Template Haskell error when using "deriving"

haskell, template-haskell

Solution

Looks like this was a bug in GHC that was fixed in version 7.6.1:

$ ~/bin/ghc-7.6.1/bin/ghci
GHCi, version 7.6.1: http://www.haskell.org/ghc/  :? for help
[...]
Prelude> :set -XTemplateHaskell
Prelude> import Language.Haskell.TH
Prelude Language.Haskell.TH> runQ [d| data X = X deriving (Show) |]
[...]
[DataD [] X_0 [] [NormalC X_1 []] [GHC.Show.Show]]

Problem

A little confused with the results I'm getting. Using the following: ``` GHCi, version 7.4.2: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Prelude> :m + Language.Haskell.TH ``` I get this successful result: ``` Prelude Language.Haskell.TH> runQ [d| data X = X |] [DataD [] X_0 [] [NormalC X_1 []] []] ``` But I get this error: ``` Prelude Language.Haskell.TH> runQ [d| data X = X deriving Show |] <interactive>:4:30: The exact Name `X' is not in scope Probable cause: you used a unique name (NameU) in Template Haskell but did not bind it In the Template Haskell quotation [d| data X = X deriving (Show) |] In the first argument of `runQ', namely `[d| data X = X deriving (Show) |]' In the expression: runQ [d| data X = X deriving (Show) |] ``` I can't derive Show when using Template Haskell?

Original source