Good examples of Not a Functor/Functor/Applicative/Monad?

A type constructor which is not a Functor:

newtype T a = T (a -> Int)

You can make a contravariant functor out of it, but not a (covariant) functor. Try writing fmap and you'll fail. Note that the contravariant functor version is reversed:

fmap      :: Functor f       => (a -> b) -> f a -> f b
contramap :: Contravariant f => (a -> b) -> f b -> f a

A type constructor which is a functor, but not Applicative:

I don't have a good example. There is Const, but ideally I'd like a concrete non-Monoid and I can't think of any. All types are basically numeric, enumerations, products, sums, or functions when you get down to it. You can see below pigworker and I disagreeing about whether Data.Void is a Monoid;

instance Monoid Data.Void where
    mempty = undefined
    mappend _ _ = undefined
    mconcat _ = undefined

Since _|_ is a legal value in Haskell, and in fact the only legal value of Data.Void, this meets the Monoid rules. I am unsure what unsafeCoerce has to do with it, because your program is no longer guaranteed not to violate Haskell semantics as soon as you use any unsafe function.

See the Haskell Wiki for an article on bottom (link) or unsafe functions (link).

I wonder if it is possible to create such a type constructor using a richer type system, such as Agda or Haskell with various extensions.

A type constructor which is an Applicative, but not a Monad:

newtype T a = T {multidimensional array of a}

You can make an Applicative out of it, with something like:

mkarray [(+10), (+100), id] <*> mkarray [1, 2]
  == mkarray [[11, 101, 1], [12, 102, 2]]

But if you make it a monad, you could get a dimension mismatch. I suspect that examples like this are rare in practice.

A type constructor which is a Monad:

[]

About Arrows:

Asking where an Arrow lies on this hierarchy is like asking what kind of shape "red" is. Note the kind mismatch:

Functor :: * -> *
Applicative :: * -> *
Monad :: * -> *

but,

Arrow :: * -> * -> *

My style may be cramped by my phone, but here goes.

newtype Not x = Kill {kill :: x -> Void}

cannot be a Functor. If it were, we'd have

kill (fmap (const ()) (Kill id)) () :: Void

and the Moon would be made of green cheese.

Meanwhile

newtype Dead x = Oops {oops :: Void}

is a functor

instance Functor Dead where
  fmap f (Oops corpse) = Oops corpse

but cannot be applicative, or we'd have

oops (pure ()) :: Void

and Green would be made of Moon cheese (which can actually happen, but only later in the evening).

(Extra note: Void, as in Data.Void is an empty datatype. If you try to use undefined to prove it's a Monoid, I'll use unsafeCoerce to prove that it isn't.)

Joyously,

newtype Boo x = Boo {boo :: Bool}

is applicative in many ways, e.g., as Dijkstra would have it,

instance Applicative Boo where
  pure _ = Boo True
  Boo b1 <*> Boo b2 = Boo (b1 == b2)

but it cannot be a Monad. To see why not, observe that return must be constantly Boo True or Boo False, and hence that

join . return == id

cannot possibly hold.

Oh yeah, I nearly forgot

newtype Thud x = The {only :: ()}

is a Monad. Roll your own.

Plane to catch...