Wow looks like this Monoid instance isn't included in Control.Monad... My mistake.


On Tue, Apr 16, 2013 at 8:47 PM, Lyndon Maydwell <maydwell@gmail.com> wrote:
You could do:

runKleisli . mconcat . map Kleisli :: Monoid (Kleisli m a b) => [a -> m b] -> a -> m b

Would that work for you?


On Tue, Apr 16, 2013 at 8:35 PM, Christopher Howard <christopher.howard@frigidcode.com> wrote:
So, I'm doing something like this

foldl (>>=) someA list :: Monad m => m a

where
  list :: Monad m => [a -> m a],
  someA :: Monad m => m a

Is there a more concise way to write this? I don't think foldM is what I
want -- or is it?

--
frigidcode.com


_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe