Bug #18121 Updated: Ternary Conditional Operator does not allow references as values.

ID: 18121 Updated by: sniper@php.net Reported By: russ@zerotech.net -Status: Open +Status: Verified -Bug Type: Variables related +Bug Type: Scripting Engine problem Operating System: All Platforms PHP Version: 4.2.1 Previous Comments: ------------------------------------------------------------------------ [2002-07-02 18:41:12] russ@zerotech.net Additional Note: This bug was tested on PHP 4.2.1 AND PHP CVS as of Today, July 2nd, 2002. Using References inside a Ternary Conditional Operator generates a general parse error. Consider the following example that tests usage with both Superglobals and regular variables: <?PHP error_reporting(E_ALL); /* This demonstrates a bug in the ternary conditional operator. It does not seem to allow references to be values. */ // Tested Using Superglobals // ---- // This works $request = ($_SERVER['REQUEST_METHOD'] == 'POST') ? $_POST : $_GET; // This however does not. $request = ($_SERVER['REQUEST_METHOD'] == 'POST') ? &$_POST : &$_GET; // Tested Using Regular Variables // ---- $foo = "Foo"; $bar = "Bar"; // This works $foobar = 1 ? $foo : $bar; // This however does not. $foobar = 1 ? &$foo : &$bar; ?> When this is run, it generates a parse error similar this this .. Parse error: parse error in /home/russ/repository/qaweb/test.php on line 19 .. on the lines in which references are used in the ternary conditional operator. Since a reference is considered a type, and can be in an assignment operator, such as $var = &$ref, there is no reason it shouldn't work inside the ternary conditional operator which is basically an assignment operator with control. ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=18121&edit=1