note 83734 added to language.references.pass

madmax, the answer is that passing $b = $a does not pass $b. "$b = $a" is an expression, expressions can not be passed by reference (E_NOTICE will tell you that.) I sure think that the compiler should "do what I mean" for foo($b = $a), but it doesn't. ---- Server IP: 69.147.83.197 Probable Submitter: 76.105.247.92 ---- Manual Page -- http://www.php.net/manual/en/language.references.pass.php Edit -- https://master.php.net/note/edit/83734 Del: integrated -- https://master.php.net/note/delete/83734/integrated Del: useless -- https://master.php.net/note/delete/83734/useless Del: bad code -- https://master.php.net/note/delete/83734/bad+code Del: spam -- https://master.php.net/note/delete/83734/spam Del: non-english -- https://master.php.net/note/delete/83734/non-english Del: in docs -- https://master.php.net/note/delete/83734/in+docs Del: other reasons-- https://master.php.net/note/delete/83734 Reject -- https://master.php.net/note/reject/83734 Search -- https://master.php.net/manage/user-notes.php