Bug #54740 [Com]: Ternary operator will not work with return by reference

Edit report at https://bugs.php.net/bug.php?id=54740&edit=1 ID: 54740 Comment by: marrch dot caat at gmail dot com Reported by: dukeofgaming at gmail dot com Summary: Ternary operator will not work with return by reference Status: Not a bug Type: Bug Package: Scripting Engine problem PHP Version: Irrelevant Block user comment: N Private report: N New Comment: Thank you, Nikic! You've removed scales from my eyes. To my great pity and shame, I didn't understand that through all my working experience and really though & is a "take reference" operator, as one as exists in C/C++ etc :( Thank you once again for your explanation! Now I see this is not really a bug... Previous Comments: ------------------------------------------------------------------------ [2013-10-02 11:20:20] nikic@php.net @marrch: PHP has no concept of a general & operator that takes the "reference" of a variable(whatever that's supposed to be). PHP only has a =& operator (which is really "=&" and not a combination of "=" and "&") which expects a variable on the right hand side. PHP also has a & modifier for function arguments and return values. If you want to do yourself and others a favor, write $foo =& $bar rather than $foo = &$bar to make sure that there are no misunderstandings regarding this ;) ------------------------------------------------------------------------ [2013-10-02 11:12:00] marrch dot caat at gmail dot com Mike, I understand that. The second note tells I caanot return a reference to an expression result, such as &$object->method() or &(new StdClass()) - I can understand that. But the code sample I provided doesn't try to do that. To make things even simplier, the following code still fails to compile: $link = $flag ? &$a : &$b; It doesn't try to return a reference to an expression, just a reference to a viriable; It doesn't try doing anything that the following code doesn't: if ($flag) $link = &$a; else $link = &$b; And maybi I'm really stupid, but after 10 years in PHP development I still don't understand why the first code cannot be compiled :( ------------------------------------------------------------------------ [2013-10-02 05:27:05] mike@php.net I meant the documentation "Note:" (warning) not the user-contributed note. ------------------------------------------------------------------------ [2013-10-01 20:35:33] marrch dot caat at gmail dot com I thoroughly read the article you mentioned, Mike, but still don't understand why the following code fails to compile: $link = isset($i) ? (& $arr[$i]) : null; - while the following works fine: $link = &$arr[$i]; In this case, &$arr[$i] is a legal reference assignment, so the first code should behave equal to if (isset($i)) { $link = &$arr[$i]; } else { $link = null; } - but this code works fine, and mentioned above isn't even compiled. What's wrong with it? ------------------------------------------------------------------------ [2013-10-01 14:42:02] mike@php.net Thank you for taking the time to write to us, but this is not a bug. Please double-check the documentation available at http://www.php.net/manual/ and the instructions on how to report a bug at http://bugs.php.net/how-to-report.php Check the second note here: http://php.net/manual/en/language.references.return.php ------------------------------------------------------------------------ The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online at https://bugs.php.net/bug.php?id=54740 -- Edit this bug report at https://bugs.php.net/bug.php?id=54740&edit=1