Ooops.. I made the same mistake in my private reply.. hah.. = = = Original message = = = I took me a little while to notice that it wasn't a function declaration :p ~It's just as Jay said, it's calling "function" with those arguments, where '!$var2' extends to '!(bool)$var2' or "treat/cast this value as/to boolean, and pass it's inverted value" or "NOT $var2". Jay Blanchard wrote: > [snip] > I've got some code from someone else I've inherited and need to sort out > some problems with. The programmer that wrote it originally was much > better than I and programmed a little over my head to say the least. > > One function that I've come across that has 5 variables as input: > > function($var1,$var2,!$var2,$var3->cc,$var3->bcc); > > The question I have is on the 3rd input variable, what does the "!" in > front of $var2 do to that variable? > [/snip] > > Hmmm. Looks like a weird mistake, to be sure. If $var2 is boolean it should > pass the opposite of its current state. Is $var2 a boolean? It is being > passed twice here, once AS and once AS NOT. -- Atentamente, J. Rafael Salazar Maga~a Innox - Innovaci~n Inteligente Tel: +52 (33) 3615 5348 ext. 205 / 01 800 2-SOFTWARE rsalazar@xxxxxxxxxxxx http://www.innox.com.mx -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php ___________________________________________________________ Sent by ePrompter, the premier email notification software. Free download at http://www.ePrompter.com. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
