note 32222 deleted from language.references.pass by sniper

Note Submitter: ---- I believe this is covered in the previous section: http://www.php.net/manual/en/language.references.arent.php. The "$dest =& $src" command is re-aliasing the replace() function's private $dest variable, which has no effect on the variable it was originally aliased to. There is no way for a function to re-alias a variable passed in as its parameter (no clean way, anyway.) Just using "=" is the correct syntax here, if your goal is to copy the global $src onto the global $dest. Remember you're not really "passing in a reference", but "passing by reference". A variable cannot "contain" a reference in PHP. References are first-class (assignable) entities in Perl, just like pointers are in C, but not so with references in PHP.