Re: [PHP] Bitwise operators and check if an bit is NOT within the flag.
On Tue, July 4, 2006 7:35 am, Mathijs wrote: //Do if VALIDATE_CHECK1 is set BUT NOT when VALIDATE_CHECK3 is set. if ($flag2 self::VALIDATE_CHECK1 $flag2 ~self::VALIDATE_CHECK3) Did you check operator precedence for versus ? Perhaps you just need parentheses... I'm also not at all sure the ~self::VALIDATE_CHECK3 is doing what you want... Echo that out and see if it's the number you would expect... -- Like Music? http://l-i-e.com/artists.htm -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Bitwise operators and check if an bit is NOT within the flag.
Mathijs wrote:
Hello there.
I am working with some bitwise Operators for validating some variables.
Now i need to know if an certain bit is NOT set and an other bit IS set.
Example.
?php
const VALIDATE_CHECK1 = 1;
const VALIDATE_CHECK2 = 2;
const VALIDATE_CHECK3 = 4;
const VALIDATE_ALL= 7;
//--Example 1 - This works nice.
$flag1 = self::VALIDATE_CHECK1;
//Do if VALIDATE_CHECK1 is set
if ($flag1 self::VALIDATE_CHECK1) {
print 'Validate 1';
}
//Do if VALIDATE_CHECK2 is set
if ($flag1 self::VALIDATE_CHECK2) {
print 'Validate 2';
}
//--Example 2 - I want to check if VALIDATE_CHECK3 is not set and then
continue.
$flag2 = self::VALIDATE_ALL;
//Do if VALIDATE_CHECK1 is set BUT NOT when VALIDATE_CHECK3 is set.
if ($flag2 self::VALIDATE_CHECK1 $flag2 ~self::VALIDATE_CHECK3) {
class Test {
const VALIDATE_CHECK1 = 1;
const VALIDATE_CHECK2 = 2;
const VALIDATE_CHECK3 = 4;
const VALIDATE_ALL= 7;
static function check($flag2)
{
if (($flag2 self::VALIDATE_CHECK1)
!($flag2 self::VALIDATE_CHECK3)) print Only Validate 1;
}
}
echo First attempt: ;
Test::check( Test::VALIDATE_ALL );
echo \nSecond attempt: ;
Test::check( Test::VALIDATE_CHECK1 );
echo \n;
print 'Only Validate 1';
}
//etc...
?
This last example i can't seem to get to work.
I Want to only do that when for example VALIDATE_CHECK3 is not within
the $flag2.
How can i do this?
Thx in advanced.
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Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Jochem Maas wrote:
Mathijs wrote:
Hello there.
I am working with some bitwise Operators for validating some variables.
Now i need to know if an certain bit is NOT set and an other bit IS set.
Example.
?php
const VALIDATE_CHECK1 = 1;
const VALIDATE_CHECK2 = 2;
const VALIDATE_CHECK3 = 4;
const VALIDATE_ALL= 7;
//--Example 1 - This works nice.
$flag1 = self::VALIDATE_CHECK1;
//Do if VALIDATE_CHECK1 is set
if ($flag1 self::VALIDATE_CHECK1) {
print 'Validate 1';
}
//Do if VALIDATE_CHECK2 is set
if ($flag1 self::VALIDATE_CHECK2) {
print 'Validate 2';
}
//--Example 2 - I want to check if VALIDATE_CHECK3 is not set and then
continue.
$flag2 = self::VALIDATE_ALL;
//Do if VALIDATE_CHECK1 is set BUT NOT when VALIDATE_CHECK3 is set.
if ($flag2 self::VALIDATE_CHECK1 $flag2 ~self::VALIDATE_CHECK3) {
class Test {
const VALIDATE_CHECK1 = 1;
const VALIDATE_CHECK2 = 2;
const VALIDATE_CHECK3 = 4;
const VALIDATE_ALL= 7;
static function check($flag2)
{
if (($flag2 self::VALIDATE_CHECK1)
!($flag2 self::VALIDATE_CHECK3)) print Only Validate 1;
}
}
echo First attempt: ;
Test::check( Test::VALIDATE_ALL );
echo \nSecond attempt: ;
Test::check( Test::VALIDATE_CHECK1 );
echo \n;
print 'Only Validate 1';
}
//etc...
