parsing - 在 Haskell 中创建一个 Read 实例

Question

我有一个数据类型

data Time = Time {hour :: Int,
                  minute :: Int
                 }

我已将 Show 实例定义为

instance Show Time where
  show (Time hour minute) = (if hour > 10
                             then (show hour)
                             else ("0" ++ show hour))
                            ++ ":" ++
                            (if minute > 10
                             then (show minute)
                             else ("0" ++ show minute))

以 07:09 的格式打印时间。

现在，Show 和 Read 之间应该是对称的，所以在阅读之后(但不是真正(我认为)理解)this和 this ，并阅读 documentation ，我想出了以下代码:

instance Read Time where
  readsPrec _ input =
    let hourPart = takeWhile (/= ':')
        minutePart = tail . dropWhile (/= ':')
    in (\str -> [(newTime
                  (read (hourPart str) :: Int)
                  (read (minutePart str) :: Int), "")]) input

这可行，但 "" 部分使它看起来是错误的。所以我的问题是:

任何人都可以向我解释实现 Read 将 "07:09" 解析为 newTime 7 9 和/或向我展示的正确方法吗？

Answer 1

我将使用 isDigit 并保留您对时间的定义。

import Data.Char (isDigit)

data Time = Time {hour :: Int,
                  minute :: Int
                 }

您使用了但没有定义 newTime，所以我自己编写了一个以便我的代码可以编译!

newTime :: Int -> Int -> Time
newTime h m | between 0 23 h && between 0 59 m = Time h m
            | otherwise = error "newTime: hours must be in range 0-23 and minutes 0-59"
     where between low high val = low <= val && val <= high

<小时/>

首先，您的 show 实例有点错误，因为 show $ Time 10 10 给出 "010:010"

instance Show Time where
  show (Time hour minute) = (if hour > 9       -- oops
                             then (show hour)
                             else ("0" ++ show hour))
                            ++ ":" ++
                            (if minute > 9     -- oops
                             then (show minute)
                             else ("0" ++ show minute))

让我们看一下readsPrec:

*Main> :i readsPrec
class Read a where
  readsPrec :: Int -> ReadS a
  ...
    -- Defined in GHC.Read
*Main> :i ReadS
type ReadS a = String -> [(a, String)]
    -- Defined in Text.ParserCombinators.ReadP

这是一个解析器 - 它应该返回不匹配的剩余字符串，而不仅仅是 ""，所以你是对的，"" 是错误的:

*Main> read "03:22" :: Time
03:22
*Main> read "[23:34,23:12,03:22]" :: [Time]
*** Exception: Prelude.read: no parse

它无法解析它，因为您在第一次读取时丢弃了 ,23:12,03:22]。

让我们重构一下，以便在继续过程中吃掉输入:

instance Read Time where
  readsPrec _ input =
    let (hours,rest1) = span isDigit input
        hour = read hours :: Int
        (c:rest2) = rest1
        (mins,rest3) = splitAt 2 rest2
        minute = read mins :: Int
        in
      if c==':' && all isDigit mins && length mins == 2 then -- it looks valid
         [(newTime hour minute,rest3)]
       else []                      -- don't give any parse if it was invalid

举个例子

Main> read "[23:34,23:12,03:22]" :: [Time]
[23:34,23:12,03:22]
*Main> read "34:76" :: Time
*** Exception: Prelude.read: no parse

但是，它确实允许“3:45”并将其解释为“03:45”。我不确定这是一个好主意，所以也许我们可以添加另一个测试长度小时== 2。

<小时/>

如果我们这样做的话，我将放弃所有这些分割和跨度的东西，所以也许我更喜欢:

instance Read Time where
  readsPrec _ (h1:h2:':':m1:m2:therest) =
    let hour   = read [h1,h2] :: Int  -- lazily doesn't get evaluated unless valid
        minute = read [m1,m2] :: Int
        in
      if all isDigit [h1,h2,m1,m2] then -- it looks valid
         [(newTime hour minute,therest)]
       else []                      -- don't give any parse if it was invalid
  readsPrec _ _ = []                -- don't give any parse if it was invalid

对我来说，这实际上看起来更干净、更简单。

这次不允许“3:45”:

*Main> read "3:40" :: Time
*** Exception: Prelude.read: no parse
*Main> read "03:40" :: Time
03:40
*Main> read "[03:40,02:10]" :: [Time]
[03:40,02:10]