来自2种不同语言的哈希函数没有给出相同的结果

Question

I have this function in the scripting language "PAWN":

stock hash(Hash[])
{
    new
        hStr1 = 1, hStr2 = 0;

    for(new i = 0, l = strlen(Hash); i < l; i ++)
    {
        hStr1 = (hStr1 + Hash[i]) % 65521;
        hStr2 = (hStr2 + hStr1)   % 65521;
    }
    return (hStr2 << 16) + hStr1;
}

This will output:

printf("%d", hash("Test")); = 64815521 returns it as an integer

This is the PHP function:

function hashEX($Hash)
{
    $hStr1 = 1; $hStr2 = 0;

    for($i = 0, $l = strlen($Hash); $i < $l; $i ++)
    {
        $hStr1 = ($hStr1 + $Hash[$i]) % 65521;
        $hStr2 = ($hStr2 + $hStr1)   % 65521;
    }
    return ($hStr2 << 16) + $hStr1;
}

This gives me:

echo intval(hashEX("Test")); = 262145

echo hashEX("Test"); = 262145

Anyone know why is this and how to fix it? To make it clear, I want to get the same value as the PAWN function in the PHP function.

Answer 1

Found my answer:

function hashEX($Hash)
{
    $hStr1 = 1; $hStr2 = 0;

    for($i = 0, $l = strlen($Hash); $i < $l; $i ++)
    {
        $hStr1 = ($hStr1 + ord($Hash[$i])) % 65521;        
        $hStr2 = ($hStr2 + $hStr1)   % 65521;
    }
    return ($hStr2 << 16) + $hStr1;
}