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

14 Apr 2010


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

Thomas Davie