You probably mean newtype Expr = Expr (ExprG Expr) Az iPademről küldve 2018. dec. 14. dátummal, 14:08 időpontban Tobias Brandt írta:

Hello,

my adhoc solution would be to add a type parameter:

data ExprG a = Var String | Enclosed String a String | Prefix a a | Ternary a String a String a deriving Show

type Expr = ExprG (ExprG ())

type ExprL = ExprG [Expr]

Cheers,

Tobias

...

On 12/14/18 1:56 PM, ducis wrote: Hello,

Say I have a certain type for parse trees: data Expr = Var String | Enclosed String Expr String | Prefix Expr Expr | Ternary Expr String Expr String Expr deriving (Show,Eq,Ord,Generic)--,Typeable)

Then I want a new type where every occurence of "Expr" definition changed to "[Expr]": data ExprL = VarL String | EnclosedL String [Expr] String | PrefixL [Expr] [Expr] | TernaryL [Expr] String [Expr] String [Expr] deriving (Show,Eq,Ord,Generic)

Sometimes, I also want to use a DList instead of list. data ExprD = VarD String | EnclosedD String (DList Expr) String | PrefixD (DList Expr) (DList Expr) | TernaryD (DList Expr) String (DList Expr) String (DList Expr) deriving (Show,Eq,Ord,Generic)

They have exactly the same structure, is there a way to unify the three definitions into one? Furthurmore, is it possible to generalise the latter two for all Functors ?

Thanks, ducis

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

---

Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

There may be some fun to be had with QuantifiedConstraints. I've thought about solving similar parser/formatter problems this way, extending parsers via functor composition. Consider:

data F f = F (f (F f))

-- Sadly the below does not work. It seems like maybe it should be able to work?

instance (forall a . Show a => Show (f a)) => Show (F f) where show (F f) = "(F " ++ show f ++ ")"

<interactive>:23:10: error: • The constraint ‘Show (f a)’ is no smaller than the instance head ‘Show (F f)’ (Use UndecidableInstances to permit this) • In the instance declaration for ‘Show (F f)’

-- We can get something almost as good

class (forall a . Show a => Show (f a)) => Show1 f

instance Show1 f => Show (F f) where show (F f) = "(F " ++ show f ++ ")"

instance Show1 Maybe

show (F $ Just $ F $ Nothing)

"(F Just (F Nothing))" On Fri, Dec 14, 2018 at 10:52 PM MarLinn wrote:

Hi Ducis,

you can parametrise over type variables of other kinds than just *.

If what you write is really what you want, the most straightforward answer is simply

data ExprG f = Var VarName | Enclosed VarName (f Expr) VarName | Prefix (f Expr) (f Expr) | Ternary (f Expr) VarName (f Expr) VarName (f Expr)

type Expr = ExprG Identity -- From Data.Functor.Identity type ExprL = ExprG [] type ExprD = ExprG DList

There is no mention of the word "functor" because you will have to add that constraint to the usage sites.

Downside: notice that the deriving clauses are gone because the instances aren't as easy to derive any more. Even the simplest and most harmless way I know to get that possibility back involves two language extensions: StandaloneDeriving and FlexibleInstances. With those you can write

deriving instance Show (ExprG Identity) deriving instance Show (ExprG []) deriving instance Show (ExprG DList) deriving instance Eq (ExprG Identity) :

I suspect though that what you actually want, but didn't write, is more along the lines of

data ExprL = … | EnclosedL VarName [ExprL] VarName | … -- using ExprL instead of Expr on the right side

data ExprD = … | EnclosedD VarName (DList ExprD) VarName | … -- using ExprD instead of Expr on the right side

The good news is that if you have the first solution, this step is rather simple. Because you can just use replace Expr with ExprG f again:

data ExprG f = Var VarName | Enclosed VarName (f (ExprG f)) VarName | Prefix (f (ExprG f)) (f (ExprG f)) | Ternary (f (ExprG f)) VarName (f (ExprG f)) VarName (f (ExprG f))

The better news is that although this looks repetitive and hard to read, it's well on the way to discovering the magic of the Free Monad.

Hope this helps.

Cheers, MarLinn

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.