A simple example showing that IO doesn't satisfy the monad laws?

Question

I've seen mentioned that IO doesn't satisfy the monad laws, but I didn't find a simple example showing that. Anybody knows an example? Thanks.

Edit: As ertes and n.m. pointed out, using seq is a bit illegal as it can make any monad fail the laws (combined with undefined). Since undefined may be viewed as a non-terminating computation, it's perfectly fine to use it.

So the revised question is: Anybody knows an example showing that IO fails to satisfy the monad laws, without using seq? (Or perhaps a proof that IO does satisfy the laws if seq is not allowed?)

Answer 1

All monads in Haskell are only monads if you exclude the weird seq combinator. This is also true for IO. Since seq is not actually a regular function but involves magic, you have to exclude it from checking the monad laws for the same reason you have to exclude unsafePerformIO. Using seq you can prove all monads wrong, as follows.

In the Kleisli category the monad gives rise to, return is the identity morphism and (<=<) is composition. So return must be an identity for (<=<):

return <=< x = x 

Using seq even Identity and Maybe fail to be monads:

seq (return <=< undefined :: a -> Identity b) () = () seq (undefined            :: a -> Identity b) () = undefined  seq (return <=< undefined :: a -> Maybe b) () = () seq (undefined            :: a -> Maybe b) () = undefined