Unfortunately, you can't derive Show on Chain as defined, because it contains a function:
> data Chain = Link Int (Int -> Chain) You can write this: > instance Show Chain where > show (Link n _) = "Link " ++ show n ++ " <fn>" Or you can make a dummy "Show" instance for functions: > instance Show (a -> b) where show _ = "<fn>" > data Chain = Link Int (Int -> Chain) deriving Show One question: Do you expect to ever call the function with a different value? For example: otherChain :: Chain otherChain = case (ints 0) of Link _ f -> f 100 If not, you can replace Chain entirely by [Int], due to laziness, something that's not possible in ML. (Although you can get the same result in ML by using (int * (() -> chain)) instead. -- ryan On Mon, May 18, 2009 at 6:36 PM, Brandon S. Allbery KF8NH <allb...@ece.cmu.edu> wrote: > On May 18, 2009, at 21:19 , michael rice wrote: > > *Main> :t ints 0 > ints 0 :: Chain > *Main> ints 0 > > <interactive>:1:0: > No instance for (Show Chain) > > In general, you want to append > deriving Show > to your types. You may also want to be able to input them in ghci, so > instead say > deriving (Show, Read) > -- > brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allb...@kf8nh.com > system administrator [openafs,heimdal,too many hats] allb...@ece.cmu.edu > electrical and computer engineering, carnegie mellon university KF8NH > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe