Re: example of derive using Template Haskell?

I understand that '[d|...|]' creates a Q [Dec], 
but in practice GHC 6.2.2 interprets it as a 
syntax error.  Do I need to do something to enable 
it?

I've read the user_guide.  The printf example is 
obsolete.  What I would really like is an example 
somewhere of deriving a class using Template 
Haskell.

If I can't figure this out, I'll try to learn 
Drift.

-Alex-

On Tue, 11 Jan 2005, Lemmih wrote:

> Hey Alex.
>
> You write sugared Haskell inside [| |]. It's just an easier way of
> constructing the various data structures.
>
> '[| |]' creates a structure of type 'ExpQ'
> Prelude Language.Haskell.TH> runQ [| 10 |] >>= print
> LitE (IntegerL 10)
>
> '[t| |]' creates a structure of type 'TypeQ'
> Prelude Language.Haskell.TH> runQ [t| Maybe Int |] >>= print
> AppT (ConT Data.Maybe.Maybe) (ConT GHC.Base.Int)
>
> '[d| |]' creates a structure of type 'Q [Dec]'
> Prelude Language.Haskell.TH> runQ [d| foo = "bar" |] >>= print
> [ValD (VarP foo) (NormalB (LitE (StringL "bar"))) []]
>
> This information is more or less available at
> http://www.haskell.org/ghc/docs/latest/html/users_guide/template-haskell.html
>
> On Mon, 10 Jan 2005 21:45:20 -0500 (Eastern Standard Time), S.
> Alexander Jacobson <[email protected]> wrote:
> [...]
>>    fooFunc =  [d|funD "foo" [clause [] (normalB $ litE $ StringL "bar") [] ] |]
>>    goo = [|2|]
>>
>>    ---
>>
>>    {-# OPTIONS -fglasgow-exts #-}
>>    import Tth
>>    foo = $(goo)
>>    $(fooFunc)
>>
>> If I eliminate the d in "[d|" I get an error about
>> Dec conflicting with Exp.
> [...]
> When you do a toplevel splice then the variable you're splicing must
> have type :: Q [Dec].
>
> -- 
> Friendly,
>  Lemmih
>

______________________________________________________________
S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com