你如何在 Haskell 中添加两个大的随机字节串？

Question

I have two lazy bytestrings that were generated randomly using the DRBG package. I want to interpret these bytestrings as numbers and be able to perform addition on them. So something like a function with signature ByteString -> ByteString -> ByteString .

In JavaScript, I was using the math-buffer package How can I do this in Haskell?

Answer 1

1. Solution 1

It seems that there is no math-buffer -like packages in Haskell, so you would be forced to implement these byte-wise operations yourself and use math-buffer for inspiration. So you would unpack your bytestring to a list of Word8 s and then implement addition of a base-256 number, like so:

byteSum :: [Word8] -> [Word8] -> [Word8]
byteSum = byteSum' 0
  where
    byteSum' 0 [] [] = []
    byteSum' 1 [] [] = [1]
    byteSum' carry (x:xs) (y:ys) =
        let v = x + y + carry
        in v : byteSum' (if v < x || v < y then 1 else 0) xs ys

main = do
    let bs1 = ...
    let bs2 = ...
    let sum = pack $ byteSum (unpack bs1) (unpack bs2)

Note this is assuming little-endianness (least significant byte first).

2. Solution 2

Alternatively, assume this bytestring is a big-endian unsigned integer, convert from ByteString -> Integer and let the Integer to the math, then convert it back:

bstoi :: ByteString -> Integer
bstoi = fromDigits 0 . unpack
  where
    fromDigits n [] = n
    fromDigits n (x:xs) = fromDigits (n * 256 + (toInteger x)) xs

itobs :: Integer -> ByteString
itobs = pack . reverse . toDigits
  where
    toDigits 0 = []
    toDigits x = (fromIntegral $ x `mod` 256) : toDigits (x `div` 256)

main = do
    let bs1 = ...
    let bs2 = ...
    let sum = itobs $ (bstoi bs1) + (bstoi bs2)

Answer 2

import qualified Data.ByteString.Lazy as BL

pizdec bl = BL.pack . BL.zipWith (+) bl