Is there a way to create a module that will export other modules?
For example, I have a list of modules: A, B, C. And I want them to be imported to module D.
So, I have to write:
import A
import B
import C
It works. But may be not very convenient sometimes.
Is there a way to create a Collection module that exports the content of A, B and C?
With this feature, instead of previous instructions, I'd only have to write:
import Collection -- Importing A, B, C.
Yes, but you need to use an explicit export list, specifying all functions, types, classes and modules to export from this module.
module Foo (module A, module B, myid) where
import A
import B
myid :: a -> a -- For example
Related
Is it possible to have haddock generating documentation for modules that are re-exported with an aliased identifier?
I have the following two modules:
Module A:
module A where
data A = A
Module B:
module B where
data B = B
Then if I re-export the A and B through a third module Lib as follows:
module Lib
( someFunc
, module A
, module B
) where
import A
import B
Haddock generates documentation for A and B. However, if I change the code of Lib to this:
module Lib
( someFunc
, module X
) where
import A as X
import B as X
While causes Lib to export the definitions of A and B, haddock will complain about it not being able to find the documentation for X:
Warning: Lib: Could not find documentation for exported module: X
Is there a way to generate documentation for A and B with an alias?
I have this code
module M where
a = 1
b = 2
x = 10
y = 20
...but as the module grows it's difficult to deal with duplicate names.
Is it possible to have namespaces like this?
module M where
module A where
a = 1
b = 2
module X where
x = 10
y = 20
..and then
...
import M
s = A.a + X.y
What you proposed isn't currently supported in Haskell AFAIK. That said, nothing's stopping you from creating seemingly namespaced modules. For example:
module Top where
myTopFunc :: a -> Int
myTopFunc _ = 1
and in another file:
module Top.Sub where
mySubFunc :: a -> String
mySubFunc _ = "Haskell"
In addition to this, you have a few more tricks up your sleeves to arrange your modules. If you import a module A into B, you can export A's visible entities from B as if they were its own. Subsequently, on importing B, you'll be able to use those functions/datatypes etc. being oblivious to where they originally came from. An example of this using the modules from above would be:
module Top (
myTopFunc,
TS.mySubFunc
) where
import qualified Top.Sub as TS
myTopFunc :: a -> Int
myTopFunc _ = 1
Now you can use both functions just by importing Top.
import Top (myTopFunc, mySubFunc)
There are hierarchical module names. You can have modules named M and M.A and M.X, but modules themselves don't nest, and M is unrelated to M.A as far as the language is concerned.
If you want M to export everything M.A and M.X export, you have to do this explicitly:
module M (module M.A, module M.X) where
import M.A
import M.X
-- rest of M
No you cannot. But your comment mentions you're just looking for a way to avoid naming collisions. For this, you can use the {-# LANGUAGE DuplicateRecordFields #-} extension and the compiler will allow duplicate record field names.
Is it possible to write a module in Haskell, which re-exports a module in addition to exporting everything visible inside?
Lets consider following module:
module Test where
import A
f x = x
This module exports everything defined inside, so it exports f but does not re-export anything imported from A.
On the other hand, if I want to re-export the module A:
module Test (
module A,
f
) where
import A
f x = x
Is there a way to re-export A and export everything defined in Test without needing to explicitly write every function defined within Test?
There is a simple solution, just export the module from the module:
module Test
( module Test
, module A
) where
import Prelude()
import A
f x = x
I am trying to implement a ray data type using the Vector type found here: http://www.haskell.org/haskellwiki/Numeric_Haskell:_A_Vector_Tutorial#Importing_the_library
The Vector will only hold doubles, so I would like to use the Unboxed version of the Vector type.
Here is the code I am trying to compile:
module Main where
import qualified Data.Vector.Unboxed as Vector
data Ray = Ray Data.Vector.Unboxed Data.Vector.Unboxed
The error I get is
Not in scope: type constructor or class `Data.Vector.Unboxed'
Failed, modules loaded: none.
The module Data.Vector.Unboxed exports a type constructor Vector which takes as an argument the type you want to store. Since you renamed this module to Vector as well, the qualified name for this type is Vector.Vector. Assuming you want two vectors of doubles, you should therefore use it like this:
data Ray = Ray (Vector.Vector Double) (Vector.Vector Double)
Normally, when you import something, you do it like this:
import Data.Foo -- A module that contains "data Bar = Bar ..."
myfunction = Bar 3 2 4 -- Use Bar
As you can see, you can access all of the stuff in the Data.Foo module directly, as if you were writing code in that same module.
You can instead import something with qualification, which means that you must specify the full module "path" to the thing that you are referring to every time you access it:
import qualified Data.Foo -- A module that contains "data Bar = Bar ..."
myfunction = Data.Foo.Bar 3 2 4 -- Use Bar
Here, you must specify the full "path" to the data type that you're accessing, because the module has been imported as qualified.
There is another way to import something with qualification; you can specify an alias for the module "path" like so:
import qualified Data.Foo as Foo -- A module that contains "data Bar = Bar ..."
myfunction = Foo.Bar 3 2 4 -- Use Bar
We have renamed the Data.Foo part to simply Foo. This way, we can write Foo.Bar when referring to the data constructor.
You imported the module Data.Vector.Unboxed with the alias Vector. This means that when you want to access the Vector data type, you must use Vector.Vector. I'd recommend that you import vectors like this instead:
import Data.Vector.Unboxed (Vector)
import qualified Data.Vector.Unboxed as Vector
This way, you import the Vector type directly, so that you can access it without any module qualifiers, but when you want to use Vector functions, you need to add the Vector prefix (For example Vector.null ...).
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?
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
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.
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.