在Haskell中创建一个Read实例 [英] Making a Read instance in Haskell
问题描述
我有一个数据类型
data Time = Time {hour :: Int，
minute :: Int
}
我已将Show的实例定义为
instance显示时间其中
显示（时间小时分钟）=（如果小时> 10
则（显示小时数）
else（0++ show hour））
++：++
（if minute> 10
then（show minute）
else 0++显示分钟））
打印时间格式为 07:09 。
现在， 这可以工作，但 任何人都可以向我解释实现Read解析 我将使用 您使用过但未定义 首先，您的show instance有点不对，因为 Show 和阅读，所以在阅读后（但不是真的（我认为）理解）这个和 this ，然后阅读
实例读取时间在哪里
readsPrec_input =
let hourPart = takeWhile（/ ='：'）
minutePart = tail。在（\str - > [（newTime
（read（hourPart str）:: Int）
（read（minutePart str）:: Int）中添加dropWhile（/ ='：'）
），）]）input
部分使它看起来不对。所以我的问题最终是：
07:09转换为 newTime 7 9 和/或展示我？
isDigit 并保留Time的定义。
import Data.Char（isDigit）
data Time = Time {hour :: Int，
minute :: Int
}
newTime ，所以我自己编写了一个代码，以便编译代码！
newTime :: Int - > Int - >时间
newTime h m |在0 23 h&&在0 59 m =时间h m
|之间否则=错误新时间：小时必须在范围0-23和分钟0-59
之间，其中低高值=低<= val&& val <= high
show $ Time 10 10 给出010：010
instance显示时间其中
显示（时间小时分钟）=（如果小时> 9 - oops
then（show hour）
else（0++ show hour））
++：++
（如果分钟> 9 - oops
then（show minute）$ b $我们来看看 readsPrec ：
* Main> ：我读取了Prec
类读取一个
readsPrec :: Int - >读取
...
- 在GHC.Read中定义
* Main> ：我ReadS
类型ReadS a = String - > [（a，String）]
- 在Text.ParserCombinators.ReadP中定义
这就是一个解析器 - 它应该返回不匹配的剩余字符串，而不是，所以你正确的是是错误的：
* Main>阅读03:22:: Time
03:22
* Main>阅读[23：34,23：12,03：22]:: [时间]
***例外：Prelude.read：不解析
它无法解析它，因为您丢弃了，23：12,03：22] 首先阅读。
让我们重复一下，以便随时了解输入：
实例读取时间其中
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 =='：'&&全部是数字&&长度分钟== 2然后 - 它看起来有效
[（newTime hour minute，rest3）]
else [] - 如果它是无效的，不给任何解析
举例
Main> ;阅读[23：34,23：12,03：22]:: [时间]
[23：34,23：12,03：22]
* Main>读34:76::时间
***例外：Prelude.read：不分析
然而，它确实允许3:45并将其解释为03:45。我不确定这是个好主意，所以也许我们可以添加另一个测试长度小时== 2 。
如果我们这样做，我会把所有这些分割和跨度的东西都去掉， d比较喜欢：
$ $ p $ $ code $实例读取时间
readsPrec _（h1：h2：'：'：m1：m2：那么）=
let hour = read [h1，h2] :: Int - 懒惰不会被计算，除非有效
minute =读取[m1，m2] :: Int
in
如果全部是数字[h1，h2，m1，m2]，那么 - 它看起来有效
[（newTime hour minute，therest）]
else [] - 不给出任何解析if它是无效的
readsPrec _ _ = [] - 如果它是无效的，不给任何解析
这实际上对我来说看起来更清洁和简单。
这一次它不允许3:45：
* Main>阅读3:40:: Time
***例外：Prelude.read：不分析
* Main>阅读03:40:: Time
03:40
* Main>阅读[03：40,02：10]:: [时间]
[03：40,02：10]
I have a data type
data Time = Time {hour :: Int,
minute :: Int
}
for which i have defined the instance of Show as being
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))
which prints out times in a format of 07:09.
Now, there should be symmetry between Show and Read, so after reading (but not truly (i think) understanding) this and this, and reading the documentation, i have come up with the following code:
instance Read Time where
readsPrec _ input =
let hourPart = takeWhile (/= ':')
minutePart = tail . dropWhile (/= ':')
in (\str -> [(newTime
(read (hourPart str) :: Int)
(read (minutePart str) :: Int), "")]) input
This works, but the "" part makes it seem wrong. So my question ends up being:
Can anyone explain to me the correct way to implement Read to parse "07:09" into newTime 7 9 and/or show me?
I'll use isDigit and keep your definition of Time.
import Data.Char (isDigit)
data Time = Time {hour :: Int,
minute :: Int
}
You used but didn't define newTime, so I wrote one myself so my code compiles!
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
Firstly, your show instance is a little wrong because show $ Time 10 10 gives "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))
Let's have a look at 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
That's a parser - it should return the unmatched remaining string instead of just "", so you're right that the "" is wrong:
*Main> read "03:22" :: Time
03:22
*Main> read "[23:34,23:12,03:22]" :: [Time]
*** Exception: Prelude.read: no parse
It can't parse it because you threw away the ,23:12,03:22] in the first read.
Let's refactor that a bit to eat the input as we go along:
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
Gives for example
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
It does, however, allow "3:45" and interprets it as "03:45". I'm not sure that's a good idea, so perhaps we could add another test length hours == 2.
I'm going off all this split and span stuff if we're doing it this way, so maybe I'd prefer:
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
Which actually seems cleaner and simpler to me.
This time it doesn't allow "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]
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