5

I have something like the following:

{-# LANGUAGE TypeFamilies #-}

class Configuration c where
    data Pig c
    data Cow c

    parsePig :: GenParser Char st (Pig c)
    parseCow :: GenParser Char st (Cow c)

data Farm c =
    { pigs :: [Pig c]
    , cows :: [Cow c]
    } deriving Show

This fails because of the deriving Show line. I don't know how to force all Configuration instances to ensure that their data Pig and data Cow implementations are all instances of Show.

I know I could make it have showPig and showCow methods and the write out the whole complex show instance, but in reality things are more complex than this and that would be quite a pain.

Is there an easy, elegant way to guarantee that type family instances are themselves instances of certain classes?

4

2 に答える 2

8

StandaloneDerivingインスタンスに対してのみ手動で制約を指定するために使用できShowます。

{-# LANGUAGE StandaloneDeriving, FlexibleContexts, UndecidableInstances #-}
deriving instance (Show (Pig c), Show (Cow c)) => Show (Farm c)

これにより、Configurationインスタンスを実装しない場合でも、インスタンスを実装しようCowとしない限り、インスタンスを実装できます。PigShowshow

于 2012-09-13T18:59:50.333 に答える
1

Since you said you wanted to force all instances of Configuration to have Pig c and Cow c implement Show, an even simpler way to do this is to simply constrain the type families in the class's context, like so:

{-# LANGUAGE TypeFamilies, FlexibleContexts #-}

class (Show (Pig c), Show (Cow c)) => Configuration c where
    data Pig c
    data Cow c

data Farm c = Farm { pigs :: [Pig c],
                     cows :: [Cow c] } deriving (Show)

EDIT:

As @hammar pointed out in his comment, the previous code won't compile. One way to fix this is to use StandaloneDeriving, as he suggested. The other way is this:

{-# LANGUAGE TypeFamilies, FlexibleContexts, GADTSyntax #-}

class (Show (Pig c), Show (Cow c)) => Configuration c where
    data Pig c
    data Cow c

data Farm c where
    Farm :: Configuration c => { pigs :: [Pig c],
                                 cows :: [Cow c] } -> Farm c deriving (Show)

These two approaches get you slightly different results, in that @hammar's approach will require a Configuration constraint if you call show, whereas my approach will make available said constraint.

于 2012-09-14T20:00:54.313 に答える