小编uv.*_*ita的帖子

通常,当我使用Haskell编写内容时,我需要具有多个构造函数的记录.例如,我想开发某种逻辑方案建模.我想到了这样的类型:

data Block a = Binary {
      binOp  :: a -> a -> a
    , opName :: String
    , in1 :: String
    , in2 :: String
    , out :: String
} | Unary {
      unOp  :: a -> a
    , opName :: String
    , in_ :: String
    , out :: String
}

它描述了两种类型的块:二进制(如和,或等)和一元(如不是).它们包含核心功能,输入和输出信号.

另一个例子:类型来描述控制台命令.

data Command = Command { info :: CommandInfo
                       , action :: Args -> Action () }
             | FileCommand { info :: CommandInfo
                           , fileAction :: F.File …

我实现了读取ByteString并以十六进制格式转换它的函数.例如,给定"AA10",它将其转换为[170,16]

import qualified Data.ByteString.Lazy as BSL
import qualified Data.ByteString.Internal as BS (w2c)

rHex :: BSL.ByteString -> BSL.ByteString
rHex bs
    | BSL.null bs = BSL.empty
    | BSL.null rest' = fromHex c1 `BSL.cons` BSL.empty
    | otherwise = rChunk c1 c2 `BSL.cons` rHex rest
          where (c1, rest') = (BSL.head bs, BSL.tail bs)
                (c2, rest) = (BSL.head rest', BSL.tail rest')
                rChunk c1 c2 = (fromHex c1) * 16 + fromHex c2
fromHex = fromIntegral . digitToInt . BS.w2c

但是我意识到我需要相同的功能但是对于简单的ByteString,而不是Lazy.我遇到的唯一方法是这样的:

import qualified Data.ByteString.Lazy as BSL …

写这样的东西是可以的:

head $ foldr (:) [] [1..]
-- 1

但是当我尝试处理元组时,它会进入无限循环:

head . fst $ foldr (\ x (ls, _) -> (x : ls, 0)) ([], 0) [1..]

我需要这个的原因是因为我想在内部函数中传递生成元素的计数.像那样:

foldr go ([], 0) [1..]
go num (ls, cnt) = -- use cnt to get l and produce new pair (l : ls, cnt + 1)