Functional dependencies question
Marcin 'Qrczak' Kowalczyk
Mon, 19 May 2003 17:02:48 +0200
Dnia wto 13. maja 2003 09:58, Simon Peyton-Jones napisał:
> Think of it like this. Should this be acceptable?
> f :: a -> a -> a
> f x y = x && y
> No, because the type (forall a. a->a->a) is plainly more general than
> the actual function.
It's not the same. The type doesn't constrain a by any class, so it would look
like any type fits, which is not true.
> bar :: (Foo Char t) => t
In this case the type seems OK: for any type t, *if* Foo Char t, then bar can
be used as type t.
__("< Marcin Kowalczyk