Hi Paul, Paul Brauner wrote:
Hi,
I'm trying to get a deep feeling of Functors (and then pointed Functors, Applicative Functors, etc.). To this end, I try to find lawless instances of Functor that satisfy one law but not the other.
I've found one instance that satisfies fmap (f.g) = fmap f . fmap g but not fmap id = id: [...] But I can't come up with an example that satifies law 1 and not law 2. I'm beginning to think this isn't possible but I didn't read anything saying so, neither do I manage to prove it.
I'm sure someone knows :)
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 -- Steffen