Template Haskell: Stage Error

Question

I'm trying to understand how the Quasi Quoter generates TH structures. So I'm trying to convert the first example from Template Meta-programming for Haskell, from a quoted format to just the types.

gen :: [Format] -> ExpQ -> ExpQ
gen []       x = x
gen (D:xs)   x = [| \n -> $(gen xs [| $x ++ show n |]) |]
gen (S:xs)   x = [| \s -> $(gen xs [| $x ++ s |]) |]
gen (L s:xs) x = gen xs [| $x ++ $(THS.lift s) |]

gen (D:xs) x would convert to

gen (D:xs)  x = lamE [varP $ mkName "n"]
                    (appE (appE (varE 'gen) (varE 'xs))
                        (uInfixE (x) (varE '(Prelude.++))
                            (appE (varE 'Prelude.show) (varE $ mkName "n"))))

However, this fragment will not compile the reference to 'xs.

GHC spits out the following error:

src/Print/Default.hs:28:51:
    Stage error: the non-top-level quoted name 'xs
    must be used at the same stage at which is is bound
    In the Template Haskell quotation 'xs
    In the first argument of `varE', namely 'xs
    In the second argument of `appE', namely `(varE 'xs)'

Is there some other way that you need to reference the name of a function parameter for Template Haskell to be able to use it?

That conversion is not correct. Notice that in the first version, locally bound names which are quoted are also spliced back in. So if `x` is locally bound, you can't have `$x`, but you can have `[| ... $x ... |]`. In the 2nd version, you haven't done this, and it doens't make sense to do this, since `lamE [varP ..]` expects a `ExpQ` but the `\n ->` inside of the splice expects a `String`. — user2407038, Jun 02 '14 at 20:18

score 1 · Accepted Answer · answered Jun 02 '14 at 23:13

user2407038 is right that the translation is off. This is a more correct translation:

    gen (D:xs)   x = lamE [varP (mkName "n")] $
                        gen xs $
                              varE '(++)
                                    `appE` x
                                    `appE` (varE 'show `appE` varE (mkName "n"))

You don't reference the name of xs or gen here: you just recursively call the function to generate a bit more AST. That keeps everything in the same stage.

Template Haskell: Stage Error

1 Answers1