Parameterized type gives Template Haskell error

[GoogleCodeExporter]

What steps will reproduce the problem?
1. Create a datatype with a parameter; the RHS can apparently be anything (the 
parameter needn't even be used at all)
2. Attempt to $(derive [''YourParameterizedType])

What is the expected output? What do you see instead?

Compilation fails with a message such as:

UnboundTest.hs:9:1:
    The exact Name ‘a_a1gU’ is not in scope
      Probable cause: you used a unique Template Haskell name (NameU), 
      perhaps via newName, but did not bind it
      If that's it, then -ddump-splices might be useful

What version of the product are you using? On what operating system?

RepLib 0.5.3.3; tried on both GHC 7.6.3 and GHC 7.8.3.

Please provide any additional information below.

See attached test case.

Original issue reported on code.google.com by jyr...@gmail.com on 12 Oct 2014 at 11:50

Attachments:

• UnboundTest.hs

[GoogleCodeExporter]

Deriving with Template Haskell requires a few language extensions for the 
generated code.  Changing the language pragma in UnboundTest.hs to read

{-# LANGUAGE TemplateHaskell, 
   ScopedTypeVariables, FlexibleInstances,MultiParamTypeClasses, 
   FlexibleContexts #-}

should do the trick.

Original comment by stephanie.weirich on 14 Oct 2014 at 1:48

• Changed state: WontFix

[GoogleCodeExporter]

!! I *thought* I'd gotten it to work before :-\ Pardon the spam. (I don't 
suppose it's possible to generate the code so it fails more helpfully in this 
case?)

Original comment by jyr...@gmail.com on 14 Oct 2014 at 3:18

[GoogleCodeExporter]

I don't think it is possible. It would useful if TH could generate the 
appropriate LANGUAGE pragmas, but that also seems a bit strange too.

Alternatively, there's a recent reimplementation of unbound using GHC.Generics 
instead of RepLib. In that case, no TH is required.
https://github.com/lambdageek/unbound-generics

Original comment by stephanie.weirich on 14 Oct 2014 at 11:48