On Tue, Jan 5, 2010 at 8:22 AM, Brent Yorgey
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
As an aside, for clarity, this class does NOT correspond to the categorical notion of "endofunctor." I don't think any such identification was Brent's intent, I just want to avoid potential confusion.