Tomasz Zielonka wrote:
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 = ().
Oops, you're right. I spoke too fast.
It's only a monoid for (). Otherwise you can't hope to have a right
identity.
Jules
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