The following function executes each action from the list until a predicate returns True. I have a feeling that this function can be rewritten using fold and without the do construct but I don't know how.
execUntil :: (Eq a, Monad m) => [m a] -> (a -> Bool) -> m ()
execUntil [] p = return ()
execUntil (a:as) p = do
r <- a
if p r then return () else execUntil as p
Not totally sure what you're looking for, but maybe something like this?
import Control.Monad (void)
import Data.Bool (bool)
execUntil :: Monad m => [m a] -> (a -> Bool) -> m ()
execUntil = flip $ \p -> void . foldl (\ma mb -> ma >>= bool (p <$> mb) (pure True)) (pure False)
I wouldn't recommend it, though. foldl doesn't have any way to short-circuit, so this implementation will traverse the entire [m a], even beyond when the a -> Bool predicate is satisfied.
IMO, the code you've provided is totally fine as-is! Except perhaps for the unnecessary Eq a constraint ;p
foldMis a left fold, I don't think you can use it for this. You also can't use a non-monadic fold.execUntilis calledwhileMinControl.Monad.ExtraandControl.Monad.Loops, implemented withwhenrather thanif/then/elsebut otherwise the same. It will still do recursion under the hood but there's no way around this. \$\endgroup\$