Re: [Haskell-cafe] lawless instances of Functor

On Tue, Jan 5, 2010 at 7:49 AM, Steffen Schuldenzucker wrote:

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

And what is the type of f here?