[Haskell-cafe] Re: instance Eq (a -> b)

[Thomas Davie]

Your instances of Finite are not quite right:

bottom :: a
bottom = doSomethingToLoopInfinitely.

instance Finite () where
 allValues = [(), bottom]

instance Finite Nothing where
 allValues = [bottom]

Though at a guess an allValuesExculdingBottom function is also useful, perhaps the class should be

class Finite a where
 allValuesExcludingBottom :: [a]

allValues :: Finite a => [a]
allValues = (bottom:) . allValuesExcludingBottom

Bob

On 14 Apr 2010, at 08:01, Ashley Yakeley wrote:

> Joe Fredette wrote:
>> this is bounded, enumerable, but infinite.
> 
> The question is whether there are types like this. If so, we would need a new class:
> 
> class Finite a where
>   allValues :: [a]
> 
> instance (Finite a,Eq b) => Eq (a -> b) where
>    p == q = fmap p allValues == fmap q allValues
> 
> instance (Finite a,Eq a) => Traversable (a -> b) where
>    sequenceA afb = fmap lookup
>      (sequenceA (fmap (\a -> fmap (b -> (a,b)) (afb a)) allValues))
>     where
>      lookup :: [(a,b)] -> a -> b
>      lookup (a,b):_ a' | a == a' = b
>      lookup _:r a' = lookup r a'
>      lookup [] _ = undefined
> 
> instance Finite () where
>   allValues = [()]
> 
> data Nothing
> 
> instance Finite Nothing where
>   allValues = []
> 
> -- 
> Ashley Yakeley
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