Past exam of the computer science course of the University of Oxford / 2015 / Principles of programming languages / 1 / g

Fix an answer type and let the continuation return its answer inside the base computational monad:

type ContT m a = (a -> m Answer) -> m Answer

result :: Monad m => a -> ContT m a
result x k = k x

bind :: Monad m => ContT m a -> (a -> ContT m b) -> ContT m b
bind p f k = p (\x -> f x k)

The continuation monad transformer changes the continuation supplied to p: it first runs f on the produced value, then uses the original final continuation k. Base sequencing is unnecessary in these two definitions because p already accepts a continuation returning m Answer. The customary lift, if desired, is lift q k = bindM q k.