Re[2]: forall a (Ord a => a-> a) -> Int is an illegal type???

Hello Brian, Thursday, February 09, 2006, 9:38:35 AM, you wrote:

...

the past few months (!) and still can't understand why the following is an illegal type:

forall a. ((Ord a => a-> a) -> Int)

i don't know right answer burt may be because "Ord a" restriction and "forall a" )"dseclaration" of type variable) should be at the same "level". imagine the following declaration: forall a. (Int -> (Ord a => a)) -> Int) it is not good to write restriction on some deep level instead of right together with "declaration" -- Best regards, Bulat mailto:bulatz@HotPOP.com