Re-export qualified? - haskell

Question

suppose you have two modules like
module Foo.A where
foo = 42
and
module Foo.B where
foo = 12
and you want to write a super module
module Foo (
module Foo.A
, module Foo.B
) where
import Foo.A
import Foo.B
which re-exports those modules, you would get a name clash.
Is there a solution for this?

Answer 1

Basically, no. This has been a long-standing feature request by people like the authors of Gtk2hs. Gtk2hs has a very broad module hierarchy where it might make sense to both:
Use the same name in several different modules (e.g. newButton in both Graphics.UI.Gtk.Buttons.Button and Graphics.UI.Gtk.Buttons.CheckButton)
Provide the convenience to the user of importing all these modules with a single import statement
For now, if you want to reexport several modules together all you can do is:
Avoid reusing names in the modules you wish to reexport
Where appropriate, use type classes to allow the same name to be used for several different purposes

Answer 2

Good question. The Haskell Report addresses this:
Exports lists are cumulative: the set of entities exported by an export list is the union of the entities exported by the individual items of the list.
[...]
The unqualified names of the entities exported by a module must all be distinct (within their respective namespace).
According to my limited Haskell knowledge I'd say it's not possible.

Answer 3

You can't export both foos without the name clash unless you use a typeclass (which I shall elaborate on), but there are other options.
You could opt to hide one of the versions of foo:
module Foo
(
module Foo.A,
module Foo.B
)
where
import Foo.A
import Foo.B hiding (foo)
This isn't ideal if you genuinely need both, but if one is rarerly used then it may be simpler to just hide it and expect people to manually reinclude (e.g. by import qualified Foo.B as B (foo), providing B.foo) it if they need it.
You could opt to hide both versions of foo and document that you expect the end user to manually import them if they need them. Under such a scheme, the user could do something like the following:
import Foo.A as A (foo)
import Foo.B as B (foo)
main :: IO ()
main = do
print A.foo
print B.foo
This is actually a fairly common thing to have to do when dealing with multiple container types since many different container types export functions like empty, null and singleton. (To the point where I wish there were actually typeclasses for those, since they are such common operations.)
If you really need both and you're prepared to do a bit of export gymnasics, you can reexport both under different names from within Foo itself:
module Foo (module Foo) where
-- Note that in this case it is important
-- that the module alias matches the
-- name of the exporting module ('Foo')
import Foo.A as Foo hiding (foo)
import Foo.B as Foo hiding (foo)
import qualified Foo.A
import qualified Foo.B
fooA = Foo.A.foo
fooB = Foo.B.foo
Thus any module importing Foo will get Foo.A.foo as fooA (with a value of 42) and Foo.B.foo as fooB (with a value of 12).
If you only have a few clashes to resolve then this is probably the most balanced solution
As another answer has already mentioned, you can use a typeclass, though depending on the context this may be a misuse of the typeclass system.
You could achieve that like so:
module Foo.C where
-- The 'a' dummy argument is necessary to disambiguate
-- which version of 'foo' you intend to use.
class FooProvider a where
foo :: a -> Int
module Foo.B where
import Foo.C
data FooB = FooB
instance FooProvider FooB where
foo _ = 12
module Foo.A where
import Foo.C
data FooA = FooA
instance FooProvider FooA where
foo _ = 42
module Foo (module Exports) where
-- In this case the name of the
-- module alias is unimportant
-- as long as all modules use the same alias.
import Foo.A as Exports
import Foo.B as Exports
import Foo.C as Exports
import Foo
-- 'FooA' and 'FooB' exist solely for
-- the purpose of disambiguating the
-- different versions of 'foo'.
main :: IO ()
main = do
print (foo FooA)
print (foo FooB)
Generally this is less desirable than the aforementioned solutions, particularly for something as simple as a named constant, but sometimes it makes sense to introduce a typeclass. For example, a typeclass such as this:
class Container c where
empty :: c a
singleton :: a -> c a
null :: c a -> Bool
Would work for a number of container types, including [] ('List'), Set, Maybe. Without singleton, it would apply to even more types (e.g. Map, Seq, IntMap). (Note that this could also have included some kind of length function since an empty container would have a length of 0 and a singleton container would have a length of 1. It would however clash with the length in Prelude, which is based on Foldable.)
There may be other possibilities that make use of typeclasses and possibly compiler extensions, but I suspect they'll only increase in complexity from herein.