Re: I'm confused about..reffering vars &$

It's not a backwards assignment. It's more like this: $bar = &$foo; You now have essentially two variable names 'foo' and 'bar' pointing to the same chunk of memory inside the computer. If you change one, you change the other: They are really not two distinct variables -- They are the same variables, and are aliases for each other. -- Visit the Zend Store at http://www.zend.com/store/ Wanna help me out? Like Music? Buy a CD: http://l-i-e.com/artists.htm Volunteer a little time: http://chatmusic.com/volunteer.htm ----- Original Message ----- From: Noel Akins <nakins@arkansas.net> Newsgroups: php.general Sent: Friday, January 26, 2001 10:52 AM Subject: [PHP] I'm confused about..reffering vars &$ > Ok, I'm reading the manual for the new php4 stuff. Now, I'm not new to php. > I started with php/fi with a little script to add links to a database. I > did a little work in php3. I just don't make enough php stuff to get a frim > grip on everything. That is changing as I am building a second pc to serve > as a dev box for php4. Here is my first question. > > This was lifted from the Zend site from the php manual there. > > To assign by reference, simply prepend an ampersand (&) to the beginning of > the variable which is being assigned (the source variable). For instance, > the following code snippet outputs 'My name is Bob' twice: > <?php > $foo = 'Bob'; // Assign the value 'Bob' to $foo > $bar = &$foo; // Reference $foo via $bar. > $bar = "My name is $bar"; // Alter $bar... > echo $foo; // $foo is altered too. > echo $bar; > ?> > > What I don't see here is how this prints (echo) "My name is Bob". The > way I see this is that $bar = &$foo; is a backwards assignment, so that > when the script cycles through, it changes $foo from Bob to "My name is > $bar", and, to my error prone thinking, it should print out "My name is My > name is My name is...". > > It would make sense to me if it were: > > <?php > $foo = 'Ed'; // Assign the value 'Ed' to $foo > $bar = &$foo; // Reference $foo via $bar > $bar = 'Bob'; // changing $foo to 'Bob' > $this = "My name is $bar"; > $that = "My name is $foo"; > echo $this; / / prints out My name is Bob > echo $that; / / prints out My name is Bob > Maybe I'm not seeing how php rereads or cycles through the script once it > sees a "assign by reference" Could someone explain please? > Thank you. > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >