On 9/24/10 5:35 AM, Axel Benz wrote:

Can anybody explain why this happens and how I can compose f and g?

Hint: It works fine if f is defined as an unary function.

As already mentioned: (g . f) x y = (\z-> g (f z)) x y = g (f x) y In order to get it to work you need to say that you want to pass two arguments to f. The immediate answer is ((g .) . f) but that doesn't really give you a general pattern to use. The general pattern is, -- | Binary composition. (...) :: (c -> d) -> (a -> b -> c) -> (a -> b -> d) (...) = (.) . (.) {-# INLINE (...) #-} infixl 8 ... and then (g ... f) x y = g (f x y). Note that the fixity is set up so that (...) plays nicely with (.). You may also be interested in, -- | Compose on second arg. (.^) :: (a -> c -> d) -> (b -> c) -> (a -> b -> d) (.^) = flip ... (.) . flip {-# INLINE (.^) #-} infix 9 .^ -- | Function composition which calls the right-hand -- function eagerly. (.!) :: (b -> c) -> (a -> b) -> a -> c (.!) = (.) . ($!) {-# INLINE (.!) #-} infixr 9 .! -- Live well, ~wren