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 Magaña
Innox - Innovación Inteligente
Tel: +52 (33) 3615 5348 ext. 205 / 01 800 2-SOFTWARE
[EMAIL PROTECTED]
http://www.innox.com.mx
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