Re: Problem with pass-by-ref

foo() should return a reference. Defint it as: function &foo() { }

I did check this too of course, but it does not work. Btw, variable is variable and your check should not (can it?) care if referenced or not.(?) Doing... function foo() { return $foo; } $foo=&foo(); is legal too although it makes no sense... andré -- · André Langhorst · t: +49 571 3201801 · · waldschrott@php.net · m: +49 173 9558736 · · PHP Quality Assurance · http://qa.php.net ·