Re: [Haskell-cafe] Solved but strange error in type inference

I expected the type of 'x' to be universally quantified, and thus can be unified with 'forall a. a' with no problem

As I get it. 'x' is not universally quantified. f is. [1] x would be universally quantified if the type of f was : f :: (forall a. a) -> (forall a. a) [1] Yet here I'm not sure this sentence is correct. Some heads-up from a type expert would be good. Would you try: f :: a -> a f x = undefined :: a And tell me if it works? IMO it doesn't. 2012/1/4 Yucheng Zhang

On Wed, Jan 4, 2012 at 7:58 PM, Yves Parès wrote:

...

f :: forall a. a -> a f x = x :: forall a. a

Which is obviously wrong: when you have entered f, x has been instatiated to a specific type 'a', and then you want it to x to be of any type? That doesn't make sense.

I did not expect the type variables to be scoped.

I expected the type of 'x' to be universally quantified, and thus can be unified with 'forall a. a' with no problem.