ScotlandPHP

What References Are

References in PHP are a means to access the same variable content by different names. They are not like C pointers; for instance, you cannot perform pointer arithmetic using them, they are not actual memory addresses, and so on. See What References Are Not for more information. Instead, they are symbol table aliases. Note that in PHP, variable name and variable content are different, so the same content can have different names. The closest analogy is with Unix filenames and files - variable names are directory entries, while variable content is the file itself. References can be likened to hardlinking in Unix filesystem.

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User Contributed Notes 3 notes

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273118949 at qq dot com
17 days ago
it just likes a person who has two different names.
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anon
9 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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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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