haskell - 导入数据构造函数而不导入类型

Question

numeric-prelude 执行此操作，其中每个数据类型都被命名为 T，每个类型类都被命名为 C。为了......一致性，我想我会一起玩:

{-# LANGUAGE NoImplicitPrelude #-}
module Number.SqrtRatio (T(..), ratioPart) where

import qualified Number.Ratio as Ratio
import Number.Ratio ((:%))

import qualified Algebra.Ring as Ring
import NumericPrelude.Base

-- | A number whose square is rational, canonicalized as a rational
--   times the square root of a squarefree integer.
data T x = T {
    numerator :: !x,
    denominator :: !x,
    rootNum :: !x
    } deriving (Eq, Show)

ratioPart :: T x -> Ratio.T x
ratioPart (T n d _) = n :% d

fromRatio :: (Ring.C x) => Ratio.T x -> T x
fromRatio (n :% d) = T n d Ring.one

ghc 没有留下深刻的印象:

Number/SqrtRatio.hs:5:22:
    In module ‘Number.Ratio’:
      ‘(:%)’ is a data constructor of ‘T’
    To import it use
      ‘import’ Number.Ratio( T( (:%) ) )
    or
      ‘import’ Number.Ratio( T(..) )

<小时/>

当然，伙计，我可以遵守:

{-# LANGUAGE NoImplicitPrelude #-}
module Number.SqrtRatio (T, ratioPart) where

import qualified Number.Ratio as Ratio
import Number.Ratio (T((:%)))
--       newly added ^

...但这最终也会导入 Ratio.T，这与我的 T 冲突!

ratioPart :: T x -> Ratio.T x
{-           ^-- Ambiguous occurrence ‘T’
    It could refer to either ‘Number.SqrtRatio.T’,
                             defined at Number/SqrtRatio.hs:11:1
                          or ‘Number.Ratio.T’,
                             imported from ‘Number.Ratio’ at Number/SqrtRatio.hs:5:22-28
-}

<小时/>

好吧，那么 import Number.Ratio (T((:%))) 隐藏 T 怎么样？

Number/SqrtRatio.hs:5:31: parse error on input ‘hiding’

我有点不知所措，盖斯。 :/

Answer 1

事实证明是执行此操作的正确方法:

{-# LANGUAGE NoImplicitPrelude, PatternSynonyms #-}
module Number.SqrtRatio (T(..), ratioPart) where

import qualified Number.Ratio as Ratio
import Number.Ratio (pattern (:%))

请注意，我使用 -XPatternSynonyms 扩展并不是为了实际定义任何模式同义词，只是为了启用 pattern 关键字，因此很明显我想导入 < em>值构造函数 :% 单独。