用例:一个应用程序,可通过多个消息传递服务跟踪用户。有一个TwitterAccount
数据类型,一个FacebookAccount
数据类型等。可以很容易地将它们与Account
求和类型结合在一起,但是层次结构的下一个层次导致了我的问题。
ATwitterAccount
具有TwitterPost
s的列表,aFacebookAccount
具有FacebookPost
s的列表,依此类推。
我的任务:我希望能够将所有帐户中最近10天的所有帖子放到一个列表中,并从中提取常见的时间和邮件正文字段以进行显示。
我没办法:我想,如果每个类的Post
实现像一个类型类SimplePost
暴露的功能messageBody
,并messageTime
认为这可能解决我的问题,但我不能创建列表[SimpleMessage]
。
我想保持aTwitterAccount
只能包含TwitterPost
s的不变性,依此类推,因此我不能使用sum-types。我宁愿不创建对象的副本来执行此操作。
解决此问题的最佳,最干净,最Haskell式设计是什么?
更新这不是一个答案,但是作为recursion.ninja和Helder Pereira提供的四种解决方案的替代方案,我一直在考虑是否可以使用幻像类型满足我的不变量,Account
并且这些Post
类型可以容纳所有人所需的所有可能信息。提供者。但是,使用元组和笨拙的逻辑意味着它不能很好地扩展。也许这应该是一个不同的问题。
{-# LANGUAGE EmptyDataDecls #-}
-- Some FSharpisms
(|>) = flip ($)
(<|) = ($)
infixr 0 <|
data Twitter
data Facebook
data LinkedIn
data Post a = Post{
postBody :: String,
postDate :: UTCTime,
postForwarded :: Bool,
postFriendMentions :: [UserName]
} deriving (Show, Eq)
data Account a = Account {
accountName :: String,
accountPosts :: [Post a]
} deriving (Show, Eq)
data User = User {
userName :: String,
userTweets :: Account Twitter,
userFaces :: Account Facebook,
userLinks :: Account LinkedIn
}
prettyShowUtc :: UTCTime -> String
prettyShowUtc utc = ...
prettyShow :: Post a -> String
prettyShow p = prettyShowUtc (postDate p) ++ " : " ++ show (postBody p)
showOrderedOf2 :: ([Post a], [Post b]) -> [String]
showOrderedOf2 ([], []) = []
showOrderedOf2 (ls, []) = map prettyShow ls
showOrderedOf2 ([], rs) = map prettyShow rs
showOrderedOf2 ((l:ls), (r:rs)) =
if postDate l < postDate r
then prettyShow l : showOrderedOf2 (ls, (r:rs))
else prettyShow r : showOrderedOf2 ((l:ls), rs)
showOrderedOf3 :: ([Post a], [Post b], [Post c]) -> [String]
showOrderedOf3 ([], [], []) = []
showOrderedOf3 (as, [], []) = map postBody as
showOrderedOf3 ([], bs, []) = map postBody bs
showOrderedOf3 ([], [], cs) = map postBody cs
showOrderedOf3 (as, bs, []) = showOrderedOf2 (as, bs)
showOrderedOf3 ([], bs, cs) = showOrderedOf2 (bs, cs)
showOrderedOf3 (as, [], cs) = showOrderedOf2 (as, cs)
showOrderedOf3 ((a:as), (b:bs), (c:cs)) =
let (adate, bdate, cdate) = (postDate a, postDate b, postDate c)
minDate = minimum [adate, bdate, cdate]
in
if adate == minDate
then prettyShow a : showOrderedOf3 (as, (b:bs), (c:cs))
else (if bdate == minDate
then prettyShow b : showOrderedOf3 ((a:as), bs, (c:cs))
else prettyShow c : showOrderedOf3 ((a:as), (b:bs), cs))
createAndShowSample :: IO ()
createAndShowSample =
let faceAc = Account {...} :: Account Facebook
twitAc = Account {...} :: Account Twitter
linkAc = Account {...} :: Account LinkedIn
in
showOrderedOf3 (accountPosts faceAc, accountPosts twitAc, accountPosts linkAc)
|> intercalate "\n"
|> putStrLn
您应该抽象FaceBookAccount
和的TwitterAccount
实例SocialMediaAccount
Haskell代码:
import Control.Applicative ((<$>))
import Data.List
import Data.Ord
import Data.Time
data FaceBookAccount = FaceBookAccount [FaceBookPost]
data TwitterAccount = TwitterAccount [TwitterPost]
data FaceBookPost = FaceBookPost String UTCTime
data TwitterPost = TwitterPost String UTCTime
data SocialMediaAccount
= SocialMediaAccount
{ accountPosts :: [SocialMediaPost]
}
data SocialMediaPost
= SocialMediaPost
{ postBody :: String
, postTime :: UTCTime
}
class SocialMedia a where
simpleAccount :: a -> SocialMediaAccount
instance SocialMedia FaceBookAccount where
simpleAccount (FaceBookAccount xs) = SocialMediaAccount $ f <$> xs
where
f (FaceBookPost text time) = SocialMediaPost text time
instance SocialMedia TwitterAccount where
simpleAccount (TwitterAccount xs) = SocialMediaAccount $ f <$> xs
where
f (TwitterPost text time) = SocialMediaPost text time
getAllMessages :: (SocialMedia a, SocialMedia b) => a -> b -> [SocialMediaPost]
getAllMessages xs ys = sortBy (comparing postTime)
$ extract xs
++ extract ys
where
extract :: SocialMedia a => a -> [SocialMediaPost]
extract = accountPosts . simpleAccount
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