Re: [Haskell-cafe] Re: Re: Re: monad subexpressions

Antoine Latter wrote:

On 8/3/07, Chris Smith wrote:

...

Yes, unless of course you did:

instance (Monad m, Num n) => Num (m n)

or some such nonsense. :)

I decided to take this as a dare - at first I thought it would be easy to declare (Monad m, Num n) => m n to be an instance of Num (just lift or return the operators as necessary), but I ran into trouble once I realized I needed two things I wasn't going to get:

An instance of Eq (m n), and an instance of Show (m n) for all monads m. Eq would need a function of the form:

(==) :: Monad m => m a -> m a -> Bool

and Show would need a function of type m a -> String

What about Eq1 and Show1 classes? In the same vein as Typeable1:

class Eq1 f where eq1 :: Eq a => f a -> f a -> Bool neq1 :: Eq a => f a -> f a -> Bool

class Show1 f where show1 :: Show a => f a -> String showsPrec1 :: Show a => Int -> f a -> ShowS

Now you can declare the Num instance:

instance (Monad m, Eq1 m, Show1 m, Num n) => Num (m n) where (+) = liftM2 (+) (-) = liftM2 (-) (*) = liftM2 (*) abs = liftM abs signum = liftM signum negate = ligtM negate fromInteger = return . fromInteger

And just to show that such instances can exist:

instance Eq1 [] where eq1 = (==) neq1 = (/=)

instance Show1 [] where show1 = show showsPrec1 = showsPrec

Note: All of this is untested code. Twan