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 <maydw...@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.how...@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 >> >> >
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