Question

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

Answer 1

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。

使用通用派生与记录Haskell

1 个答案: