Yitzchak Gale wrote:
Can anyone explain the following behavior (GHCi 6.6):
Prelude Control.Monad.ST> runST (return 42) 42 Prelude Control.Monad.ST> (runST . return) 42
<interactive>:1:9: 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 means 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. Brian. -- http://www.metamilk.com