SHA2 512 的改编给出了不正确的结果

我正在尝试调整 SecureHashAlgorithm 的纯 Lua 实现here http://lua-users.org/wiki/SecureHashAlgorithm对于 SHA2 512 而不是 SHA2 256。当我尝试使用改编时，它没有给出正确的答案。

这是改编：

--
-- UTILITY FUNCTIONS
--
-- transform a string of bytes in a string of hexadecimal digits
local function str2hexa (s)
    local h = string.gsub(s, ".", function(c)
        return string.format("%02x", string.byte(c))
    end)
    return h
end

-- transforms number 'l' into a big-endian sequence of 'n' bytes
--(coded as a string)
local function num2string(l, n)
    local s = ""
    for i = 1, n do
        --most significant byte of l
        local remainder = l % 256
        s = string.char(remainder) .. s
        --remove from l the bits we have already transformed
        l = (l-remainder) / 256
    end
    return s
end

-- transform the big-endian sequence of eight bytes starting at
-- index 'i' in 's' into a number
local function s264num (s, i)
    local n = 0
    for i = i, i + 7 do
        n = n*256 + string.byte(s, i)
    end
    return n
end

--
-- MAIN SECTION
--

-- FIRST STEP: INITIALIZE HASH VALUES
--(second 32 bits of the fractional parts of the square roots of the first 9th through 16th primes 23..53)

local HH = {}

local function initH512(H)
    H = {0x6a09e667f3bcc908, 0xbb67ae8584caa73b, 0x3c6ef372fe94f82b, 0xa54ff53a5f1d36f1, 0x510e527fade682d1, 0x9b05688c2b3e6c1f, 0x1f83d9abfb41bd6b, 0x5be0cd19137e2179}
    return H
end

-- SECOND STEP: INITIALIZE ROUND CONSTANTS
--(first 80 bits of the fractional parts of the cube roots of the first 80 primes 2..409)

local k = {
    0x428a2f98d728ae22, 0x7137449123ef65cd, 0xb5c0fbcfec4d3b2f, 0xe9b5dba58189dbbc, 0x3956c25bf348b538,
    0x59f111f1b605d019, 0x923f82a4af194f9b, 0xab1c5ed5da6d8118, 0xd807aa98a3030242, 0x12835b0145706fbe,
    0x243185be4ee4b28c, 0x550c7dc3d5ffb4e2, 0x72be5d74f27b896f, 0x80deb1fe3b1696b1, 0x9bdc06a725c71235,
    0xc19bf174cf692694, 0xe49b69c19ef14ad2, 0xefbe4786384f25e3, 0x0fc19dc68b8cd5b5, 0x240ca1cc77ac9c65,
    0x2de92c6f592b0275, 0x4a7484aa6ea6e483, 0x5cb0a9dcbd41fbd4, 0x76f988da831153b5, 0x983e5152ee66dfab,
    0xa831c66d2db43210, 0xb00327c898fb213f, 0xbf597fc7beef0ee4, 0xc6e00bf33da88fc2, 0xd5a79147930aa725,
    0x06ca6351e003826f, 0x142929670a0e6e70, 0x27b70a8546d22ffc, 0x2e1b21385c26c926, 0x4d2c6dfc5ac42aed,
    0x53380d139d95b3df, 0x650a73548baf63de, 0x766a0abb3c77b2a8, 0x81c2c92e47edaee6, 0x92722c851482353b,
    0xa2bfe8a14cf10364, 0xa81a664bbc423001, 0xc24b8b70d0f89791, 0xc76c51a30654be30, 0xd192e819d6ef5218,
    0xd69906245565a910, 0xf40e35855771202a, 0x106aa07032bbd1b8, 0x19a4c116b8d2d0c8, 0x1e376c085141ab53,
    0x2748774cdf8eeb99, 0x34b0bcb5e19b48a8, 0x391c0cb3c5c95a63, 0x4ed8aa4ae3418acb, 0x5b9cca4f7763e373,
    0x682e6ff3d6b2b8a3, 0x748f82ee5defb2fc, 0x78a5636f43172f60, 0x84c87814a1f0ab72, 0x8cc702081a6439ec,
    0x90befffa23631e28, 0xa4506cebde82bde9, 0xbef9a3f7b2c67915, 0xc67178f2e372532b, 0xca273eceea26619c,
    0xd186b8c721c0c207, 0xeada7dd6cde0eb1e, 0xf57d4f7fee6ed178, 0x06f067aa72176fba, 0x0a637dc5a2c898a6,
    0x113f9804bef90dae, 0x1b710b35131c471b, 0x28db77f523047d84, 0x32caab7b40c72493, 0x3c9ebe0a15c9bebc,
    0x431d67c49c100d4c, 0x4cc5d4becb3e42b6, 0x597f299cfc657e2a, 0x5fcb6fab3ad6faec, 0x6c44198c4a475817
}

-- THIRD STEP: PRE-PROCESSING (padding)

local function preprocess(toProcess, len)
    --append a single '1' bit
    --append K '0' bits, where K is the minimum number >= 0 such that L + 1 + K = 896mod1024
    local extra = 128 - (len + 9) % 128
    len = num2string(8 * len, 8)
    toProcess = toProcess .. "\128" .. string.rep("\0", extra) .. len
    assert(#toProcess % 128 == 0)
    return toProcess
end

local function rrotate(rot, n)
    return (rot >> n) | ((rot << 64 - n))
end

local function digestblock(msg, i, H)
    local w = {}
    for j = 1, 16 do w[j] = s264num(msg, i + (j - 1)*4) end
    for j = 17, 80 do
        local v = w[j - 15] 
        local s0 = rrotate(v, 1) ~ rrotate(v, 8) ~ (v >> 7)
        v = w[j - 2] 
        w[j] = w[j - 16] + s0 + w[j - 7] + ((rrotate(v, 19) ~ rrotate(v, 61)) ~ (v >> 6))
    end

    local a, b, c, d, e, f, g, h = H[1], H[2], H[3], H[4], H[5], H[6], H[7], H[8]
    for i = 1, 80 do
        a, b, c, d, e, f, g, h = a , b , c , d , e , f , g , h 
        local s0 = rrotate(a, 28) ~ (rrotate(a, 34) ~ rrotate(a, 39))
        local maj = ((a & b) ~ (a & c)) ~ (b & c)
        local t2 = s0 + maj
        local s1 = rrotate(e, 14) ~ (rrotate(e, 18) ~ rrotate(e, 41))
        local ch = (e & f) ~ (~e & g)
        local t1 = h + s1 + ch + k[i] + w[i]
        h, g, f, e, d, c, b, a = g, f, e, d + t1, c, b, a, t1 + t2
    end

    H[1] = (H[1] + a) 
    H[2] = (H[2] + b) 
    H[3] = (H[3] + c) 
    H[4] = (H[4] + d) 
    H[5] = (H[5] + e) 
    H[6] = (H[6] + f) 
    H[7] = (H[7] + g) 
    H[8] = (H[8] + h) 
end

local function finalresult512 (H)
    -- Produce the final hash value:
    return
    str2hexa(num2string(H[1], 8)..num2string(H[2], 8)..num2string(H[3], 8)..num2string(H[4], 8)..
            num2string(H[5], 8)..num2string(H[6], 8)..num2string(H[7], 8)..num2string(H[8], 8))
end


-- Returns the hash512 for the given string.
local function hash512 (msg)
    msg = preprocess(msg, #msg)
    local H = initH512(HH)

    -- Process the message in successive 1024-bit (128 bytes) chunks:
    for i = 1, #msg, 128 do
        digestblock(msg, i, H)
    end

    return finalresult512(H)
end

给定 hash512("a")：

Expect: 1f40fc92da241694750979ee6cf582f2d5d7d28e18335de05abc54d0560e0f5302860c652bf08d560252aa5e74210546f369fbbbce8c12cfc7957b2652fe9a75

Actual: e0b9623f2194cb81f2a62616a183edbe390be0d0b20430cadc3371efc237fa6bf7f8b48311f2fa249131c347fee3e8cde6acfdab286d648054541f92102cfc9c

我知道我正在创建正确位大小（1024 位）的消息，并且也在 1024 位块中工作，或者至少我相信我是这样。

我不确定这是否与整数的处理有关（标准要求无符号整数），或者我是否在实用函数之一中犯了错误，或两者兼而有之。如果确实是整数处理的问题，我将如何处理该问题。在处理 256 位版本的改编时，我能够通过在处理摘要块方法中的数字时使用 mod 2^32 来解决此问题。我尝试使用 512 位版本执行 mod 2^64 和 2^63，但它无法解决问题。我很困惑。

我应该提到的是，我无法使用众多库实现之一，因为我使用的沙盒 Lua 不提供此访问权限，这就是为什么我需要一个纯 lua 实现。提前致谢。

不幸的是，在 Lua 5.3 中引入整数之后，为 Lua 编写脚本变得更加复杂。
你必须总是想关于整数和浮点数之间的转换。
总是。是的，那很无聊。

你的错误之一就是这个“Lua 的黑暗角落”的一个很好的例子。

  local remainder = l % 256
  s = string.char(remainder) .. s
  --remove from l the bits we have already transformed
  l = (l-remainder) / 256

你的价值l最初是一个64位整数。
切断其第一个字节后l包含 (64-8) =56位，但现在它是浮点数 (with 53位精度，当然）。

可能的解决方案：使用l = l >> 8 or l = l // 256代替l = (l-remainder) / 256

另一个错误是使用s264num(msg, i + (j - 1) * 4)代替s264num(msg, i + (j - 1) * 8)

下面这一行还有一个错误：

local extra = 128 - (len + 9) % 128

正确的代码是

local extra = - (len + 17) % 128 + 8

（请注意-a%m+b不等于b-a%m由于运算符优先级）

修复这 3 个错误后，您的代码可以正常工作。