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I need to generate code for the function signature with type class constraints, for example:

fun :: (Ord a) => a -> a

I'm using the following signature constructor:

SigD Name Type

So, I need to generate a type. My best guess is to use the following constructor:

ForallT [TyVarBndr] Cxt Type

It corresponds to the following declaration (as seen at Temaplate Haskell documentation): 

forall <vars>. <ctxt> => <type>

But Cxt is just a synonym for the list of Type's, and I can't find an appropriate constructor of Type for generating a type class constraint. What should I do to generate code for a type class constraint?

O.Phillips
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    I haven't tried but I think as you want `Ord a`, you need something along `AppT (ConstT (mkName "Ord")) (VarT (nkName "a"))` ... – mb14 Jun 18 '18 at 10:46
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    If I recall correctly, from TH's perspective a constraint is just a `Type`. (Constraints and types have more or less the same syntax, and with `ConstraintKinds` turned on a constraint can appear syntactically pretty much anywhere a `Type` can.) – Benjamin Hodgson Jun 18 '18 at 18:33

1 Answers1

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I believe that, as was mentioned in the comments, TH does not draw a distinction between Types and Constraints (which kind of makes sense, especially now that ConstraintKinds is around). I've generally found inspecting the quasiquoted AST for a specific, concrete, example of whatever I'm trying to implement to be useful when working on generating code with TH that the quasiquoters aren't good enough for. For example, in GHCI:

λ> :set -XTemplateHaskell
λ> import Language.Haskell.TH
λ> runQ $ [d|f :: Ord a => a -> a; f = id|]
[SigD f_1 (ForallT [] [AppT (ConT GHC.Classes.Ord) (VarT a_0)] (AppT (AppT ArrowT (VarT a_0)) (VarT a_0))),ValD (VarP f_1) (NormalB (VarE GHC.Base.id)) []]

This shows that the constraint portion is just the Name for Ord applied to the name that you use for the type variable. Note that, at least at this level, the ForallT parameter list is empty, since the a in the type signature wasn't written as an explicit forall a..

Lucy Maya Menon
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