forall quantifier

[Tomasz Zielonka]

On Wed, Jun 04, 2003 at 04:43:10PM +0200, Markus.Schnell@infineon.com wrote:
> Yes, this would work, thanks.
> 
> But let me extent my question: what if all the types would be in a 
> class FakeClass which has function specialID :: a -> ID and
> I would like to do
> 
> > foo $ map specialID [a,b,c,d]
> ?

Well, in Template Haskell you can do:

  {-# OPTIONS -fglasgow-exts #-}

  module B where

  import Language.Haskell.THSyntax

  data AnyShow = forall s. Show s => AnyShow s

  instance Show AnyShow where
      show (AnyShow s) = show s

  metaMap :: ExpQ -> ExpQ -> ExpQ
  metaMap qf ql = do
      l <- ql
      case l of
	  Infix (Just x) (Con "GHC.Base::") (Just xs) -> do
	      let ys = metaMap qf (return xs)
	      [| ($qf $(return x) : $ys) |]
	  Con "GHC.Base:[]" -> do
	      [| [] |]
	  ListExp xs -> do
	      ys <- sequence [ [| $qf $(return x) |] | x <- xs ]
	      return $ ListExp ys
	  other -> do
	      fail $ "metaMap: unexpected syntax: " ++ show other

and then:

    map show $( metaMap [| AnyShow |] [| [1, 1.3, [1,2,3,4]] |]

and

    map show $( metaMap [| AnyShow |] [| [1, 1.3, 'a'] |]

will work, but this won't

    map show $( metaMap [| AnyShow |] [| [1, 1.3, 'a', [1,2,3,4]] |]

Is there a way to turn of typing within meta quotations?

Best regards,
Tom

-- 
.signature: Too many levels of symbolic links