Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Using Generic Deriving with a Record Haskell

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

like image 236
mdedetrich Avatar asked Jul 02 '13 02:07

mdedetrich


1 Answers

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.

like image 146
felix-eku Avatar answered Nov 15 '22 22:11

felix-eku