[Haskell] recursive deriving

[Duncan Coutts]

On Tue, 2007-11-20 at 19:18 -0500, Alex Jacobson wrote:
> When you want automated deriving of show/read etc., you need all the 
> components of your type also to be instances of show/read but you won't 
> want to *require* them to be automatically generated verions.
> 
> Standalone deriving does the wrong thing here.  Standalone deriving 
> should not cause an overlapping instance error if someone derives an 
> instance manually.  Instead, the manually derived instance should be 
> treated as more specific and win out.
> 
> The other part of this problem is that you can't do automatic recursive 
> deriving and this results in a ridiculous amount of boilerplate.  I know 
> some people have a theory that they want to avoid accidentally creating 
> instances for things that shouldn't have them, but the solution to that 
> is probably to add some declaration for types that prohibits automatic 
> deriving for those types.  The 99% case is that automatic deriving is ok.
> 
> Proposed syntax:
> 
>    derive instance Show T recursively
>    data T = T no-deriving (Ord,Eq)

I would expect that if the data constructor for T is not exported then
standalone deriving should not work. However this appears not to be the
case which breaks module abstraction.

Foo.hs:
module Foo ( T, t ) where
data T = T
t = T

Bar.hs:
import Foo
deriving instance Eq T

$ ghci Bar.hs -XStandaloneDeriving
[1 of 2] Compiling Bar              ( Bar.hs, interpreted )
[2 of 2] Compiling Main             ( Baz.hs, interpreted )
Ok, modules loaded: Bar, Main.
*Main> t == t
True

You could write that Eq instance by hand since they do not have access
to the T constructor, then StandaloneDeriving should not be able to so
either. I think it's a design flaw in standalone deriving.

Does anyone else agree? Should we file a bug report?

Duncan