Quoting "Matthew Steele" <mdste...@alum.mit.edu>:
1) bar ifn = case ifn of IntFn fn -> foo fn
2) bar ifn = foo (case ifn of IntFn fn -> fn)
I can't help feeling like maybe I am missing some small but
important piece from my mental model of how rank-2 types work.
As SPJ suggested, translation to System F with explicit type
applications makes the issue clearer:
1) bar = \(ifn :: forall a. IntFn a).
case ifn _a of IntFn (fn :: _a -> Int) -> foo (/\a. ???)
2) bar = \(ifn :: forall a. IntFn a).
foo (/\a. case ifn a of IntFn (fn :: a -> Int) -> fn)
Lauri
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
