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有没有办法在haskell中做更多的“动态”数据构造器?

constructor haskell data-structures

4

Rafael S. Calsaverini · 技术社区 · 16 年前

是否有一些haskell扩展能够创建比gadt更复杂的数据构造函数?

假设我想要创建一个数据结构,它是一个有序的列表,并且有一个类似于 (:) 使用列表和类型签名:

data MyOrdList a where
    (>>>) :: (Ord a) -> a -> MyOrdList a -> MyOrdList a

但我想要 (>>>) 要有一个特定的行为,比如:

(>>>) :: (Ord a) => a -> [a] -> [a]
x >>> [] = [x] 
x >>> xs = low ++ [x] ++ high 
  where low  = filter (<x) xs
      high = filter (>x) xs

所以这个结构总是一个有序的结构。(如果这是一个很好的实践,我现在不想,我只是举一个发生在我身上的最简单的例子来说明我想要的行为类型)。

当然我可以用函数 (& gt;& gt;& gt;) ,但我将没有模式匹配和其他好处,我会有它的 >>> 是数据构造函数。

有什么办法可以做这种事吗?

2 回复 | 直到 16 年前

1

3

Greg Bacon 16 年前

你可以做 :>>> 一个数据构造函数,但是为了保持不变量,必须隐藏它。注意,您可以根据它进行模式匹配,如 render :

module MyOrdList (mkMyOrdList,MyOrdList,render) where

import Data.List

import qualified Data.ByteString as BS

data MyOrdList a
  = EmptyDataList
  | (:>>>) a (MyOrdList a)
  deriving (Show)

mkMyOrdList [] = EmptyDataList
mkMyOrdList xs = y :>>> mkMyOrdList ys
  where y = minimum xs
        ys = delete y xs 

render :: (Show a) => MyOrdList a -> String
render EmptyDataList = "<empty>"
render (x :>>> xs) = (show x) ++ " -> " ++ render xs

那么你可以用 MyOrdList 模块如

module Main where

import Control.Applicative
import System.IO

import qualified Data.ByteString as BS

import MyOrdList

main = do
  h <- openBinaryFile "/dev/urandom" ReadMode 
  cs <- readBytes 10 h
  -- but you cannot write...
  -- let bad = 3 :>>> 2 :>>> 1 :>>> EmptyOrdList
  putStrLn (render $ mkMyOrdList cs)
  where
    readBytes 0 _ = return []
    readBytes n h = do c <- BS.head <$> BS.hGet h 1 
                       cs <- readBytes (n-1) h
                       return (c:cs)

样品输出:

54 -> 57 -> 64 -> 98 -> 131 -> 146 -> 147 -> 148 -> 190 -> 250 -> <empty>

2

6

Rüdiger Hanke shraddha hattimare 16 年前

你可以做 MyOrdList 抽象类型和 (>>>) 一个函数并使用视图模式。为了简单起见,我在这里使用标准列表作为“后端”。

module MyOrdList
  (MyOrdList,
   MyOrdListView (OrdNil, OrdCons),
   (>>>),
   emptyOrdList,
   ordview
  ) where

import Data.List (sort)

newtype MyOrdList a = List [a]
  deriving Show

data MyOrdListView a = OrdNil | OrdCons a (MyOrdList a)

infixr 5 >>>

(>>>) :: (Ord a) => a -> MyOrdList a -> MyOrdList a
x >>> (List xs) = List (sort $ x:xs)

emptyOrdList = List []

ordview :: MyOrdList a -> MyOrdListView a
ordview (List []) = OrdNil
ordview (List (x:xs)) = OrdCons x (List xs)

你可以这样使用它:

{-# LANGUAGE ViewPatterns #-}

import MyOrdList

ordlength :: MyOrdList a -> Int
ordlength (ordview -> OrdNil) = 0
ordlength (ordview -> OrdCons x xs) = 1 + ordlength xs

作品:

*Main> ordlength $ 2 >>> 3 >>> 1 >>> emptyOrdList 
3
*Main> 2 >>> 3 >>> 1 >>> emptyOrdList 
List [1,2,3]

所以您的类型是抽象的,列表只能由 emptyOrdList 和 (& gt;& gt;& gt;) 但是您仍然有一些模式匹配的便利性。