cont_monad_example_unroll.hs

k0Plusk15 = \k -> k 0 + k 15

myCont = Cont k0Plusk15 -- Extracted the function for ease of notation later below.

myCont2 = myCont >>= (\x -> return (x * 2))
        = (Cont k0Plusk15) >>= (\x -> Cont ($ (x * 2)))
        = Cont $ \k' -> k0Plusk15 (\x -> runCont (Cont ($ (x * 2))) k')
        = Cont $ \k' -> k0Plusk15 (\x -> ($ (x * 2)) k')
        = Cont $ \k' -> k0Plusk15 (\x -> k' (x * 2))
        = Cont $ \k' -> k0Plusk15 (\x -> k' (x * 2))
        = Cont $ \k' -> (\k -> k 0 + k 15) (\x -> k' (x * 2))
        = Cont $ \k' -> (\x -> k' (x * 2)) 0 + (\x -> k' (x * 2)) 15
        = Cont $ \k' ->  k' (0 * 2) +  k' (15 * 2)
        = Cont $ \k' ->  k' 0 +  k' 30 -- Ah-ha!