使用记录Haskell进行通用派生 [英] Using Generic Deriving with a Record Haskell

问题描述

我基本上试图看看我是否可以在Haskell中模拟一个ORM框架，这样如果用户想创建一个数据库模型，他们会这样做。

  data Car = Car {
 company :: String，
 model :: String，$ b $ year :: Int 
}派生（Model ）
  

如果表格是Car，列将是公司，模型，年份

要在Haskell中执行此操作，您必须使用类和泛型的组合，这就是我陷入困境的地方。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（这只是基本上复制和重命名，所以我可以得到代码的工作）

  { - ＃LANGUAGE DeriveGeneric，TypeOperators，TypeSynonymInstances，FlexibleInstances＃ - } 
 
模块Main其中
 import GHC.Generics 
 
 class SModel b其中
 s_new :: b  - > IO（）
 
实例SModel Int其中
 s_new s = putStrLn s ++：Int
 
实例SModel整数其中
 s_new s = putStrLn s ++ ：整数
 
实例SModel字符串其中
 s_new s = putStrLn s ++：String
 
 class模型m其中
 new :: ma  - > ; IO（）
 
实例模型U1其中
新U1 = putStrLn单元
 
实例（模型a，模型b）=>模型（a：*：b）其中
新（a：*：b）= do 
新a 
新b 
 
实例（模型a，模型b ）=>模型（a：+：b）其中
 new（L1 x）= new x 
 new（R1 x）= new x 
 
 instance（Model a）=>模型（M1 i c a）其中
 new（M1 x）= new x 
 
 instance（SModel a）=>模型（K1 ia）其中
 new（K1 x）= s_new x 
 
 data Car = Car {
 company :: String，
 model :: String，
 year :: Int 
}派生（模型）
  

上述代码将生成错误

 无法派生形式为`Model（Car ...）'的良好主观实例
 Class`Model '期待类似的论点`*  - > *'
在`Car'的数据声明中
  

而且我有点卡住这一点，我相信我已经实例化了所有必需的泛型类型来覆盖记录 解决方案

As kosmikus 在他的评论中说，你不能直接派生 Model 。首先，您需要一个用于 Model 的'前端'类，以提供一个通用的默认类型，如下所示：

  class FModel a where 
 fnew :: a  - > IO（）
 
 default new ::（Generic a，Model（Rep a））=> a  - > IO（）
 fnew = new。从
  

然后你可以这样做：

  Car = ...推导泛型
实例FModel Car 
  

并且您拥有所需的实例。

I am basically attempting to see if I can emulate an ORM framework within Haskell, so that if a user wants to make a database model, they would do something like this

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

Where the table would be "Car", and the columns would be the company,model,year

To do this within Haskell, you have to use a combination of classes and generics, and this is where I am getting stuck. Using this tutorial (http://www.haskell.org/ghc/docs/7.4.1/html/users_guide/generic-programming.html), I came up with this (which was just basically copying and renaming so I can get the code working)

{-# 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)

The above code will produces the error

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

And I am kind of stuck at this point, I believe I have already instantiated all of the required Generic types to cover a record

解决方案

As kosmikus said in his comment, you cannot derive Model directly. First you need a 'front-end' class for Model providing a generic default, which could look like this:

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

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

Then you can just do:

Car = ... deriving Generic
instance FModel Car

And you have the desired instance.

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