静态和全局变量的修饰符是否实现了不可修改的引用?

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英文:

Do static and global modifiers for variables implement a non-modifiable reference?

问题

PHP文档中写道:

PHP使用引用来实现变量的静态和全局修饰符。

<?php
function test_global_ref() {
    global $obj;
    $new = new stdClass;
    $obj = &$new;
}

function test_global_noref() {
    global $obj;
    $new = new stdClass;
    $obj = $new;
}

test_global_ref();
var_dump($obj);
test_global_noref();
var_dump($obj);
?>

由于程序的第一个输出结果是NULL,这是否意味着实现的引用是不可修改的(因此对&$new的引用在某种程度上被取消了)?文档中指出,该实现会导致意外的行为。这背后是否有合乎逻辑的解释?

英文:

Source

The PHP doc says

> PHP implements the static and global modifier for variables in terms
> of references.

&lt;?php
function test_global_ref() {
    global $obj;
    $new = new stdClass;
    $obj = &amp;$new;
}

function test_global_noref() {
    global $obj;
    $new = new stdClass;
    $obj = $new;
}

test_global_ref();
var_dump($obj);
test_global_noref();
var_dump($obj);
?&gt;

Since the program yields NULL as the first output, is this to say that the implemented reference is non-modifiable(hence the reference to &$new is nullified somehow)? The doc says the implementation results in an unexpected behaviour. Is there a logical explanation to this?

答案1

得分: 1

这不涉及全局或静态,而是涉及引用的概念。

考虑以下代码:

$a = "a"; $b = "b";

$r = &$a;
var_dump($a, $b, $r); # a, b, a

$r = &$b;
var_dump($a, $b, $r); # a, b, b

容易理解的是,但重要的是语句 $r = &$b; 意味着将 $b引用复制给 $r,因此 $b$r 都指向相同的值

接下来如果你执行:

$r = $a;
var_dump($a, $b, $r); # a, a, a

语句 $r = $a; 意味着将 $a复制给 $r,所以 $r 的值从 "b" 变为 "a"。由于 $b$r 指向相同的值$b 的值也变为 "a"。

最后,如果你执行:

$r = "r";
var_dump($a, $b, $r); # a, a, r

仍然只有 $b 的值被改变为 $r$a 保持其原始值。


回到你的问题,你的第一个函数几乎等同于:

function test_global_ref(&$r) {
    $b = "b";
    $r = &$b;
}

$a = "a";
test_global_ref($a);
英文:

This is not about global or static, this is about the concept of reference.

Think about the following codes:

$a = &quot;a&quot;; $b = &quot;b&quot;;

$r = &amp;$a;
var_dump($a, $b, $r); # a, b, a

$r = &amp;$b;
var_dump($a, $b, $r); # a, b, b

It's easy to understand, but the important thing is the statement $r = &amp;$b; means copy the reference of $b to $r, so both $b and $r refer to the same value.

Next if you do:

$r = $a;
var_dump($a, $b, $r); # a, a, a

The statement $r = $a; means copy the value of $a to $r, so the value of $r changes from "b" to "a". Since both $b and $r refer to the same value, the value of $b also becomes "a".

Finally if you do:

$r = &quot;r&quot;;
var_dump($a, $b, $r); # a, r, r

Still only the value of $b to $r is changed, $a keeps its original value.


Back to your question, your first function is almost equivalent to:

function test_global_ref(&amp;$r) {
    $b = &quot;b&quot;;
    $r = &amp;$b;
}

$a = &quot;a&quot;;
test_global_ref($a);

I changed the variable names and values to those corresponding to the above example, hope this is easier to understand. So the global variable $a is passed to the function as a reference $r, when you copy the reference of $b to $r, the global variable $a won't be influenced.

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  • 本文由 发表于 2023年2月8日 16:27:27
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