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
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
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