?
This last example i can't seem to get to work.
I Want to only do that when for example VALIDATE_CHECK3 is not within
the $flag2.
How can i do this?
Thx in advanced.
Thank you very much.
This seems to work :).
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Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Mathijs wrote:
Jochem Maas wrote:
Mathijs wrote:
...
Thank you very much.
This seems to work :).
cool. heres's a couple of funcs that might help you to understand bitwise
operations better:
?php
/* whether there is only 1 single bit set or not */
function single_bit_set(/*int*/ $i)
{
// beware: if $i is zero !($i ~get_ls1bit($i)) returns true;
return $i ? !($i ~get_ls1bit($i)): false;
}
/* return the value of the least significant bit */
function get_ls1bit(/*int*/ $i)
{
for ($j = 1; $i !($i $j); $j = 1);
return $i ? $j : 0;
}
/* return the value of the most significant bit */
function get_ms1bit(/*int*/ $i)
{
$x = 0;
for ($j = $i; $i !($j == 1); $j = 1) { $x++; }
return $i ? $j = $x: 0;
}
/* doesn't break when exponents are near the wordsize
* of the machine as the native xor does, here is some example code to
* illustrate:
php -r '
// include function definition here
$x = 3851235679;
$y = 43814;
echo \nThis is the value we want;
echo \n3851262585;
echo \nThe result of a native xor operation on integer values is treated
as a signed integer;
echo \n.($x ^ $y);
echo \nWe therefore perform the MSB separately;
echo \n.get_xor32($x, $y).\n;
'
*/
function get_xor32(/*int*/ $a, /*int*/ $b)
{
$a1 = $a 0x7FFF;
$a2 = $a 0x;
$a3 = $a 0x8000;
$b1 = $b 0x7FFF;
$b2 = $b 0x;
$b3 = $b 0x8000;
$c = ($a3 != $b3) ? 0x8000 : 0;
return (($a1 ^ $b1) |($a2 ^ $b2)) + $c;
}
function get_bit_str($var, $safety = 0)
{
$rtn = '';
$var = intval($var);
if ($var 0) { $var = 0 - $var; }
while ($var != 0 /* $safety 31*/) {
$rtn .= ($var 1);
$var = 1;
$safety++;
}
return $rtn;
}
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Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Jochem Maas wrote:
Mathijs wrote:
Jochem Maas wrote:
Mathijs wrote:
...
Thank you very much.
This seems to work :).
cool. heres's a couple of funcs that might help you to understand bitwise
operations better:
?php
/* whether there is only 1 single bit set or not */
function single_bit_set(/*int*/ $i)
{
// beware: if $i is zero !($i ~get_ls1bit($i)) returns true;
return $i ? !($i ~get_ls1bit($i)): false;
}
/* return the value of the least significant bit */
function get_ls1bit(/*int*/ $i)
{
for ($j = 1; $i !($i $j); $j = 1);
return $i ? $j : 0;
}
/* return the value of the most significant bit */
function get_ms1bit(/*int*/ $i)
{
$x = 0;
for ($j = $i; $i !($j == 1); $j = 1) { $x++; }
return $i ? $j = $x: 0;
}
/* doesn't break when exponents are near the wordsize
* of the machine as the native xor does, here is some example code to
* illustrate:
php -r '
// include function definition here
$x = 3851235679;
$y = 43814;
echo \nThis is the value we want;
echo \n3851262585;
echo \nThe result of a native xor operation on integer values is treated as a
signed integer;
echo \n.($x ^ $y);
echo \nWe therefore perform the MSB separately;
echo \n.get_xor32($x, $y).\n;
'
*/
function get_xor32(/*int*/ $a, /*int*/ $b)
{
$a1 = $a 0x7FFF;
$a2 = $a 0x;
$a3 = $a 0x8000;
$b1 = $b 0x7FFF;
$b2 = $b 0x;
$b3 = $b 0x8000;
$c = ($a3 != $b3) ? 0x8000 : 0;
return (($a1 ^ $b1) |($a2 ^ $b2)) + $c;
}
function get_bit_str($var, $safety = 0)
{
$rtn = '';
$var = intval($var);
if ($var 0) { $var = 0 - $var; }
while ($var != 0 /* $safety 31*/) {
$rtn .= ($var 1);
$var = 1;
$safety++;
}
return $rtn;
}
Thx alot :).
