使用Template Haskell,如何将相同类型拼接到多个位置？

Question

我正在从矢量空间为OpenGL类型定义类的实例,并且为了节省我的打字肌肉,我想使用Template Haskell为我编写一堆实例.

我开始时通过定义函数来派生实例AdditiveGroup:

{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE TypeFamilies #-}
module Data.VectorSpace.OpenGL.TH where

import Control.Applicative
import Control.Monad
import Data.AdditiveGroup
import Data.VectorSpace

import Language.Haskell.TH

deriveScalarAdditive ts = concat <$> forM (map conT ts) (\t -> [d| 
    instance AdditiveGroup $t where zeroV = 0; (^+^) = (+); negateV = negate
  |])

这很好,但请注意,我只是拼接$t到牛津支架一次.现在,派生VectorSpace实例的函数:

deriveScalarVectorSpace ts = concat <$> forM (map conT ts) (\t -> [d|    
    instance VectorSpace $t where type Scalar $t = $t; (*^) = (*)
  |])

但是,这个barfs:

Type indexes must match class instance head
Found `t_tt' but expected `t_ts'
In the associated type instance for `Scalar'
In the instance declaration for `VectorSpace $t'
In the Template Haskell quotation
  [d| instance VectorSpace $t where
          type instance Scalar $t = $t
          { *^ = (*) } |]

错误t_ts和t_tt错误之间的区别告诉我TH每次拼接$t时都会创建一个新的,唯一的名称,当然,只有这些类型相同时,定义才会起作用.

有没有办法通过牛津括号获得我想要的行为,还是我必须回到良好的旧词法范围和Language.Haskell.TH组合器？我知道CPP可能会更容易,但我想借此机会学习一些TH.

Answer 1

我想你必须使用Language.Haskell.TH组合器.请参阅以下门票:

• GHC 4230
• GHC 4135

这样做非常简单.我会从这开始(稍微格式化)

*Foo Language.Haskell.TH> runQ (deriveScalarAdditive [''Int] ) >>= print

[InstanceD [] (AppT (ConT Data.AdditiveGroup.AdditiveGroup) (ConT GHC.Types.Int))
  [ValD (VarP zeroV_12) (NormalB (LitE (IntegerL 0))) [],
   ValD (VarP ^+^_13) (NormalB (VarE GHC.Num.+)) [],
   ValD (VarP negateV_14) (NormalB (VarE GHC.Num.negate)) []]
]

从这里开始,看看如何使用组合器构造实例非常简单.另请注意,您可以将引号与引号表达式混合使用[| some code |] :: ExpQ,这通常对创建函数体很有用.