Am 20.05.2010 um 14:16 schrieb Tony Morris: > I've compared and clearly the former is significantly superior :) > > I'm rather interested if there are any sound suggestions to resolve the > general issue of retrospective type-class extension. >
I would like to have something like parent class Functor f <= Applicative f where fmap f x = pure f <*> x Then one could write instance Applicative MyApplicative deriving parent Functor where (<*>) = ... pure = ... as an abbreviation for instance Functor MyApplicative where fmap f x = pure f <*> x This way, we do not only save some keystrokes, but now it is clear that (fmap f x == pure f <*> x) is expected to hold for type MyApplicative. One could also write parent class Applicative a <= Monad a deriving parent Functor where (<*>) = ap pure = return fmap = liftM overriding the default definition of Functor's fmap. Then instance Monad MyMonad deriving parent Applicative where (>>=) = ... return = ... would be an abbreviation for instance Functor MyMonad where fmap = liftM instance Applicative MyMonad where (<*>) = ap pure = return Now the compiler can even conclude that (liftM f x == pure f <*> x) is expected to hold for type MyMonad. But there is an ambiguity if one also defines parent class Functor f <= Monad f where fmap f x = trace "boo!" (liftM f x) Then it might not be clear which definition of fmap should be used, because there are two possible paths: (Monad => Applicative => Functor) and (Monad => Functor). But then the programmer has to decide whether he writes 'deriving parent Applicative' or 'deriving parent Functor'. Thus, as long as every class or instance declaration contains at most one 'deriving parent' statement, there will always be one unambiguous path, so that this will not become a problem. This extension would have three advantages: - it is merely syntactic sugar, so that it can easily be implemented, - it does not involve tricky resolution of methods or types, so that it is easy to comprehend, and - it allows to encode knowledge about the laws class instances (should) follow. > > Miguel Mitrofanov wrote: >> That won't be a great idea; if I just want my monad to be declared as >> one, I would have to write >> >> instance Functor MyMonad where fmap = ... >> instance Pointed MyMonad where pure = ... >> instance Applicative MyMonad where (<*>) = ... >> instance Monad MyMonad where join = ... >> >> Compare this with >> >> instance Monad MyMonad where >> return = ... >> (>>=) = ... >> >> and take into account that (>>=) is usually easier to write than join. >> >> Limestraël wrote: >>> Then it would be: >>> >>> class Functor f where >>> fmap :: (a -> b) -> f a -> f b >>> >>> class (Functor f) => Pointed f where >>> pure :: a -> f a >>> >>> class (Pointed f) => Applicative f where >>> (<*>) :: f (a -> b) -> f a -> f b >>> >>> class (Applicative f) => Monad f where >>> join :: f (f a) -> f a >>> >>> This would be a great idea, for the sake of logic, first (a monad >>> which is not a functor doesn't make sense), and also to eliminate >>> redudancy (fmap = liftM, ap = (<*>), etc.) >>> >>> 2010/5/20 Tony Morris <tonymor...@gmail.com >>> <mailto:tonymor...@gmail.com>> >>> >>> Ivan Miljenovic wrote: >>>> On 20 May 2010 14:42, Tony Morris <tonymor...@gmail.com >>> <mailto:tonymor...@gmail.com>> wrote: >>>> >>>>> We all know that "class (Functor f) => Monad f" is preferable >>> but its >>>>> absence is a historical mistake. We've all probably tried once: >>>>> >>>>> instance (Functor f) => Monad f where >>>>> >>>> >>>> Do you mean the reverse of this (instance (Monad m) => Functor m >>> where) ? >>>> >>> Yes. >>> >>> -- >>> Tony Morris >>> http://tmorris.net/ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
