PHP 32位.如何比较uint64的字符串和二进制表示

Question

先决条件

PHP 5.3.6 32位(不可能移动到64位(。

需要比较两个值uint64(8字节无符号整数(。其中一个是字符串，另一个是二进制字符串

问题

在PHP 32位上，是否可以将uint64的字符串表示转换为8字节的array，或者使用uint64将8字节的arraytring？

插图

我尝试了base_convert函数来比较基于2的字符串表示，得到了奇怪的结果。我知道字节数组包含与相应字符串相同的uint64。但我不知道如何确保它们代表相同的数字。

这是带有一些实际值的测试代码来说明问题：

function byte_to_base2string($byte)
{
    $byte = base_convert($byte, 10, 2);
    $byte = str_pad($byte, 8, '0', STR_PAD_LEFT);
    return $byte;
}
function print_info($base10_string1, $bin_string2)
{
    $bin_string1 = null; // TODO: how to obtain it?
    $base2_string1 = base_convert($base10_string1, 10, 2);
    $base2_string1 = str_pad($base2_string1, 64, '0', STR_PAD_LEFT);
    $base2_string2 = array_map('byte_to_base2string', $bin_string2);
    $base2_string2 = implode('', $base2_string2);
    $base10_string2 = base_convert($base2_string2, 2, 10);
    echo sprintf("Wrong base-2 string:n%st%sn", $base10_string1, $base2_string1);
    echo sprintf("base-2 string matches $base10_string1, but base-10 string does notn%st%sn", $base10_string2, $base2_string2);
    echo "n";
    // Can't compare because:
    // $base2_string1 != $base2_string2
    // $base10_string1 != $base10_string2
    // $bin_string1 no idea how to convert
}
$strings = [
    '288512493108985552',
    '288512958990381002',
    '288512564016815754'
];
// obtained via unpack('C*', $binaryStr)
$bytes = [
    [4, 1, 0, 149, 121, 5, 254, 208],
    [4, 1, 1, 1, 241, 183, 239, 202],
    [4, 1, 0, 165, 251, 117, 158, 138]
];
array_map('print_info', $strings, $bytes);

输出为：

Wrong base-2 string:
288512493108985552  0000010000000001000000001001010101111001000001011111111011000000
base-2 string matches 288512493108985552, but base-10 string does not
288512493108985526  0000010000000001000000001001010101111001000001011111111011010000
Wrong base-2 string:
288512958990381002  0000010000000001000000010000000111110001101101111110111111000000
base-2 string matches 288512958990381002, but base-10 string does not
288512958990381002  0000010000000001000000010000000111110001101101111110111111001010
Wrong base-2 string:
288512564016815754  0000010000000001000000001010010111111011011101011001111010000000
base-2 string matches 288512564016815754, but base-10 string does not
288512564016815764  0000010000000001000000001010010111111011011101011001111010001010

已更新

找到了一个解决方案(见下面的答案(，但不确定这是否是最好的方法。仍然希望找到更清晰、更直接的东西。

Answer 1

好吧，感谢PHP手册注释方面的一位好人。此注释中的函数完成任务。我找不到路，只是因为我忘了数学。

因此，目前我的工作答案是(使用评论中的convBase()(：

$id1 = "..."; // string representation of uint64 value
$id2 = [...]; // array of bytes
// convert to base-2 string
$id1 = convBase($id1, '0123456789', '01');
// convert each byte to base-2 string (8 chars) and join them
$id2 = implode('', array_map(function($b) {
    $b = convBase($b, '0123456789', '01');
    $b = str_pad($b, 8, '0', STR_PAD_LEFT);
    return $b;
}, $id2));
// pad with leading zeroes to equal length
$len = max(strlen($id1), strlen($id2));
$id1 = str_pad($id1, $len, '0', STR_PAD_LEFT);
$id2 = str_pad($id2, $len, '0', STR_PAD_LEFT);
// Now its OK!
$id1 === $id2;
convBase($id1, '01', '0123456789') === convBase($id2, '01', '0123456789');

我将

在此处放置复制粘贴函数源代码。以防万一；(

function convBase($numberInput, $fromBaseInput, $toBaseInput)
{
    if ($fromBaseInput == $toBaseInput) return $numberInput;
    $fromBase = str_split($fromBaseInput, 1);
    $toBase = str_split($toBaseInput, 1);
    $number = str_split($numberInput, 1);
    $fromLen = strlen($fromBaseInput);
    $toLen = strlen($toBaseInput);
    $numberLen = strlen($numberInput);
    $retval = '';
    if ($toBaseInput == '0123456789') {
        $retval = 0;
        for ($i = 1; $i <= $numberLen; $i++)
            $retval = bcadd($retval, bcmul(array_search($number[$i - 1],  fromBase), bcpow($fromLen, $numberLen - $i)));
        return $retval;
    }
    if ($fromBaseInput != '0123456789')
        $base10 = convBase($numberInput, $fromBaseInput, '0123456789');
    else
        $base10 = $numberInput;
    if ($base10 < strlen($toBaseInput))
        return $toBase[$base10];
    while ($base10 != '0') {
        $retval = $toBase[bcmod($base10, $toLen)] . $retval;
        $base10 = bcdiv($base10, $toLen, 0);
    }
    return $retval;
}

PS：很抱歉我的代码中可能有拼写错误或错误，我写得很快，没有测试，只是为了说明解决方案。

Answer 2

GMP库在您的PHP版本中可用吗？http://php.net/manual/en/intro.gmp.php它可以在3个字符串中完成所有工作：

$a = gmp_init('892348924892894240808924308925', 10);
$b = gmp_init('111111111010101010101011110101010101001010101010101010101011', 2);
var_dump(gmp_compare($a, $b));