[Haskell-cafe] How to understand the 'forall' ?
frodo at theshire.org
Wed Sep 16 04:04:48 EDT 2009
On Tue, Sep 15, 2009 at 11:38 PM, Daniel Fischer
<daniel.is.fischer at web.de> wrote:
>> foo :: forall a. a -> a
> This is exactly the same type as
> foo :: a -> a
> (unless you're using ScopedTypeVariables and there's a type variable a in scope), since
> type signatures are implicitly forall'd.
Yep, perhaps I used the wrong example. What about foo: (forall a. a) -> Int?
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