使用通用派生与记录Haskell

时间:2013-07-02 02:38:50

标签: generics haskell record deriving

我基本上试图看看我是否可以在Haskell中模拟一个ORM框架,这样如果用户想要创建一个数据库模型,他们会做这样的事情

data Car = Car {
        company :: String, 
        model :: String, 
        year :: Int
        } deriving (Model)

表格为“Car”,列为公司,型号,年份

要在Haskell中执行此操作,您必须使用类和泛型的组合,这就是我遇到困难的地方。使用本教程(http://www.haskell.org/ghc/docs/7.4.1/html/users_guide/generic-programming.html),我想出了这个(基本上只是复制和重命名,所以我可以使代码工作)

{-# LANGUAGE DeriveGeneric, TypeOperators, TypeSynonymInstances, FlexibleInstances #-}

module Main where
import GHC.Generics

class SModel b where
        s_new :: b -> IO()

instance SModel Int where
        s_new s = putStrLn s++":Int"

instance SModel Integer where
        s_new s = putStrLn s++":Integer"

instance SModel String where
        s_new s = putStrLn s++":String"    

class Model m where
        new :: m a -> IO()

instance Model U1 where
        new U1 = putStrLn "unit"

instance (Model a, Model b) => Model (a :*: b) where
        new (a :*: b) = do
                new a
                new b

instance (Model a, Model b) => Model (a :+: b) where
        new (L1 x) = new x
        new (R1 x) = new x

instance (Model a) => Model (M1 i c a) where
        new (M1 x) = new x

instance (SModel a) => Model (K1 i a) where
        new (K1 x) = s_new x

data Car = Car {
        company :: String, 
        model :: String, 
        year :: Int
        } deriving (Model)

上面的代码会产生错误

Cannot derive well-kinded instance of form `Model (Car ...)'
      Class `Model' expects an argument of kind `* -> *'
    In the data declaration for `Car'

我有点陷入困境,我相信我已经实例化了所有必需的通用类型来覆盖记录

1 个答案:

答案 0 :(得分:7)

正如kosmikus在评论中所说,你不能直接导出Model。首先,您需要Model的“前端”类,提供通用默认值,如下所示:

class FModel a where
    fnew :: a -> IO()

    default new :: (Generic a, Model (Rep a)) => a -> IO()
    fnew = new . from

然后你可以这样做:

Car = ... deriving Generic
instance FModel Car

你有所需的实例。