Re: [PHP4BETA] Using $GLOBALS array in an object

$var = new Object; is equivalent to $var = new Object(); On Fri, 21 Apr 2000, Bob Silva wrote: > Is the constructor behavior different also? Meaning if I add a constructor > to all my classes, will I also need to modify all my current code thats like > this: $var = new Object; to $var = new Object(); Or is it implied? > > Thanks for the fast answer Zeev. > > Salty`` > > ----- Original Message ----- > From: "Zeev Suraski" <zeev@zend.com> > To: "Bob Silva" <bsilva@umesd.k12.or.us> > Cc: <php4beta@lists.php.net> > Sent: Friday, April 21, 2000 2:11 AM > Subject: Re: [PHP4BETA] Using $GLOBALS array in an object > > > > On Fri, 21 Apr 2000, Bob Silva wrote: > > > > > In one master include file at the top outside any objects I have: > > > > > > $CACHE_ROOT = "/cache"; > > > > > > then inside an object later in the same page I have: > > > > > > Line 370: var $CACHE_DIR = $GLOBALS["CACHE_ROOT"]; > > > > > > This is giving a parse error: > > > Parse error: parse error in libBravenet.php on line 370 > > > > > > I have a very similar line where I am setting a class var to a GLOBALS > > > variable and it dies there as well. > > > This code has worked fine in PHP3 for ages, so I cant figure out whats > going > > > on. > > > > > > I do have the optimizer installed, but even disabling that didnt help. > > > Any ideas? > > > > The optimizer doesn't affect the parsing engine, or the behavior of PHP in > > general, except for making it run faster. > > > > As of PHP 4.0, initialization values (for static members, class members or > > default function arguments) have to be static (i.e. numbers, static > > strings or static array()'s). In order to initialize your cache dir, you > > must move the initialization to the class's constructor. > > > > Zeev > > > > -- > > Zeev Suraski <zeev@zend.com> > > http://www.zend.com/ > > > > > -- Zeev Suraski <zeev@zend.com> http://www.zend.com/