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

[Yves Parès]

> 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 <yczhang89 at gmail.com>

> On Wed, Jan 4, 2012 at 7:58 PM, Yves Parès <limestrael at gmail.com> 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.
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.haskell.org/pipermail/haskell-cafe/attachments/20120104/892e4bc6/attachment.htm>