我编写了一个在haskell中执行SHA-1的程序，虽然它确实产生哈希，但它们与其他SHA-1程序生成的哈希不匹配
示例：>>> bool(0)
False
>>> bool(1)
True
>>> bool([])
False
>>> bool([1,2])
True
哈希到：cat但是应该哈希到b5be86bc8bccfc24b01b093228ebb96fc92fa804
我的代码是：
9d989e8d27dc9e0ec3389fc855f142c3d40f0c50我不知道出了什么问题。有人能告诉我哪里弄错了吗？
修改
我修复了指出的东西，但它仍然无法正常工作。它正常工作直到内循环。
我清理了代码，因此内部循环的函数可用(old code omitted)
，f1和f2
f3现在有趣地散布到cat。
代码：
ebe6c9fa1afa0ef5a0ca80bab251fd41cc29127e另外，一个小问题：import Data.Word
import Data.Bits
import Data.Char (ord, intToDigit)
import Data.Binary (encode, decode)
import Numeric (showHex, showIntAtBase)
import System.IO (stdin)
import Data.Sequence ((<|), (|>))
import qualified Data.Sequence as S
import qualified Data.ByteString.Lazy as B
type Quintuple32 = (Word32, Word32, Word32, Word32, Word32)
addQuintuple (a, b, c, d, e) (f, g, h, i, j) =
(a + f, b + g, c + h, d + i, e + j)
shower :: Quintuple32 -> String
shower (a, b, c, d, e) = concatMap (`showHex` "") [a, b, c, d, e]
hash :: Int -> S.Seq Word32 -> Quintuple32 -> Quintuple32
hash i w h@(a, b, c, d, e)
| i < 20 = hash (i + 1) w (newhash (f1 h + k1))
| i < 40 = hash (i + 1) w (newhash (f2 h + k2))
| i < 60 = hash (i + 1) w (newhash (f3 h + k3))
| i < 80 = hash (i + 1) w (newhash (f2 h + k4))
| otherwise = h
where (k1, k2, k3, k4) = (0x5A827999, 0x6ED9EBA1, 0x8F1BBCDC, 0xCA62C1D6)
newhash a' = (rotate a 5 + a' + e + (w `S.index` i), a, rotate b 30, c, d)
f1 :: Quintuple32 -> Word32
f1 (_, b, c, _, _) = (b .&. c) .|. (complement b .&. c)
f2 :: Quintuple32 -> Word32
f2 (_, b, c, d, _) = b `xor` c `xor` d
f3 :: Quintuple32 -> Word32
f3 (_, b, c, d, _) = (b .&. c) .|. (b .&. d) .|. (c .&. d)
starting :: Quintuple32
starting = (0x67452301
, 0xEFCDAB89
, 0x98BADCFE
, 0x10325476
, 0xC3D2E1F0)
hasher :: Quintuple32 -> S.Seq Word32 -> Quintuple32
hasher acc x = addQuintuple acc (hash 0 (extend x) acc)
process :: B.ByteString -> Quintuple32
process = foldl hasher starting . chunks . pad
extend :: S.Seq Word32 -> S.Seq Word32
extend = extend' 16
extend' :: Int -> S.Seq Word32 -> S.Seq Word32
extend' 80 a = a
extend' i a = extend' (i + 1) (a |> xored)
where xored = rotate ((a `S.index` (i - 3)) `xor`
(a `S.index` (i - 8)) `xor`
(a `S.index` (i - 14)) `xor`
(a `S.index` (i - 16))) 1
toBytes :: String -> B.ByteString
toBytes = B.pack . map (fromIntegral . ord)
splitEvery n xs
| B.null xs = S.empty
| otherwise = B.take n xs <| splitEvery n (B.drop n xs)
chunks :: B.ByteString -> [S.Seq Word32]
chunks xs
| B.null xs = []
| otherwise = x : chunks (B.drop 64 xs)
where x = fmap decode (splitEvery 4 (B.take 64 xs))
pad :: B.ByteString -> B.ByteString
pad xs = B.append (add0 $ add1 xs) length64
where length64 = encode (fromIntegral (8 * B.length xs) :: Word64)
add1 :: B.ByteString -> B.ByteString
add1 = flip B.append (B.singleton 128)
add0 :: B.ByteString -> B.ByteString
add0 xs
| modulo /= 448 = add0 $ B.append xs (B.singleton 0)
| otherwise = xs
where modulo = (B.length xs * 8) `rem` 512
是否可以设置多个变量？
答案 0 :(得分：6)
哦，另外一个！
立即向我跳出两个错误：
pad :: B.ByteString -> B.ByteString
pad xs = B.append (add0 $ add1 xs) length64
where length64 = encode (fromIntegral (B.length xs) :: Word64)
请注意，您追加的长度应该是位长度，而不是字节长度。
add1 :: B.ByteString -> B.ByteString
add1 = flip B.append (B.singleton 255)
注意255 /= 0b10000000，垫应该是后者。
一般情况下，您可以通过以下方式调试这些：1）一遍又一遍地查看规范。 2）与另一个实现相比，例如Adam Wick的SHA包，并在尽可能细粒度的水平上进行比较。
编辑：还有两个错误，基本上是转录错误。如果你仍然被困住，请四处寻找并大喊。