Thanks all, this all makes sense!
On Tue, Mar 2, 2021 at 4:34 PM Seph Shewell Brockway
Is there a way to somehow convince GHC of that fact so that g typechecks?
No.
First, it would actually be unsound to do so, because of the possibility of exotic types, built with pathological combinations of type and data families: https://gitlab.haskell.org/ghc/ghc/-/issues/14420 < https://gitlab.haskell.org/ghc/ghc/-/issues/14420>
But, actually, the bigger problem is that we need a class constraint in order to allow a function to compute at runtime. The function f actually takes two arguments at runtime: a representation of the instance which carries f's implementation (this is sometimes called a dictionary), and the normal argument of type a. `g`, on the other hand, has no access to the dictionary needed at runtime, and so it's unclear how it should compute.
Put another way: a value of type Foo carries no information (beyond the fact that it terminates), because Foo has only one data constructor. So
On Tue, Mar 02, 2021 at 03:00:55PM +0000, Richard Eisenberg wrote: there's no way that g :: Foo a -> Int could be anything but a constant function. You need the class constraint to change this fact.
Hope this helps! Richard
To elaborate a little on Richard’s answer, this is the reason for GHC.TypeLits’ KnownNat class, so that we have
natVal :: KnownNat n => proxy n -> Integer
You’d have to supplement your program with a class KnownTag, like so:
data Tag = A | B data Foo (a :: Tag) = Foo
class KnownTag (a :: Tag) where tagVal :: proxy a -> Tag
instance KnownTag A where tagVal _ = A
instance KnownTag B where tagVal _ = B
class C a where f :: a -> Int
instance KnownTag a => C (Foo a) where f _ = case tagVal (Proxy :: Proxy a) of A -> 1; B -> 2
g :: KnownTag a => Foo a -> Int g = f
Regards,
Seph
-- Seph Shewell Brockway, BSc MSc (Glas.) Pronouns: she/her