Error when ($) is used, but no error without

[Brian Hulley]

Robin Bate Boerop wrote:
> This code compiles properly (with -fglasgow-exts on GHC 6.4.1):
> 
> class CC a
> type C x = CC a => a x
> f, g :: C a -> Int
> f _ = 3
> g x = f x
> 
> But, this code:
> 
> class CC a
> type C x = CC a => a x
> f, g :: C a -> Int
> f _ = 3
> g x = f $ x  -- the only change
> 
> gives this error:
> 
>     Inferred type is less polymorphic than expected
>       Quantified type variable `a' escapes
>       Expected type: a a1 -> b
>       Inferred type: C a1 -> Int
>     In the first argument of `($)', namely `f'
>     In the definition of `g': g x = f $ x
> 
> What's going on here?

I think the type declaration is actually equivalent to:

type C x = forall a. CC a => a x

so that you are declaring:

f,g :: (forall a. CC a => a Int) -> Int -- not allowed

instead of:

f,g :: forall a. (CC a => a Int ->Int)