[Haskell-cafe] Monad fold

[Lyndon Maydwell]

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 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
>>
>> 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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>
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