Re: 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 & 0x7FFF0000;
       $a2 = $a & 0x0000FFFF;
       $a3 = $a & 0x80000000;
       $b1 = $b & 0x7FFF0000;
       $b2 = $b & 0x0000FFFF;
       $b3 = $b & 0x80000000;

       $c = ($a3 != $b3) ? 0x80000000 : 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 :).