On Mon, Jan 04, 2010 at 11:49:33PM +0100, Steffen Schuldenzucker wrote:
data Foo a = Foo a
instance Functor Foo where fmap f (Foo x) = Foo . f . f $ x
Then:
fmap id (Foo x) == Foo . id . id $ x == Foo x
fmap (f . g) (Foo x) == Foo . f . g . f . g $ x fmap f . fmap g $ (Foo x) == Foo . f . f . g . g $ x
Now consider Foo Int and
fmap ((+1) . (*3)) (Foo x) == Foo $ (x * 3 + 1) * 3 + 1 == Foo $ x * 9 + 4 fmap (+1) . fmap (*3) $ (Foo x) == Foo $ x * 3 * 3 + 1 + 1 == Foo $ x * 9 + 2
As others have pointed out, this doesn't typecheck; but what it DOES show is that if we had a type class class Endofunctor a where efmap :: (a -> a) -> f a -> f a then it would be possible to write an instance for which efmap id = id but efmap (f . g) /= efmap f . efmap g. The difference is that with the normal Functor class, once you have applied your function f :: a -> b to get a b, you can't do anything else with it, since you don't know what b is. With the Endofunctor class, once you have applied f :: a -> a, you CAN do something with the result: namely, apply f again. -Brent