[英]PHP 4: Can not understand the OOP behaviour
While practicing for certification, I came across one MCQ the question.在练习认证时,我遇到了一个 MCQ 这个问题。 It was as below.
如下所示。
Q: Consider the following script.问:考虑以下脚本。 What will it output?
它会输出什么?
<?php
$global_obj = null;
class my_class
{
var $value;
function my_class()
{
global $global_obj;
$global_obj = &$this;
}
}
$a = new my_class;
$a->my_value = 5;
$global_obj->my_value = 10;
echo $a->my_value;
?>
Choose Right One Options:选择正确的一个选项:
I chose and B: 10 as because in my_class
constructor $global_obj
is being initialized by the reference of $this
.我选择和 B: 10 因为在
my_class
构造函数中$global_obj
正在由$this
的引用初始化。 By when I cross checked my answer was wrong当我交叉检查时,我的答案是错误的
As a description they mentioned that作为描述,他们提到
This is a really tricky one.这是一个非常棘手的问题。 Upon first examination, it would seem that the constructor of
my_class
stores a reference to itself inside the $global_obj variable
.第一次检查时,似乎
my_class
的构造函数在$global_obj variable
存储了对自身的引用。 Therefore, one would expect that, when we later change the value of $global_obj->my_value
to 10, the corresponding value in $a would change as well.因此,可以预期,当我们稍后将
$global_obj->my_value
的值更改为 10 时,$a 中的相应值也会更改。 Unfortunately, the new operator does not return a reference, but a copy of the newly created object.不幸的是,new 运算符不返回引用,而是返回新创建对象的副本。 Therefore, the script will output 5 and the correct answer is A.
因此,脚本将输出 5,正确答案是 A。
Ya I agree the description is well enough but still I am not able to digest it as because we have clearly assigned $global_obj
the reference of $this then how can be this possible?是的,我同意描述已经足够好,但我仍然无法消化它,因为我们已经明确地
$global_obj
分配了$global_obj
的引用,那么这怎么可能呢? Can any one please explain in detail?哪位大神能详细解释一下吗?
we have clearly assigned $global_obj the reference of $this then how can be this possible?
我们已经明确地为 $global_obj 分配了 $this 的引用,那么这怎么可能呢?
You assign the global variable within the constructor.您在构造函数中分配全局变量。 At that time, you are referencing kind of a temporary object.
那时,您正在引用某种临时对象。 Then the constructor returns a copy of that temporary object.
然后构造函数返回该临时对象的副本。 This is that copy that $a will reference (whereas the global var still references the temporary object).
这是 $a 将引用的副本(而全局变量仍然引用临时对象)。
Going OOP with php 4 is nuts.使用 php 4 进行 OOP 是很疯狂的。 I guess you should not understand the OOP behaviour of PHP 4 but just accept it...
我想你不应该理解 PHP 4 的 OOP 行为,而只是接受它......
this script will output "5" which would be A along with the following warning Warning: Creating default object from empty value此脚本将输出“5”,这将是 A 以及以下警告警告:从空值创建默认对象
the reason for this is when you set this原因是当你设置这个
`$global_obj = &$this;`
what your assigning is reference to the current object您的分配是对当前对象的引用
this article may help这篇文章可能会有所帮助
http://www.webmaster-source.com/2010/02/25/why-do-some-php-variables-have-an-ampersand-before-them/ http://www.webmaster-source.com/2010/02/25/why-do-some-php-variables-have-an-ampersand-before-them/
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