Hmmm, I generally try to avoid using seq. It's a hack and often one can refrain from using it. Here's how I would solve the problem: First of all your monad constains an int which is passed around. Typically we want this to be computed strictly. The way to do this is to add a strictness annotation. A naive try would be this: newtype M a = M (Int -> (a,!Int)) Unfortunately we cannot have stricness annotation in this way. So to achieve this effect we make a new data type for pairs with a strict int: data PairInt a = P a !Int Secondly, your bind is lazy. This can be usedful but usually one wants it to be strict because then we get tail-recursion. The lack of tail-recursion is the second problem in this example. Strictness is achieved be converting let to case like this: M f >>= k = M $ \i -> case f i of P x i2 -> case k x of M f2 -> f2 i2 This version doesn't have any problems with spaceleaks. It can now swallow pretty large numbers, I've tried with 1000000. I hope this sheds some light on the problem. /Josef On Fri, 15 Nov 2002, Magnus Lindberg wrote:
Hi again! Thanks for your help, Malcolm. I added `seq' at two places to prevent laziness and now I never run out of memory.
instance Monad M where return x = M $ \i -> (x,i) M f >>= k = M $ \i -> let (x,i2) = i `seq' f i M f2 = x `seq' k x in f2 i2
(however, I think that just one of the two `seq' should do but if I remove any single `seq' then I still run out of memory :( )
Kind regards Magnus _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell