[Haskell-cafe] Monad fold

[Arseniy Alekseyev]

Hi!

Although foldM won't make things much nicer, it can be used here as well:

someA >>= \a -> foldM (flip id) a list


Cheers!
Arseniy

On 16 April 2013 13:35, 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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