Re: Stack usage with a state monad

On Wed, Dec 31, 2003 at 02:54:18AM +0900, Koji Nakahara wrote:

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

Ahh, right. So I didn't have to use UnboxedState. StrictState would do. Best regards, Tom -- .signature: Too many levels of symbolic links