The second value is the number of spaces to shift, dint realize that. Thanks for your time Chris. Angelo ----- Original Message ----- From: "Chris Boget" <chris.boget@xxxxxxxx> To: <cron@xxxxxxxxxx>; <php-general@xxxxxxxxxxxxx> Sent: Monday, September 26, 2005 2:02 PM Subject: Re: Bitwise operators > >I tested; I don't want to waste peoples time. Rewriting the question: > > this outputs: > > c = 32 > > d = 0 > > The question is why? > > First row is the bit's number and the second row is the bit's value: > > #8 | #7 | #6 | #5 | #4 | #3 | #2 | #1 > --------------------------------------- > 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 > > So your variable $a, which has a value of 4 starts out in the bit position > of #3 with that bit turned on.. Shift it left 3 (value of $b) spaces and > you > end up with bit #6 getting turned on giving you a value of 32, which is what > $c is echoing out. > > Shifting to the right 3 spaces from the same starting position nets you 0 > with > all the bits turned off. At least, I'm pretty sure that's right. > > thnx, > Chris > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
