I appreciate everyone who wrote, but I remain confused. It seems to me that there is a weird short-circuit in the following code:
$b = 1; $b = $b++; print $b; --> 1 I expected "2" Whereas, $b = 1 $b = ++$b; print $b; --> 2 As I expected. I translate the autoincrement ($b++) as $b + 1, but post-increment doesn't seem to work when assigned to itself, whereas pre-increment does. So I was hoping for some confirmation that it is supposed to work that way, and perhaps why... So while Connie & Dave & Chris have answered the question to a degree (and I thank them each), they haven't resolved my confusion. It seems that '$b = $b++;' is always a coding mistake. Nobody should use that syntax on purpose, apparently, because it ignores the 'side effect' auto-increment. Is this true? I'll try to write my questions better in the future. /Michael Turner -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
