16 May
2007
16 May
'07
8:38 a.m.
On Wed, May 16, 2007 at 09:28:31AM +0100, Jules Bean wrote:
Tomasz Zielonka wrote:
You mean using the (Monoid b) => Monoid (a -> b) instance ? I can see that IO () makes a perfect Monoid, but there doesn't seem to be a standard instance for that.
Indeed, all Monads are Monoids (that is, if m :: * -> * is a Monad, then m a :: * is a Monoid, for any fixed type a) by using >>.
Are you sure that (IO Int) is a monoid with mappend = (>>)? How do you define mempty, so it is an identity for mappend? It would help if type a was a Monoid, then: mempty = return mempty mappend mx my = do x <- mx y <- my return (x `mappend` y) It's easier if a = (). Regards Tomek