On Sep 13, 6:25 pm, Maciej Piechotka
I started experiment with strict functors. I come to:
import Control.Exception import Foreign import Prelude hiding (catch)
data StrictMonad a = StrictMonad a deriving Show
instance Functor StrictMonad where f `fmap` StrictMonad v = return $ f v
instance Applicative StrictMonad where pure = return (<*>) = ap
instance Monad StrictMonad where return x = unsafePerformIO $ do (return $! x) `catch` \(SomeException _) -> return x return $! StrictMonad x StrictMonad v >>= f = f v
It seems to be valid IMHO Functor and Monad (I haven't prove it) as long as functions terminates.
Here, I just believe you, and assume you mean that some non- terminating function would give problems to your strict functor, i.e. it wouldn't satisfy the functor/monad laws. Then, I also wonder if the functor you have is any different from an identity functor - I see why the monad could be strict.
Some time ago there was post stating that there is not possible strict 'interesting' functor - I guess that the above is 'interesting' (and due to halting problem I guess it is not possible to create strict Functor which would deal with that problem).
I'm no expert, but since a functor on the Hask category must work on all functions available there, it looks like you "proved" that yours is not a functor and can't be fixed; maybe, that means that no strict functor exist. Your function is probably valid in a different category, containing mostly the same objects but just total functions - if it is a valid category; I wonder what would happen when arrows in this category were applied on undefined, but maybe this means that objects in this category should not contain undefined as an element; at that point, you are in a strongly normalizing programming language and probably "strict" makes no sense. Then, I would also like to understand what exactly a strict functor is, in detail, and/or a link to the post you reference. Best regards Paolo