Re: Stack usage with a state monad

Hi, I think the problem is in the State Monad itself; State Monad is lazy to compute its state. I am not a haskell expert, and there may be better ideas. But anyhow, when I use these >>>= and >>> instead of >>= and >>, your example runs fine. I hope it becomes some help. m >>>= k = State $ \s -> let (a, s') = runState m s in s `seq` runState (k a) s' -- force evaluation of the state m >>> k = m >>>= \_ -> k -- Koji Nakahara