Re: [Haskell-cafe] Defining show for a function type.

On Jul 11, 2006, at 8:27 AM, ihope wrote:

On 7/10/06, Fritz Ruehr wrote:

...

Were you interested in "seeing" the function, you could do so, at least for finite, total functions (you can also enumerate them, compare them for equality, etc.). See my haskell-cafe message at <http://www.haskell.org/pipermail/haskell-cafe/2006-April/ 015197.html>.

Hmm, interesting. How does it handle curried functions?

The trick is to define appropriate instances of Bounded, Enum, Eq, Ord and Show for function types, like this (and similarly for Bounded (a,b) and Enum (a,b)):

instance (Enum a, Bounded a, Enum b, Bounded b) => Bounded (a -> b) where ... instance (Enum a, Bounded a, Enum b, Bounded b) => Enum (a -> b) where ... instance (Enum a, Bounded a, Eq b) => Eq (a -> b) where ... instance (Enum a, Bounded a, Enum b, Bounded b, Eq b) => Ord (a -> b) where ... instance (Enum a, Bounded a, Show a, Show b) => Show (a -> b) where ...

Then curried functions of arbitrarily high order are handled automatically, so long as the "base" types are suitably Enum, Bounded, Showable, etc. In other words, a function of type (Bool,Bool) -> (Bool -> Bool) -> Bool -> Bool (or whatever) is resolved in stages, working down to the "base" type of Bool. Note of course that the Show form for such a function gives only an *extensional* description, in terms of argument-result pairs: there's no way to get the original function definition back, short of resorting to tricks (i.e, by defining something function-like, from which a function can be extracted, but which internally "stores" its means of definition). -- Fritz PS: I can email you the source if you're really interested ... but it's more instructive to at least try it yourself :) .