Re: syntax error?

At 06:51 PM 6/20/00 +0200, Jonas Jochum wrote:

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Am 20.06.00 um 19:35 hat Andi Gutmans geschrieben: AG>>>$catid = (isset($HTTP_GET_VARS[catid])) ? &$HTTP_GET_VARS["catid"] : 0; AG>& isn't an operator in PHP. AG>It can only be used under very specific circuimstances such as assigning AG>references (= &) function parameter list (&$param). hrm. but that's what the construct should do...And I can do the same in C, eg. int *foo,bar,baz; bar = 5; baz=6; foo = (bar > baz) ? &bar : &baz; printf("%d\n",*foo); Well, anyway, PHP is not C, I guess. However, having the & work in such a conditional as the above would be "a good thing" (tm). ;) Thanks for answering!

As you said, PHP isn't C :) In any case, improving the references support in PHP might happen sometime in the future but don't expect this to happen soon. Right now, I personally think it is more important to keep fixing PHP 4 bugs and extending its support for 3rd party libraries and web servers. Andi Andi --- Andi Gutmans <andi@zend.com> http://www.zend.com/