php[tek] 2018 : Call for Speakers

引用是什么

在 PHP 中引用意味着用不同的名字访问同一个变量内容。这并不像 C 的指针:例如你不能对他们做指针运算,他们并不是实际的内存地址…… 查看引用不是什么了解更多信息。 替代的是,引用是符号表别名。注意在PHP 中,变量名和变量内容是不一样的, 因此同样的内容可以有不同的名字。最接近的比喻是 Unix 的文件名和文件本身——变量名是目录条目,而变量内容则是文件本身。引用可以被看作是 Unix 文件系统中的硬链接。

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User Contributed Notes 3 notes

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6
273118949 at qq dot com
1 month ago
it just likes a person who has two different names.
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-3
anon
10 months ago
In summary, "&$reference" means "do-not-copy-on-write the value here, in perpetuity". Assigning by reference is not assignment, it's "make &$variable a reference and its value do-not-copy-on-write, in perpetuity, and make the variable I'm assigning to use that do-not-copy-on-write value as well".

To "unreference/unalias" you have to either unset or make an explicit copy into a new variable.

Object properties that are references will survive cloning and remain references. Generally the same is true with references in arrays and PHP's array functions (combine, intersect, call_user_func, func_get_args, etc).

Calling a function that uses a reference parameter will *make* the supplied variable a reference. This is also true when using variadic array expansion for arguments; the supplier's array element will become a reference.

Generally, don't use them unless you're dealing with low-level calls, or need an accumulator, etc. For poorly designed functions that use them, give them a copy to mangle.
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-33
stilezy
1 year ago
One subtle effect of PHP's assign-by-reference is that operators which might be expected to work with args that are references usually don't.  For example:

$a = ($b ? &$c : &$d);

fails (parser error) but the logically identical

if ($b)
   $a =& $c;
else
   $a =& $cd;

works. It's not always obvious why seemingly identical code throws an error in the first case. This is discussed on a PHP bug report ( https://bugs.php.net/bug.php?id=54740 ). TL;DR version, it acts more like an assignment term ($var1) "=&" ($var2) than a function/operator ($var1) "=" (&$var2).
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