Re: [Haskell-cafe] problem using ST monad

On Sat, Oct 25, 2008 at 11:55 PM, Paul L wrote:

Tnaks for the clarification, please see my further questions below

On 10/25/08, Daniel Fischer wrote:

...

Sure, (g (flip readArray 0)) :: ST s Int, or, explicitly, forall s. ST s Int, there's nothing to restrict the s, so it's legitimate to pass it to runST.

..[snipped]..

...

What would be a generic mapST, which type should it have?

I tried this type for mapST, it doesn't work:

mapST :: (a -> ST s b) -> [a] -> [b] mapST f (x:xs) = runST (f x) : mapST f xs mapST f [] = []

By your reasoning, (f x) should have type forall s . ST s b, and should match what runST expects, but apparently GHC complains about it. Why?

From the perspective of code inside the definition of mapST, f is not

The problem is that the type variable s is determined by the caller of mapST. If you write the type with explicit foralls, you get: mapST :: forall a b s. (a -> ST s b) -> [a] -> [b] polymorphic, because a, b, and s have already been determined. The type you need is, mapST :: forall a b. (forall s. a -> ST s b) -> [a] -> [b] Now runST is able to pass an arbitrary type for s to f. (GHC can silently convert between "forall s. a -> ST s b" and "a -> forall s. ST s b".) It may be helpful to rewrite the types with a more explicit notation. For example, runST :: (a :: *) -> ((s :: *) -> ST s a) -> a mapST_wrong :: (a :: *) -> (b :: *) -> (s :: *) -> (f :: a -> ST s b) -> [a] -> [b] mapST_right :: (a :: *) -> (b :: *) -> (f :: (s :: *) -> a -> ST s b) -> [a] -> [b] -- Dave Menendez http://www.eyrie.org/~zednenem/