Message: 6 Date: Fri, 15 Sep 2006 18:36:35 +0300 From: "Anatoly Zaretsky" Subject: Re: [Haskell-cafe] Why is type 'b' forced to be type 'm a' and not possibly 'm a -> m a' To: "Vivian McPhail" Cc: Haskell Cafe Message-ID: Content-Type: text/plain; charset=ISO-8859-1; format=flowed

On 9/15/06, Vivian McPhail wrote:

...

class Forkable a where fork :: String -> a -> a -> a

... {- instance (Monad m, Forkable (m a), Forkable b) => Forkable (m a -> b)

where

...

fork n a1 a2 a = do a' <- a fork n (a1 $ return a') (a2 $ return a') -}

Let's do manual type checking. First, fork :: Forkable a => String -> a -> a -> a So for Forkable (m a -> b) fork :: String -> (m a -> b) -> (m a -> b) -> m a -> b Then fork n a1 a2 a :: b But you define it as fork n a1 a2 a = do {...} So it should be of type Monad t => t a, not just any `b'.

Instead, you can define instance (Monad m, Forkable (m b)) => Forkable (m a -> m b) where ...

Well, I can partially instantiate what I am trying to achieve by enumerating cases. Note that when the first type is a monadic type the computation gets evaluated and then forked, but when the first type is a function it merely gets passed. My problem is that there are a very large number of possible cases. So in the case Forkable (m a -> b), a number of instances of which I can implement (e.g. Forkable (m a -> m a -> m a), Forkable ((m a -> m a) -> m a), and Forkable (m a -> (m a -> m a) -> m a)), I don't see why 'b' should necessarily typecheck to 't t1'. What I would like to be able to do is differentiate between Forkable (m a -> b) and Forkable (<function type> -> b). By the way, the following code typechecks and runs correctly, my problem is that enumerating all possible types requires five factorial (120) different instances, and to a lazy functional programmer who can 'see' the pattern it seems that there must be a nicer way of achieving my end. \begin{code} instance (Monad m, Forkable (m a)) => Forkable (m a -> m a) where fork n a1 a2 a = do a' <- a fork n (a1 $ return a') (a2 $ return a') instance (Monad m, Forkable (m a)) => Forkable (m a -> m a -> m a) where fork n a1 a2 a b = do a' <- a fork n (a1 $ return a') (a2 $ return a') b instance (Monad m, Forkable (m a)) => Forkable ((m a -> m a) -> m a) where fork n a1 a2 a = do fork n (a1 a) (a2 a) instance (Monad m, Forkable (m a)) => Forkable (m a -> (m a -> m a) -> m a) where fork n a1 a2 a b = do a' <- a fork n (a1 $ return a') (a2 $ return a') b \end{code}

Note that to compile it you also need -fallow-undecidable-instances and -fallow-overlapping-instances.

-- Tolik

Thanks for your help so far!

Cheers, Vivian -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.1.405 / Virus Database: 268.12.4/449 - Release Date: 15/09/2006