Re: using composition with multiple argument functions

[Dean Herington]

There has been a recent exchange on [EMAIL PROTECTED] about expressing composition of functions where the inner function takes more than one argument.  (When the inner function takes a single argument, the (.) operator does quite nicely, of course.)  Here's a way, using Haskell language extensions (options "-98" for Hugs, "-fglasgow-exts" for GHC) to express such composition: class Composable f g r | f g -> r where   ( # ) :: f -> g -> r instance Composable (a->z) a z  where   f # g  = f g instance Composable (b->z) (a->b) (a->z)  where  (f # g) a = f (g a) instance Composable (c->z) (a->b->c) (a->b->z)  where  (f # g) a b = f (g a b) instance Composable (d->z) (a->b->c->d) (a->b->c->z)  where  (f # g) a b c = f (g a b c) instance Composable (e->z) (a->b->c->d->e) (a->b->c->d->z)  where  (f # g) a b c d = f (g a b c d) instance Composable (f->z) (a->b->c->d->e->f) (a->b->c->d->e->z)  where  (f # g) a b c d e = f (g a b c d e) notelem :: (Eq a) => a -> [a] -> Bool notelem = not # elem   However, I don't understand why the following fails to compile: instance Composable (c->d->z) (a->b->(c,d)) (a->b->z)  where  (f # g) a b = let (c,d) = g a b in f c d f1, g1 :: a -> a -> (a,a) f1 c d = (d,c) g1 a b = (a,b) h1 :: (a -> a -> (a,a)) -> (a -> a -> (a,a)) -> (a -> a -> (a,a)) h1 = f1 # g1 Hugs reports: ERROR "Composition.hs" (line 52): Cannot justify constraints in explicitly typed binding *** Expression    : h1 *** Type          : (a -> a -> (a,a)) -> (a -> a -> (a,a)) -> a -> a -> (a,a) *** Given context : () *** Constraints   : Composable (b -> b -> (b,b)) (c -> c -> (c,c)) ((a -> a -> (a,a)) -> (a -> a -> (a,a)) -> a -> a -> (a,a)) GHC's report is wordier but seems to be saying about the same thing. Anyone have any ideas? --Dean Herington   John Hughes wrote: From: Martin DeMello   I played about a bit with the (.) operator, but couldn't manage, frinstance, to express   notelem :: Eq a => a -> [a] -> Bool   notelem = \x -> not . (elem x) without the lambda. Simple!     notelem = (not.) . elem (opinions differ on whether or not this is readable...) John Hughes To unsubscribe from this group, send an email to: [EMAIL PROTECTED]