It will sure help me understanding it better :).
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Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Jochem Maas wrote:
Mathijs wrote:
Jochem Maas wrote:
Mathijs wrote:
...
Thank you very much.
This seems to work :).
cool. heres's a couple of funcs that might help you to understand bitwise
operations better:
?php
/* whether there is only 1 single bit set or not */
function single_bit_set(/*int*/ $i)
{
// beware: if $i is zero !($i ~get_ls1bit($i)) returns true;
return $i ? !($i ~get_ls1bit($i)): false;
}
/* return the value of the least significant bit */
function get_ls1bit(/*int*/ $i)
{
for ($j = 1; $i !($i $j); $j = 1);
return $i ? $j : 0;
}
/* return the value of the most significant bit */
function get_ms1bit(/*int*/ $i)
{
$x = 0;
for ($j = $i; $i !($j == 1); $j = 1) { $x++; }
return $i ? $j = $x: 0;
}
/* doesn't break when exponents are near the wordsize
* of the machine as the native xor does, here is some example code to
* illustrate:
php -r '
// include function definition here
$x = 3851235679;
$y = 43814;
echo \nThis is the value we want;
echo \n3851262585;
echo \nThe result of a native xor operation on integer values is treated as a
signed integer;
echo \n.($x ^ $y);
echo \nWe therefore perform the MSB separately;
echo \n.get_xor32($x, $y).\n;
'
*/
function get_xor32(/*int*/ $a, /*int*/ $b)
{
$a1 = $a 0x7FFF;
$a2 = $a 0x;
$a3 = $a 0x8000;
$b1 = $b 0x7FFF;
$b2 = $b 0x;
$b3 = $b 0x8000;
$c = ($a3 != $b3) ? 0x8000 : 0;
return (($a1 ^ $b1) |($a2 ^ $b2)) + $c;
}
function get_bit_str($var, $safety = 0)
{
$rtn = '';
$var = intval($var);
if ($var 0) { $var = 0 - $var; }
while ($var != 0 /* $safety 31*/) {
$rtn .= ($var 1);
$var = 1;
$safety++;
}
return $rtn;
}
Thx alot :).
It will sure help me understanding it better :).
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Jochem Maas wrote:
Mathijs wrote:
Jochem Maas wrote:
Mathijs wrote:
...
Thank you very much.
This seems to work :).
cool. heres's a couple of funcs that might help you to understand bitwise
operations better:
?php
/* whether there is only 1 single bit set or not */
function single_bit_set(/*int*/ $i)
{
// beware: if $i is zero !($i ~get_ls1bit($i)) returns true;
return $i ? !($i ~get_ls1bit($i)): false;
}
/* return the value of the least significant bit */
function get_ls1bit(/*int*/ $i)
{
for ($j = 1; $i !($i $j); $j = 1);
return $i ? $j : 0;
}
/* return the value of the most significant bit */
function get_ms1bit(/*int*/ $i)
{
$x = 0;
for ($j = $i; $i !($j == 1); $j = 1) { $x++; }
return $i ? $j = $x: 0;
}
/* doesn't break when exponents are near the wordsize
* of the machine as the native xor does, here is some example code to
* illustrate:
php -r '
// include function definition here
$x = 3851235679;
$y = 43814;
echo \nThis is the value we want;
echo \n3851262585;
echo \nThe result of a native xor operation on integer values is treated as a
signed integer;
echo \n.($x ^ $y);
echo \nWe therefore perform the MSB separately;
echo \n.get_xor32($x, $y).\n;
'
*/
function get_xor32(/*int*/ $a, /*int*/ $b)
{
$a1 = $a 0x7FFF;
$a2 = $a 0x;
$a3 = $a 0x8000;
$b1 = $b 0x7FFF;
$b2 = $b 0x;
$b3 = $b 0x8000;
$c = ($a3 != $b3) ? 0x8000 : 0;
return (($a1 ^ $b1) |($a2 ^ $b2)) + $c;
}
function get_bit_str($var, $safety = 0)
{
$rtn = '';
$var = intval($var);
if ($var 0) { $var = 0 - $var; }
while ($var != 0 /* $safety 31*/) {
$rtn .= ($var 1);
$var = 1;
$safety++;
}
return $rtn;
}
Thx alot :).
