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

[Twan van Laarhoven]

Antoine Latter wrote:

> On 8/3/07, Chris Smith <cdsmith at twu.net> 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