Re: #4189: Add (<.>) operator (generalizing (.) to Functor)

On 01/08/10 16:05, Felipe Lessa wrote:

+1

What should be its fixity?

I guess right: \a b c -> a <.> b <.> c :: (Functor f) => (a1 -> b) -> (a11 -> a1) -> (a -> f a11) -> a -> f b As opposed to: \a b c -> a <.> b <.> c :: (Functor f, Functor f1) => (a1 -> b) -> (a11 -> f a1) -> (a -> f1 a11) -> a -> f1 (f b) Regards