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对于一个网络协议,我需要能够从Source m ByteString. 有lines组合器,它将输入拆分为行,但我需要能够组合读取行和固定数量的字节。

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。在函数返回时重复并将Left 其结果累积到一个列表中。当函数返回时Right,连接累积的结果(包括最后一个)并返回它,使用leftover. Nothing如果没有可用的输入则返回。

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)

有没有更好的办法?我想这个问题一定已经在某个地方解决了。

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1 回答 1

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看起来是对的,它与 Warp 解析请求标头的方式非常相似,尽管 Warp 不会打扰任何更高级别的组合器。

于 2013-07-30T11:05:23.947 回答