,但不应该因为$b是它自己的变量.考虑当我们将$b更改为$a的引用时会发生什么

$a = array(1,3);  
$b = &$a;
$c = &$b[2];
$c = 4;
print_r($a); // prints (1,4)

这就像你描述的那样,因为$b和$a引用相同的东西(技术上$b现在是一个指向$a的符号)

如果你想更深入地探讨这个主题,这里有一篇很好的文章,深入介绍它. http://webandphp.com/how-php-manages-variables

PHP引用如何在数组引擎下工作？

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