我正在尝试用纯lua编写PBKDF2实现。我正在编写它，因为我想在不允许外部库的沙盒 lua 环境中使用它。我看了一下IETF的标准文件，并得到了它。下面是我想出的代码：
do
package.preload["pbkdf2"] = function()
local hmac = require 'hmac'
local len = string.len
local gsub = string.gsub
local format = string.format
local byte = string.byte
local char = string.char
local concat = table.concat
local ceil = math.ceil
local function toBytes(str)
local tmp = {}
for i = 1, len(str) do
tmp[i] = byte(str, i)
end
return tmp
end
local function toString(bArray)
local tmp = {}
for i = 1, #bArray do
tmp[i] = char(bArray[i])
end
tmp = concat(tmp)
return tmp
end
-- transform a string of bytes in a string of hexadecimal digits
local function asHex(s)
local h = gsub(s, ".", function(c)
return format("%02x", byte(c))
end)
return h
end
local num2string = function(l, n)
local s = {}
for i = 1, n do
local idx = (n + 1) - i
s[idx] = char(l & 255)
l = l >> 8
end
s = concat(s)
return s
end
local buildBlock = function(hFun, password, salt, c, int)
local tmp
local tmp2
for i = 1, c do
if i == 1 then
print(int)
print(salt .. int)
-- PRF(password, salt || INT_32_BE(i)
-- return result of hash as a byte string
tmp = hmac.hash(hFun, password, salt .. num2string(int, 4), true)
else
-- returns result of hash as byte string
tmp2 = hmac.hash(hFun, password, tmp, true)
-- transform to byte arrays
tmp2 = toBytes(tmp2)
tmp = toBytes(tmp)
assert(#tmp == #tmp2)
-- apply XOR over bytes in both arrays
-- save results to final array
for j = 1, #tmp do
-- perform XOR operation on both elements in the respective arrays
tmp[j] = tmp[j] ~ tmp2[j]
end
-- transform back into byte string to pass to next hash
tmp = toString(tmp)
end
end
return tmp
end
local truncate = function(str, pos)
return string.sub(str, 1, pos)
end
local deriveKey = function(hFun, message, salt, c, dLen)
local hLen = hFun.outputSize
-- the derived key cannot be larger than (2^32 * hLen)
if dLen > (2^32) * hLen then error("The derived key cannot be larger than 2^32 times the output size of the hash function.") end
-- the block size is the desired key length divided by the output size of the underlying hash function, rounded up
local blockSize = ceil(dLen/hLen)
-- to store our blocks
local final = {}
for i = 1, blockSize do
-- lets make our blocks in here
final[i] = buildBlock(hFun, message, salt, c, i)
end
local result
if #final == 1 then
result = final[1] -- we only have one block
else
result = concat(final) -- turns final into a bytestring to be outputted
end
--if #result > dLen then truncate(final, dLen) end
assert(#result == dLen)
return asHex(result) -- outputs as a hex value
end
return {deriveKey = deriveKey}
end
end
此代码未获得正确答案。使用此处提供的测试向量测试此代码，假设底层 PRF 是 HMAC-SHA256，输出如下：
key: "password"
salt: "salt"
c: 1
dkLen: 32
Got: 13463842ec330934dc124494b40d8baade465b72f3fcadad741f2d0e052fd2f5
Expected: 120fb6cffcf8b32c43e7225256c4f837a86548c92ccc35480805987cb70be17b
key: "password"
salt: "salt"
c: 2
dkLen: 32
Got: 8b82aed26f503effdbc6c14bc7f0338b2b90e387f14ac1f91f9ad74e618f9558
Expected: AE4D0C95AF6B46D32D0ADFF928F06DD02A303F8EF3C251DFD6E2D85A95474C43
我相信这可能与字符串到字节编码有关，但我无法确定究竟是什么导致了这个问题。当我测试我的HMAC代码时，我不得不依靠在线生成器，因为我找不到HMAC-SHA224和HMAC-SHA256的载体。一些计算器会为相同的键、消息组合提供完全不同的输出值。这可能是因为他们如何处理输入，但我不确定。如果更有经验的人可以帮助我解决这个问题，我将不胜感激。
编辑：此问题已解决。似乎所需要的只是将int作为长度为 4 的二进制字符串传递。我用修复程序更新了代码。
编辑2：我再次阅读标准，意识到解决方案一直在我的脸上(标准说将i编码为32位大端整数(。
解决方案是将int转换为长度为 4 的二进制字符串。感谢@EgorSkriptunoff的洞察力。