Question

我正在尝试使用发现here的SecureHashAlgorithm的纯Lua实现来修改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。

以下是改编：

--
-- 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

给出hash512（“a”）：

Expect: 1f40fc92da241694750979ee6cf582f2d5d7d28e18335de05abc54d0560e0f5302860c652bf08d560252aa5e74210546f369fbbbce8c12cfc7957b2652fe9a75

Actual: e0b9623f2194cb81f2a62616a183edbe390be0d0b20430cadc3371efc237fa6bf7f8b48311f2fa249131c347fee3e8cde6acfdab286d648054541f92102cfc9c

我知道我正在创建一个正确位大小（1024位）的消息，并且还在1024位块中工作，或者至少我相信我是。

我不确定它是否与整数的处理有关（标准需要无符号整数），或者我是否在其中一个效用函数中出错，或者两者兼而有之。如果确实是处理整数的问题，我将如何处理问题。在处理digestblock方法中的数字时，使用mod 2 ^ 32处理256位版本的自适应时，我能够解决这个问题。我尝试使用512位版本执行mod 2 ^ 64和2 ^ 63，但它不能解决问题。我很难过。

我应该提到我不能使用众多库实现中的一个，因为我使用的沙箱Lua不提供此访问权限，这就是我需要纯lua实现的原因。提前谢谢。

Answer 1

不幸的是，在Lua 5.3中引入整数后，为Lua编写脚本变得更加复杂您必须始终认为有关整数和浮点数之间的转换总是。是的，那太无聊了。

你的一个错误是这个“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

您的值l最初是 64 位整数在切断其第一个字节l包含（64-8）= 56 位后，现在它是浮点数（ 53 < / strong> -bit precision，当然）。

可能的解决方案：使用l = l >> 8或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)

以下一行中还有一个错误：

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

正确的代码是

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

（请注意，由于运营商优先级，-a%m+b与b-a%m不同）

修正这3个错误后，您的代码可以正常工作。

适应SHA2 512会产生错误的结果

1 个答案: