forall a (Ord a => a-> a) -> Int is an illegal type???
brianh at metamilk.com
Thu Feb 9 01:38:35 EST 2006
Brian Hulley wrote:
> Hi -
> I've been puzzling over section 220.127.116.11 of the ghc users manual for
> the past few months (!) and still can't understand why the following
> is an illegal type:
> 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?
> Thanks, Brian.
A better way of putting the question is that I can't see the difference
forall a. ((Ord a => a->a) -> Int)
forall a. Ord a => (a->a) -> Int
More information about the Glasgow-haskell-users