Haskell export current module with additional imported module

Question

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?

score 153 · Accepted Answer · answered Aug 03 '13 at 17:57

153

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

answered Aug 03 '13 at 17:57

Exploited [here](https://github.com/ndmitchell/tagsoup/commit/1f1fff1e9ce4bd3233a2ef9607287b5018b516a7#diff-430b750c0b6fd0e9461c82565a1345dcR8). – PyRulez Jan 26 '16 at 18:25
Also, any insight as to why this works? (Any documentation?) – PyRulez Jan 26 '16 at 18:31
1

@PyRulez The Haskell Report is the definitive source: https://www.haskell.org/onlinereport/haskell2010/haskellch5.html#x11-1000005.2 – Thomas M. DuBuisson Jan 26 '16 at 19:21

1 Answers1