forall a (Ord a => a-> a) -> Int is an illegal type???
brianh at metamilk.com
Thu Feb 9 01:29:04 EST 2006
I've been puzzling over section 22.214.171.124 of the ghc users manual for the past
few months (!) and still can't understand why the following is an illegal
forall a. ((Ord a => a-> a) -> Int)
(forall a. Ord a => a->a) -> Int
is legal. I can understand why the second one *is* legal but I can't seem to
understand why the first syntax is not just exactly the same thing even
though the parse tree is different.
Can anyone shed some light on this?
More information about the Glasgow-haskell-users