[Haskell-cafe] Monad fold
maydwell at gmail.com
Tue Apr 16 14:55:52 CEST 2013
Wow looks like this Monoid instance isn't included in Control.Monad... My
On Tue, Apr 16, 2013 at 8:47 PM, Lyndon Maydwell <maydwell at 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 at frigidcode.com> wrote:
>> So, I'm doing something like this
>> foldl (>>=) someA list :: Monad m => m a
>> 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?
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
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