[Haskell-cafe] runST readSTRef type error

[Maciej Marcin Piechotka]

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
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