[Haskell-beginners] Re: Overlapping Instances
John Smith
voldermort at hotmail.com
Mon Oct 25 16:00:34 EDT 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-extensions.html#instance-overlap
>
>
> Markus Läll
>
> On Sun, Oct 24, 2010 at 4:54 PM, John Smith<voldermort at hotmail.com> 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
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