[Haskell-cafe] Composing functions with runST
brianh at metamilk.com
Mon Jan 1 06:35:18 EST 2007
Yitzchak Gale wrote:
> Can anyone explain the following behavior (GHCi 6.6):
> Prelude Control.Monad.ST> runST (return 42)
> Prelude Control.Monad.ST> (runST . return) 42
> Couldn't match expected type `forall s. ST s a'
> against inferred type `m a1'
> In the second argument of `(.)', namely `return'
> In the expression: (runST . return) 42
> In the definition of `it': it = (runST . return) 42
Section 7.4.8 of GHC manual states that a type variable can't be
instantiated with a forall type, though it doesn't give any explanation why.
Hazarding a guess, I suggest it *might* be due to the fact that
forall s. ST s a
forall s. (ST s a)
whereas you'd need it to mean
(forall s. ST s) a
in order for it to unify with (m a).
Just a guess - I'd be interested to know the real reason as well.
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