[Haskell-beginners] Re: Overlapping Instances

25 Oct 2010


      Thank you. Why does this code succeed for a, but stills fails on instance Show B? Do they not both invoke the same Show A?

{-# LANGUAGE TypeSynonymInstances, OverlappingInstances, IncoherentInstances #-}

import Data.Array

type A = Array Int Bool

data B = B A

instance Show A where
     show a = "foo"

instance Show B where
     show (B a) = show a

a = show $ listArray (1,3) $ repeat True

On 24/10/2010 16:37, Markus Läll wrote:
...
Hi John,
from what I gather this is because Show instance for "Array a b",
which you are overlapping, is defined in a module without the
OverlappingInstances declaration. Check the last few paragraphs of
this from the GHC's user's guide:
http://www.haskell.org/ghc/docs/6.12.2/html/users_guide/type-class-extension...
Markus Läll
On Sun, Oct 24, 2010 at 4:54 PM, John Smith  wrote:
...
Following is a simplification of some code which I have written. I have an
overlapping Show instance for A, which is more specific than the general
instance for arrays, so I would expect it to be acceptable as an overlapping
instance. Nevertheless, I get the following compiler error:
Overlapping instances for Show A
      arising from a use of `show' at 13:17-22
    Matching instances:
      instance (Ix a, Show a, Show b) =>  Show (Array a b)
        -- Defined in GHC.Arr
      instance [overlap ok] Show A -- Defined at 9:9-14
    In the expression: show a
    In the definition of `show': show (B a) = show a
    In the instance declaration for `Show B'
Compilation failed.
I've tried UndecidableInstances and IncoherentInstances, but they don't seem
to help. What am I doing wrong?
Many thanks in advance for any assistance.
-John
{-# LANGUAGE TypeSynonymInstances, OverlappingInstances #-}
import Data.Array
type A = Array Int Bool
data B = B A
instance Show A where
    show a = "foo"
instance Show B where
    show (B a) = show a