CODEWITHSUNDEEP
haskelltemplate-haskell
Patrick Collins
Patrick Collins

Reputation: 10574

Why does Template Haskell add unexpected parens?

I have the following:

import Control.Applicative
import Control.Monad
import Language.Haskell.TH


mkExp :: [Name] -> ExpQ
mkExp (name:[]) = [| ZipList $(varE name) |]
mkExp (name:names) = [| ZipList $(varE name) <*> $(mkExp names) |]

zipN :: Int -> ExpQ
zipN n = do
  names <- mapM newName $ replicate n "x"
  fn <- newName "f"
  let vps = map varP (fn:names)
  lamE vps $ [| $(varE fn) <$> $(mkExp names) |]

I'd like $(zipN 2) to generate:

\f x y -> f <$> ZipList x <*> ZipList y

so that it has the type (a -> b -> c) -> [a] -> [b] -> [c]. But poking around the output from -ddump-splices and filtering out the noise, I've found that $(zipN 2) instead generates:

\f x y -> f <$> ((ZipList x) <*> (ZipList y))

with the type (a -> b) -> [a1 -> a] -> [a1] -> ZipList b. Similarly, $(zipN 3) generates:

\f x1 x2 x3 -> (f <$> ((ZipList x1) <*> ((ZipList x2) <*> (ZipList x3)))

so it looks like each instance of $([|...|]) is being replaced with (...) rather than ..., which is surprising to me, since the docs seemed to say that pairs of $( ) and [| |] "cancelled out."

Why does Template Haskell generate this AST, and what can I do to get a function from it with the correct type?

Upvotes: 1

Views: 58

Answers (1)

Cirdec
Cirdec

Reputation: 24156

Both <$> and <*> are left associative. You are right associating them.

You can build the expression so that the operators are left-associated instead.

mkExp' :: ExpQ -> [Name] -> ExpQ
mkExp' acc [] = acc
mkExp' acc (name:names) = mkExp'' [| $(acc) <$> ZipList $(varE name) |] names
    where
        mkExp'' acc [] = acc
        mkExp'' acc (name:names) = mkExp'' [| $(acc) <*> ZipList $(varE name) |] names

zipN :: Int -> ExpQ
zipN n = do
  names <- mapM newName $ replicate n "x"
  fn <- newName "f"
  let vps = map varP (fn:names)
  lamE vps $ mkExp' (varE fn) names

Upvotes: 2

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