对于一个网络协议,我需要能够灵活地从Source m ByteString
读取不同类型的块。有lines
组合子,它将输入分成线,但我需要能够组合读取线和固定数量的字节。
我目前的方法是创建一个辅助函数:
|在输入上折叠给定的函数。函数返回时重复
Left
并将其结果累积在一个列表中。当函数返回Right
时, 连接累计结果(包括最后一个)并返回它, 使用leftover
存储剩下的内容。如果没有输入,则返回Nothing
可用。
chunk :: (Monad m, Monoid a)
=> (s -> i -> Either (a, s) (a, i))
-> s
-> Consumer i m (Maybe a)
chunk f = loop []
where
loop xs s = await >>= maybe (emit xs) (go xs s)
go xs s i = case f s i of
Left (x, s') -> loop (x : xs) s'
Right (x, l) -> leftover l >> emit (x : xs)
emit [] = return Nothing
emit xs = return (Just . mconcat . L.reverse $ xs)
-- Note: We could use `mappend` to combine the chunks directly. But this would
-- often get us O(n^2) complexity (like for `ByteString`s) so we keep a list of
-- the chunks and then use `mconcat`, which can be optimized by the `Monoid`.
使用此功能,我创建了特定的消费者:
bytes :: (Monad m) => Int -> Consumer ByteString m (Maybe ByteString)
bytes = chunk f
where
f n bs | n' > 0 = Left (bs, n')
| otherwise = Right $ BS.splitAt n bs
where n' = n - BS.length bs
line :: (Monad m) => Consumer ByteString m (Maybe ByteString)
line = chunk f ()
where
f _ bs = maybe (Left (bs, ()))
(\i -> Right $ BS.splitAt (i + 1) bs)
(BS.findIndex (== '\n') bs)
有更好的方法吗?我想这个问题肯定已经解决了。
答案 0 :(得分:1)
看起来正确,它与Warp解析请求标头的方式非常相似,但Warp并不打扰任何更高级别的组合器。