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我需要用 attoparsec 解析固定长度的字段，但我现在正在努力使用编译器。我还是个新手，下面的代码是我最接近的解决方案：

> {-# LANGUAGE OverloadedStrings #-}
> import Control.Applicative
> import Data.Text as T
> import Data.Attoparsec.Combinator
> import Data.Attoparsec.Text hiding (take)
> import Data.Char
> import Prelude hiding (take)
>
> type City = Text
> type Ready = Bool
> data CityReady = CR City Ready deriving Show
>
> input = T.unlines ["#London              1",
>                    "#Seoul               0",
>                    "#Tokyo               0",
>                    "#New York            1"]
>
> parseCityReady = many $ CR <$> cityParser <*> readyParser <* endOfLine
>
> cityParser = char '#' *>
>              takeTill isSpace <*
>              skipWhile isHorizontalSpace
>
>
> readyParser = char '1' *> pure True  <|> char '0' *> pure False
>
> main =
>   case parseOnly parseCityReady input of
>      Left err  -> print err
>      Right xs  -> mapM_ print xs
>

这一切都很好，但它只是返回没有空间的城市。

CR "London" True
CR "Seoul" False
CR "Tokyo" False

我尝试使用 applicative 为 City 文本字符串取 20 个字符

> cityParser = char '#' *>
>              take 20

乃至do syntax

> cityParser = do char '#'
>                 city <- take 20
>                 return city

但两种方法都无法编译并出现此错误：

Couldn't match expected type `attoparsec-0.10.4.0:Data.Attoparsec.Internal.Types.Parser
                                Text b0'
            with actual type `Text -> Text'
In the return type of a call of `take'
Probable cause: `take' is applied to too few arguments
In the second argument of `(*>)', namely `take 20'
In the expression: char '#' *> take 20

是什么导致 ghc 在有 类型Text -> Text时询问？takeInt -> Text -> Text

如何在 applicative 和 do-syntax 中解决它？

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1 回答 1

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因此，您的问题是您隐藏了该take 功能的多个版本。特别是，您隐藏了takefromattoparsec而不是模块中的take功能Text。您需要做的就是像这样更改您的导入

> import Control.Applicative
> import Data.Attoparsec.Combinator
> import Data.Attoparsec.Text
> import Data.Char
> import Data.Text as T hiding (take)
> import Prelude hiding (take)
于 2013-02-05T19:30:03.147 回答