I believe I found a simple sample for a ccc, based on the answers from Tom Leinster, Finn Lawler and John Bourke, and http://opus.bath.ac.uk/23104[1]

I also used the fact I found in Moggi's "Computational lambda-calculus and monads" - that a category should be [well-pointed][2].

Take category **2** (two objects, three arrows), and a topos Set<sup>**2**</sup>. This topos is obviously not well-pointed, so we can proceed. Take a monad <code>M</code> that is similar to the one described in http://opus.bath.ac.uk/23104/[1]. Namely, <code>(a: a<sub>0</sub> &rarr; a<sub>1</sub>)</code> maps to <code>a + (0 &rarr; a<sub>0</sub>)</code>, with obvious unit and multiplication.
Now this monad is not strong. Suppose it were, then for <code>a: a<sub>0</sub> &rarr; a<sub>1</sub></code> and <code>b: b<sub>0</sub> &rarr; b<sub>1</sub></code>, the strength <code>a &times; M[b] &rarr; M[a &times; b]</code> would involve specifying a map from <code>a<sub>1</sub></code> to <code>a<sub>0</sub> &times; b<sub>0</sub> + a<sub>1</sub> &times; b<sub>1</sub></code>. This map cannot be <code>a &times; b<sub>1</sub></code>, because in this case it won't be preserving the tensor product's (which is Cartesian in our case) unit. And what if <code>b<sub>0</sub></code> is empty.

I believe this kind of topos would be a good testing area for the favorite Haskell constructs. Some of them won't hold, I believe.

Now I wonder... can we prove that if all monads over a topos are strong, then the topos is Boolean? Will post it in another question.

  [1]: http://opus.bath.ac.uk/23104/
  [2]: http://en.wikipedia.org/wiki/Well-pointed_category