I generally have 2 questions. I am writing some functions to derive typeclass instances with a package called derive. I just want to get rid of using Derivation type with a stupid reason. I want use just Name type.
How can I defined a function with type
derivings :: Name -> Name -> Q [Dec]
by using function derive :: Derivation -> Name -> Q [Dec]
so that user can just write derivings ''Eq ''D instead of derive makeEq ''D?
> {-# LANGUAGE TemplateHaskell #-}
> import Data.DeriveTH -- in derive package
> import qualified Language.Haskell.TH.Syntax as S
> data D = D
> derivings :: Name -> String -> Q Exp
> derivings cla typ = do
> let makeClassName = mkName $ "make" ++ nameBase cla
> a <- [| derive makeClassName (mkName typ) |]
> return a
> instance S.Lift Name where
> lift x = varE x
Main> :t derivings ''Eq "D"
derivings ''Eq "D" :: Q Exp
Main> :t $(derivings ''Eq "D")
$(derivings ''Eq "D") :: Q [Dec]
There is no problem to quote twice in other file in order to get the declaration. Maybe it should called second order meta programming. But the following has problems:
> derivings'' :: Name -> Name -> Q Exp
> derivings'' cla typ = do
> let makeClassName = mkName $ "make" ++ nameBase cla
> a <- [| derive makeClassName typ |]
> return a
Main> :t derivings''
derivings'' :: Name -> Name -> Q Exp
Main> :t derivings'' ''Eq ''D
derivings'' ''Eq ''D :: Q Exp
Main> :t $(derivings'' ''Eq ''D)
<interactive>:1:3:
Illegal variable name: ‘D’
When splicing a TH expression:
Data.DeriveTH.derive makeEq (Language.Haskell.TH.Syntax.mkName Main.D)
In the splice: $(derivings'' ''Eq ''D)
In the first case I used String, it works. however, it is not workiing if I use D with a Name type. Cannot figure it out.
My second question is that if my return type is Q [Exp] while the [Exp] can be further expanded into [Dec]. How can I do it to expand [Exp] into [Dec]? Do we have a function for $ in $(thcode_with_Q_Exp) so that I do not need to splice each Exp value?
Best wishes
Song