Re: PHP newbie question

If you don't assign a value for $a why would PHP assign it one? The statement if(!$a) should not assign a value to $a. Thus when print $a is executed, $a still has no value. BTW: 0 is false. Dave M. Online Services, USA ----- Original Message ----- From: "Neil Zanella" <nzanella@cs.mun.ca> To: "Toby Butzon" <php-lists@imawebdesigner.com> Cc: "PHP General Mailing List" <php-general@lists.php.net> Sent: Wednesday, January 10, 2001 10:33 AM Subject: Re: [PHP] PHP newbie question > > On Wed, 10 Jan 2001, Toby Butzon wrote: > > > : <?php if (! $a) print "Hello, World!"; // script 2 ?> > > > > $a evaluates to false, the ! reverses it, and it prints "Hello, World!" > > What is bothering me is the following: if variables that are not assigned > a value were to evaluate to false then since false is the same as the > number 1 the following PHP script should print the number 1 but instead > prints nothing: > > <?php if (! $a) print $a; ?> > > How is this behavior justified? > I could not find anything on this in the PHP manual. > > Thanks, > > -- Neil > > > -- > 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 > >