| bitwise left shift differing results
32-bit vs 64-bit |
Nuttzy99
msg:1274713 | 10:03 am on Jun 28, 2006 (gmt 0) |
My dev environment is WinXP (32 bit) running PHP 4.3.10. Running this simple compand.... echo (192<<24) ; ...yields a value of -1073741824 which is what I want since that's what most browsers running javascript will yield. My production/staging environment is a 64-bit AMD Opteron on FreeBSD running PHP 5.0.4. For the same command it yields a value of 3221225472. I'm guessing it has to do with the 64-bit system. Can anyone tell me how to get my server to match 32-bit systems for this result? I really don't know much about bitwise left shift, but I'm trying to get TEA encryption going. Thanks,
-Nuttzy
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Nuttzy99
msg:1274714 | 12:31 pm on Jun 28, 2006 (gmt 0) |
While writing my post I think I figured it out... define('BIT32_MAX_SIGNED', 2147483648) ; // this is 2 to 31st power (need one bit for sign) //$val = 3221225472 ;
$val = (192<<24) ;
echo $val . "<br/>" ;
if ($val > BIT32_MAX_SIGNED)
{
$val = BIT32_MAX_SIGNED - $val ;
}
echo $val . "<br/>" ; ...basically a 32-bit system won't exceed BIT32_MAX_SIGNED, so if we do, then we know it's a 64-bit system and need to adjust. I'll probably need to do the same thing in my javascript encrypt function in case there are any 64-bit browsers out there. I'm still having problem with my TEA encryption in one case, so I might not have it entirely solved yet. -Nuttzy
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Nuttzy99
msg:1274715 | 1:25 pm on Jun 28, 2006 (gmt 0) |
Blah, when val is.... $val = (192<<24) ; ...it doesn't normalize to the same results as my 32-bit system. I have no idea why. If someone better versed in this stuff has the answer, I'll be happy to learn it. However, for my needs, I think if I just set my javascript to use unsigned shift (<<<) then I won't have a problem on my 64-bit production system. My 32-bit system is just for development and I can work around it. Cookie for anyone that solves this ;-)
-Nuttzy
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Nuttzy99
msg:1274716 | 2:24 pm on Jun 28, 2006 (gmt 0) |
Sorry for all the replies. I made a bad typo in my previous message. It should read... Blah, when val is.... $val = (152<<24) ...makes a BIG difference ;-) -Nuttzy
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rizal_a
msg:3154938 | 9:32 am on Nov 13, 2006 (gmt 0) |
System: The following 2 messages were spliced on to this thread from: http://www.webmasterworld.com/php/3154936.htm [webmasterworld.com] by coopster - 9:30 am on Nov. 14, 2006 (utc -6)
This is regarding Nuttzy99's posts on June 28 2006. The following function will do a 32-bit left shift:
function _32bitleftshift($number, $steps)
{
$result = 0;
if (strlen(decbin($number)) + $steps <= 31)
{
$result = $number << $steps;
}
else
{
// shift left
$binary = decbin($number).str_repeat("0", $steps);
// get the last 32 digits
$binary = substr($binary, strlen($binary) - 32);
// is the leftmost digit 1, i.e., is it a negative number?
if ($binary{0} == "1")
{
// get the last 31 digits
$binary = substr($binary, strlen($binary) - 31);
// get the 1's complement
$binary_1s_complement = "";
for ($i = 0; $i <= 30; $i++)
{
$binary_1s_complement .= ($binary{$i} == "0"? "1" : "0");
}
// get the 2's complement
$binary_2s_complement = decbin(bindec($binary_1s_complement) + 1);
$result = -1 * bindec($binary_2s_complement);
}
else
{
$result = bindec($binary);
}
}
return $result;
}
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rizal_a
msg:3155703 | 11:55 pm on Nov 13, 2006 (gmt 0) |
Whoops ... there's actually a much faster way to find out the 2's complement of any number, which is -(pow(2, n) - N) where n is the number of bits and N is the number for which to find out its 2's complement. Therefore:
function leftshift32($number, $steps)
{
$binary = decbin($number).str_repeat("0", $steps);
$binary = str_pad($binary, 32, "0", STR_PAD_LEFT);
$binary = substr($binary, strlen($binary) - 32);
if ($binary{0} == "1")
{
return -(pow(2, 31) - bindec(substr($binary, 1)));
}
else
{
return bindec($binary);
}
}
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