As I investigated the code more carefully, I found that the type unification
failure may not be related to the suspected class constraint on data
constructor.
I have made minor changes to the original code to remove the Ord constraint,
including introducing a FakedMap with no requirement on Ord. The type
unification
failure continues:
> Couldn't match type `nt1' with `nt'
> `nt1' is a rigid type variable bound by
> the type signature for
> subsome :: [RRule nt1 t1 s1] -> Either String ([t1], s1)
> at xx.hs:34:19
> `nt' is a rigid type variable bound by
> the type signature for
> legSome :: LegGram nt t s -> nt -> Either String ([t], s)
> at xx.hs:29:1
> Expected type: [Symbols nt1 t1]
> Actual type: [Symbols nt t]
> In the first argument of `makeWord', namely `r'
> In the expression: makeWord r
The complete changed code follows:
data Symbols nt t = NT nt -- ^ non terminal
| T t -- ^ terminal
deriving (Eq, Ord)
type Sem s = [s]->s
data Rule nt t s = Rule { refined :: nt
, expression :: [Symbols nt t]
, emit :: Sem s
}
type RRule nt t s = ([Symbols nt t], Sem s)
data FakedMap a b = FakedMap
delete :: k -> FakedMap k a -> FakedMap k a
delete a b = b
lookup :: k -> FakedMap k a -> Maybe a
lookup a b = Nothing
data LegGram nt t s = LegGram (FakedMap nt [RRule nt t s])
legSome :: LegGram nt t s -> nt -> Either String ([t], s)
legSome (LegGram g) ntV =
case Main.lookup ntV g of
Nothing -> Left "No word accepted!"
Just l -> let sg = legSome (LegGram (Main.delete ntV g))
subsome :: [RRule nt t s] -> Either String ([t], s)
subsome [] = Left "No word accepted!"
subsome ((r,sem):l) =
let makeWord [] = Right ([],[])
makeWord ((NT nnt):ll) =
do (m, ss) <- sg nnt
(mm, sss) <- makeWord ll
return (m++mm, ss:sss)
makeWord ((T tt):ll) =
do (mm, sss) <- makeWord ll
return (tt:mm, sss)
in
case makeWord r of
Right (ll, mm) -> Right (ll, sem mm)
Left err -> subsome l
in subsome l