[Haskell-cafe] Why aren't there anonymous sum types in Haskell?

[Alexander Solla]

On Wed, Jun 22, 2011 at 1:25 PM, pipoca <eliyahu.ben.miney at gmail.com> wrote:
>
>
> Also, I don't think that the formulation of (:|:) above is
> sufficient.  Suppose:
>
> foo :: Foo -> Bar
> baz :: Baz -> Quux
>
> foobaz :: [Foo :|: Baz]
>
> -- map foo and baz over foobaz
> barquux :: [Bar :|: Quux]
> barquux = map f foobaz
> how would  f be implemented?
>
> It seems to me that you'd need an additional function:
> either' :: (a -> c) -> (b -> d) -> Either a b -> Either c d
>
>
Sure, you would want to treat Either as a bifunctor.  Your either' function
is usually called bimap.  So the call would look like:

map (bimap f g) foobaz

for some f and g.


> However, it still seems to me that that isn't sufficient.  Suppose
> instead:
>
> a :: A -> B :|: C
> d :: D -> E :|: F
>
> ad :: [A :|: D]
>
> if we want to map a and d over ad using either' to get
> bcef :: [B :|: C :|: E :|: F]
> it wouldn't work, we'd get
> bcef :: [(B :|: C) :|: (E :|: F)]
> i.e.
> bcef :: [Either (Either B C) (Either E F)]
> instead, which is presumably not what we wanted...
>


You need is a proof that :|: is associative.  In other words, we need to
pick a "normal form" for expressions like (a :|: (b :|: c)) and ((a :|: b)
:|: c) and show that they can all be normalized into the normal form.  I use
a typeclass:

class Monic a b where inject :: a -> b

for that kind of work.  I always have an instance

instance Monic a a where inject = id

so I might pick to write:

instance Monic  ((a :|: b) :|: c) (a :|: (b :|: c)) where
    inject (Left (Left a)) = Left a
    inject (Left (Right b)) = (Right (Left b))
    inject (Right c) = (Right (Right c))

If you do that, you will soon run into two other problems I can think of
(relating to MacLane's coherence conditions, which I think would be fixed by
judicious composition with "inject").

You're building up (Either a b) into a monoidal category.  There used to be
a package called category-extras for this kind of stuff.  I think it has
been broken up.  Does anybody know the status of its replacement(s)?
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