Template Haskell compile error

Question

Consider the following code:

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5

data Hello1 = Hello1
data Hello2 = Hello2

record = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord

f1 = $([| (\r1 -> (r1 .!. Hello1)) |]) 

main = print $ f1 record

This compiles fine and prints out "Hello1" as expected.

However, adding the following line (GHC 7.4.1) gives a compile error:

f2 = $([| (\r2 -> (r2 .!. Hello2)) |])

The error given is:

error.hs:16:1:
    Could not deduce (Data.HList.Record.HasField Hello2 r0 v0)
      arising from the ambiguity check for `main'
    from the context (Data.HList.Record.HasField Hello2 r v)
      bound by the inferred type for `main':
                 Data.HList.Record.HasField Hello2 r v => IO ()
      at error.hs:(16,1)-(20,38)
    Possible fix:
      add an instance declaration for
      (Data.HList.Record.HasField Hello2 r0 v0)
    When checking that `main'
      has the inferred type `forall r v.
                             Data.HList.Record.HasField Hello2 r v =>
                             IO ()'
    Probable cause: the inferred type is ambiguous

error.hs:16:1:
    Could not deduce (Data.HList.Record.HasField Hello2 r0 v0)
      arising from the ambiguity check for `f1'
    from the context (Data.HList.Record.HasField Hello2 r v)
      bound by the inferred type for `f1':
                 Data.HList.Record.HasField Hello2 r v =>
                 Data.HList.Record.Record
                   (Data.HList.HListPrelude.HCons
                      (Data.HList.Record.LVPair Hello1 [Char])
                      (Data.HList.HListPrelude.HCons
                         (Data.HList.Record.LVPair Hello2 [Char])
                         Data.HList.HListPrelude.HNil))
                 -> [Char]
      at error.hs:(16,1)-(20,38)
    Possible fix:
      add an instance declaration for
      (Data.HList.Record.HasField Hello2 r0 v0)
    When checking that `f1'
      has the inferred type `forall r v.
                             Data.HList.Record.HasField Hello2 r v =>
                             Data.HList.Record.Record
                               (Data.HList.HListPrelude.HCons
                                  (Data.HList.Record.LVPair Hello1 [Char])
                                  (Data.HList.HListPrelude.HCons
                                     (Data.HList.Record.LVPair Hello2 [Char])
                                     Data.HList.HListPrelude.HNil))
                             -> [Char]'
    Probable cause: the inferred type is ambiguous

Why does adding the f2 line result in a compile error?

Note: The Template Haskell parts may look silly here, but they are a simplification of more complex Template Haskell which does work on tuples. I've posted the simplest example I could construct that still exhibited the error. I realise removing the Template Haskell fixes the issue in this case, but that isn't an option in my real code.

Edit:

In addition, the following fails to compile. Why is this the case:

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5

data Hello1 = Hello1
data Hello2 = Hello2
data Hello3 = Hello3

record1 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord
record2 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. (Hello3 .=. "Hello3") .*. emptyRecord

f1 = $([| (\r1 -> (r1 .!. Hello1)) |]) 

main = print $ (f1 record1, f1 record2)

After adding the definition for f2, do you use it anywhere? If not, does it work if you give f2 an explicit type annotation? — Daniel Wagner
– Daniel Wagner, Commented Aug 27, 2012 at 12:20
@DanielWagner: Using f2 seems to help. However, I've found other issues. See stackoverflow.com/questions/12144250/… an example of the issue that doesn't involve HList. — Clinton
– Clinton, Commented Aug 27, 2012 at 14:48
@DanielWagner: On second thoughts, perhaps that question is a red herring, though it may be related. I've found a workaround and put it as an answer to my own question, it's not ideal, but workable solution for my problem. — Clinton
– Clinton, Commented Aug 27, 2012 at 15:53

huon · Accepted Answer · 2012-08-27 17:06:12Z

I've found giving your top level functions type signatures fixes any issues. See the code below:

{-# LANGUAGE TemplateHaskell #-}

module X where
  import Data.HList.GhcSyntax((.!.))

  f = [| (\x r -> (r .!. x)) |]

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}
{-# LANGUAGE FlexibleContexts #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5
import X
import Data.HList.Record (HasField)

data Hello1 = Hello1
data Hello2 = Hello2
data Hello3 = Hello3

record1 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord
record2 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. (Hello3 .=. "Hello3") .*. emptyRecord

g1 :: (HasField Hello1 a b) => a -> b -- Type signature here
g1 = $(f) Hello1

g2 :: (HasField Hello2 a b) => a -> b -- Type signature here
g2 = $(f) Hello2

main = print $ (g1 record1, g2 record1, g1 record2, g2 record2)

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