Re: [Haskell-cafe] Re: How to "Show" an Operation?

On Friday 11 June 2010 11:50:55, Luke Palmer wrote:

On Thu, Jun 10, 2010 at 2:10 PM, Maciej Piechotka wrote:

...

data Named a = Named String a

instance Functor Named where    f `fmap` (Named s v) = Named s (f v)

instance Applicative Named where    pure x = Named "" x    (Named s f) <*> (Named t v) = Named (s ++ "(" ++ t ++ ")") (f v)

This is not technically a legal applicative instance, because it is not associative.

Good spot. I think (Named s f) <*> (Named t v) = Named (s ++ " $ " ++ t) (f v) fixes it.

This can be seen when you try to clean up the usage as we have been discussing:

g <.> f = liftA2 (.) g f

f = Named "f" (+1) g = Named "g" (*2) h = Named "h" (^3)

ghci> f <*> (g <*> (h <*> namedPure 42)) f(g(h(42))) ghci> (f <.> g <.> h) <*> namedPure 42 f(g)(h)(42)

The Applicative laws are supposed to guarantee that this refactor is legal. Of course, the latter answer is nonsense.

Luke