PHP“通过引用赋值"古怪 [英] PHP "Assign by reference" oddity
问题描述
我遇到了一个代码片段，其中包含 $a = &$b; 但还没有测试 $b 是否真的存在(if (isset($b))).我不确定 PHP 是如何处理这个问题的，所以我进行了一个快速的裸测试，现在我更加感兴趣了.
I came across a code snippet which included $a = & $b; but hadn't tested whether $b actually existed (if (isset($b))). I wasn't sure how PHP handled this so I knocked up a quick bare test and now I'm even more intrigued.
$a = array('a'=>'b', 'x'=>'y');
$b = array();
$b[10] = &$a['a'];
$b[11] = &$a['ppp'];
var_dump($a);
var_dump($b);
echo (isset($a['ppp']) ? "SET" :" NOT SET") . "\n";
echo (isset($b[11]) ? "SET" :" NOT SET") . "\n";
这是裸代码，但输出显示的是:
It's bare code but what the output shows is:
只需
$b[11] = &$a['ppp']的裸赋值就足够了，var_dump($a)被报告因为有 3 个成员而不是 2 个，即使没有为$a['ppp']赋值.$a['ppp']被报告为有一个值 (NULL) 但也有isset()=FALSE.
Just the bare assignment of
$b[11] = &$a['ppp']is enough,var_dump($a)is reported as having 3 members not 2, even though no assignment was made for$a['ppp'].$a['ppp']is reported as having a value (NULL) but alsoisset()=FALSE.
同时，$b[11] 显示值 NULL 和 isset()=FALSE 即使它的指涉对象(显然)确实存在(！)
Meanwhile at the same time, $b[11] shows a value NULL and isset()=FALSE even though its referent (apparently) does exist (!)
我很欣赏首先检查可以解决问题"，但我主要是想获得更深入的了解.发生了什么?
I appreciate that checking first fixes the 'problem', but I'm mainly looking for a deeper understanding. What's happening?
推荐答案
解释就这么简单
如果您通过引用分配、传递或返回未定义的变量，它将被创建.
If you assign, pass, or return an undefined variable by reference, it will get created.
(强调我的)
这就是你正在做的；通过引用分配一个未定义的索引以便它被创建.
That's what you're doing; Assigning an undefined index by reference so it gets created.
示例 #1 使用未定义变量的引用
Example #1 Using references with undefined variables
<?php
function foo(&$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($c, 'd')); // bool(true)
?>
那么你还有一个问题:
与此同时，$b[11] 显示了一个值 NULL 和 isset()=FALSE，即使它的所指对象(显然)确实存在(！)
Meanwhile at the same time, $b[11] shows a value NULL and isset()=FALSE even though its referent (apparently) does exist (!)
说明书上也有说明
isset — 确定是否设置了变量并且未设置NULL
isset() 如果测试已设置为 NULL 的变量将返回 FALSE
isset() will return FALSE if testing a variable that has been set to NULL
由于它是 NULL，isset() 返回 FALSE.
Since it is NULL, isset() returns FALSE.
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