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.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 > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
