runST uses forall. It messes up type inference (and yes, type inference is at play even if top-level function has type signature). See, if you write "runST $ f x", then GHC deduces this: ($) :: (a -> b) -> a -> b runST :: (forall s. ST s z) -> z therefore (a -> b) ~ (forall s. ST s z) -> z therefore a ~ forall s. ST s z, b ~ z and also f x :: a therefore f x :: forall s. ST s z Then it goes on checking that f x :: forall s. ST s z actually makes sense, which it does by first stripping away forall and checking that f x :: ST s z is sound. If you write (runST . f) x, then what GHC sees is (.) :: (b -> c) -> (a -> b) -> (a -> c) runST :: (forall s. ST s z) -> z therefore b -> c ~ (forall s. ST s z) -> z therefore b ~ (forall s. ST s z), c ~ z and also f :: a -> b therefore f :: a -> forall s. ST s z Unfortunately, f is of type a -> ST s z, or, if you quantify explicitly, forall s. a -> ST s z. These two types don't match. If you write type arguments explicitly, like you'll do in Agda, you'd have (s :: Type) -> a -> ST s z vs a -> (s :: Type) -> ST s z So they are different. I don't think there is an easy way to fix that, apart from using "$". On 27 Apr 2014, at 12:27, Kai Zhang <kai@kzhang.org> wrote:
Hi Cafe,
I don't understand why this doesn't work:
import qualified Data.Vector.Generic as G import Control.Monad.ST
test :: G.Vector v a => v a -> v a test x = runST . (>>= G.freeze) . G.thaw $ x
But if I replace "." with "$", it can compile.
test x = runST $ (>>= G.freeze) . G.thaw $ x
I originally though f $ g x can always be written as f . g $ x _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe