Try a bitwise and (I think the operator is '&')... <?php $bitMask = 16; $value1 = 24; $value2 = 36;
?>$value1 has bit 5 set? <?php if($bitMask & $value1) { //should get here echo 'true\r\n'; } else { echo 'false\r\n'; } ?>$value2 has bit 5 set? <?php if($bitMask & $value1) { echo 'true\r\n'; } else { //should get here echo 'false\r\n'; } ?> - Theo -----Original Message----- From: Miguel Cruz [mailto:[EMAIL PROTECTED]] Sent: Sunday, April 07, 2002 5:04 PM To: Charlie Killian Cc: [EMAIL PROTECTED] Subject: Re: [PHP] Test for one bit set? On Sun, 7 Apr 2002, Charlie Killian wrote: > How can I test if a number only has on bit set? > > So testing different numbers will return TRUE or FALSE: > > testing 00010000 would return TRUE. > testing 00000011 would return FALSE. Think back to math class when you were 14! function isOneBitSet($n) { $x = log($n)/log(2); return ($x == intval($x)); } Cheesy alternative: function isOneBitSet($n) { return (1 == substr_count(base_convert($n, 10, 2), '1')); } miguel -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php