It will sure help me understanding it better :).
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Bitwise operators and check if an bit is NOT within theflag.
Jochem Maas wrote:
Mathijs wrote:
Jochem Maas wrote:
Mathijs wrote:
...
Thank you very much.
This seems to work :).
cool. heres's a couple of funcs that might help you to understand bitwise
operations better:
?php
/* whether there is only 1 single bit set or not */
function single_bit_set(/*int*/ $i)
{
// beware: if $i is zero !($i ~get_ls1bit($i)) returns true;
return $i ? !($i ~get_ls1bit($i)): false;
}
/* return the value of the least significant bit */
function get_ls1bit(/*int*/ $i)
{
for ($j = 1; $i !($i $j); $j = 1);
return $i ? $j : 0;
}
/* return the value of the most significant bit */
function get_ms1bit(/*int*/ $i)
{
$x = 0;
for ($j = $i; $i !($j == 1); $j = 1) { $x++; }
return $i ? $j = $x: 0;
}
/* doesn't break when exponents are near the wordsize
* of the machine as the native xor does, here is some example code to
* illustrate:
php -r '
// include function definition here
$x = 3851235679;
$y = 43814;
echo \nThis is the value we want;
echo \n3851262585;
echo \nThe result of a native xor operation on integer values is treated as a
signed integer;
echo \n.($x ^ $y);
echo \nWe therefore perform the MSB separately;
echo \n.get_xor32($x, $y).\n;
'
*/
function get_xor32(/*int*/ $a, /*int*/ $b)
{
$a1 = $a 0x7FFF;
$a2 = $a 0x;
$a3 = $a 0x8000;
$b1 = $b 0x7FFF;
$b2 = $b 0x;
$b3 = $b 0x8000;
$c = ($a3 != $b3) ? 0x8000 : 0;
return (($a1 ^ $b1) |($a2 ^ $b2)) + $c;
}
function get_bit_str($var, $safety = 0)
{
$rtn = '';
$var = intval($var);
if ($var 0) { $var = 0 - $var; }
while ($var != 0 /* $safety 31*/) {
$rtn .= ($var 1);
$var = 1;
$safety++;
}
return $rtn;
}
Thx alot :).
It will sure help me understanding it better :).
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RE: [PHP] Bitwise operators
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] on Monday, September 26, 2005 9:18 AM said: So i ask what this output? $a = 4; $b = 3; echo $a $b; echo $a $b; You just spent 3-5 minutes writing an email and now almost 10 minutes waiting for a reply to something that would have taken you 2 minutes to test on your own. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Bitwise operators
On 9/26/05, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: From php manual: $a $b Shift leftShift the bits of $a $b steps to the left (each step means multiply by two) $a $b Shift rightShift the bits of $a $b steps to the right (each step means divide by two) So i ask what this output? $a = 4; $b = 3; echo $a $b; 32 echo $a $b; 0 So your program will output '320'. What was the problem again? -robin -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Bitwise operators
I tested; I don't want to waste peoples time. Rewriting the question: $a = 4; $b = 3; $c = $a $b; $d = $a $b; echo c = . $c . br; echo d = . $d . br; this outputs: c = 32 d = 0 The question is why? Angelo - Original Message - From: Chris W. Parker [EMAIL PROTECTED] To: php-general@lists.php.net Sent: Monday, September 26, 2005 1:28 PM Subject: RE: [PHP] Bitwise operators [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] on Monday, September 26, 2005 9:18 AM said: So i ask what this output? $a = 4; $b = 3; echo $a $b; echo $a $b; You just spent 3-5 minutes writing an email and now almost 10 minutes waiting for a reply to something that would have taken you 2 minutes to test on your own. -- 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
Re: [PHP] 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
Re: [PHP] Bitwise operators
The second value is the number of spaces to shift, dint realize that. Thanks for your time Chris. Angelo - Original Message - From: Chris Boget [EMAIL PROTECTED] To: [EMAIL PROTECTED]; php-general@lists.php.net Sent: Monday, September 26, 2005 2:02 PM Subject: Re: [PHP] 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
