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
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] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH
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