On Wed, 2011-05-04 at 02:00 -0400, Ken Takusagawa II wrote:

I run into the following type error:

foo :: ST s (STRef s Int) -> Int foo p = (runST (p >>= readSTRef))

with ghc 6.12.1 st.hs:8:16: Couldn't match expected type `s1' against inferred type `s' `s1' is a rigid type variable bound by the polymorphic type `forall s1. ST s1 a' at st.hs:8:9 `s' is a rigid type variable bound by the type signature for `foo' at st.hs:7:10 Expected type: ST s1 (STRef s Int) Inferred type: ST s (STRef s Int) In the first argument of `(>>=)', namely `p' In the first argument of `runST', namely `(p >>= readSTRef)'

However, if I add {-# LANGUAGE RankNTypes #-}

and change the type signature to foo :: (forall s.ST s (STRef s Int)) -> Int

it works. I don't fully understand what's going on here.

Is this the "right" way to fix the problem? Are there other options? My gut feeling is, for such a simple use case of the ST monad, I shouldn't need such a big hammer as RankNTypes.

--ken

To make the interface of ST works - i.e. to keeps it pure the signature of runST is:

runST :: (forall s. ST s a) -> a

Otherwise consider following code:

incST :: Num a => STRef s a -> ST s a incST r = readSTRef r >>= \v -> writeSTRef r (v + 1) >> return v

add :: STRef s Int -> Int -> Int add r x = runST (incST r >>= \v -> return (v + x))

test :: [Int] test = runST (newSTRef 0) >>= \r -> map (add r) [1,2,3]

What is the result? And what is the result of:

test2 :: [Int] test2 = runST (newSTRef 0) >>= \r -> map (add r) (map (add r) [1,2,3])

Regards