Question

我一直在试用Vinyl package，它使用类型级别种类来创建具有字段级别多态性的记录结构并自动提供镜头。这两个特性对我的项目非常方便，因为前者允许记录结构是彼此的子类型而没有名称冲突，后者简化了嵌套结构的更新。

问题在于序列化结果。通常我使用Data.DeriveTH自动派生二进制实例，但它似乎无法应对这些结构。以下代码

{-# LANGUAGE DataKinds, TypeOperators #-}
{-# LANGUAGE FlexibleContexts, NoMonomorphismRestriction #-}
{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}
{-# LANGUAGE TemplateHaskell #-}

import Data.Vinyl

import Data.Binary
import Data.DeriveTH

eID          = Field :: "eID"      ::: Int
location     = Field :: "location" ::: (Double, Double)

type Entity = Rec 
    [   "eID"      ::: Int
    ,   "location" ::: (Double, Double)
    ]

$(derive makeBinary ''Entity)

导致GHCI出现此错误

Exception when trying to run compile-time code:
  Could not convert Dec to Decl
TySynD Main.Entity [] (AppT (ConT Data.Vinyl.Rec.Rec) (AppT (AppT PromotedConsT (AppT (AppT (ConT Data.Vinyl.Field.:::) (LitT (StrTyLit "eID"))) (ConT GHC.Types.Int))) (AppT (AppT PromotedConsT (AppT (AppT (ConT Data.Vinyl.Field.:::) (LitT (StrTyLit "location"))) (AppT (AppT (TupleT 2) (ConT GHC.Types.Double)) (ConT GHC.Types.Double)))) PromotedNilT)))
Language/Haskell/Convert.hs:(37,14)-(40,8): Non-exhaustive patterns in case

  Code: derive makeBinary ''Entity
Failed, modules loaded: none.

这似乎与Derive Convert模块中的这段代码有关：

instance Convert TH.Dec HS.Decl where
    conv x = case x of
        DataD cxt n vs con ds -> f DataType cxt n vs con ds
        NewtypeD cxt n vs con ds -> f NewType cxt n vs [con] ds
        where
            f t cxt n vs con ds = DataDecl sl t (c cxt) (c n) (c vs) (c con) []

现在我真的不知道怎么读模板Haskell所以我不能在这里取得很大进展。在我看来，我正在喂食一个类型的同义词，而不是一个数据类型，这可能会破坏它，所以我尝试将它包装成一个新类型：

newtype Entity2 = Entity2 {entity :: Entity}

$(derive makeBinary ''Entity2)

导致这个更加迟钝的错误：

Exception when trying to run compile-time code:
    Could not convert Type to Type
AppT (AppT PromotedConsT (AppT (AppT (ConT Data.Vinyl.Field.:::) (LitT (StrTyLit "eID"))) (ConT GHC.Types.Int))) (AppT (AppT PromotedConsT (AppT (AppT (ConT Data.Vinyl.Field.:::) (LitT (StrTyLit "location"))) (AppT (AppT (TupleT 2) (ConT GHC.Types.Double)) (ConT GHC.Types.Double)))) PromotedNilT)
Could not convert Type to Type
AppT PromotedConsT (AppT (AppT (ConT Data.Vinyl.Field.:::) (LitT (StrTyLit "eID"))) (ConT GHC.Types.Int))
Could not convert Type to Type
PromotedConsT
Language/Haskell/Convert.hs:(71,5)-(80,26): Non-exhaustive patterns in function conv

查看Convert.hs，我们有

instance Convert TH.Type HS.Type where
    conv (ForallT xs cxt t) = TyForall (Just $ c xs) (c cxt) (c t)
    conv (VarT x) = TyVar $ c x
    conv (ConT x) | ',' `elem` show x = TyTuple Boxed []
                  | otherwise = TyCon $ c x
    conv (AppT (AppT ArrowT x) y) = TyFun (c x) (c y)
    conv (AppT ListT x) = TyList $ c x
    conv (TupleT _) = TyTuple Boxed []
    conv (AppT x y) = case c x of
        TyTuple b xs -> TyTuple b $ xs ++ [c y]
        x -> TyApp x $ c y

现在我猜测出错的是GHC 7.6引入了新的语言结构，导致模板Haskell没有考虑到这种结构，导致非详尽的模式。

所以我的问题是，是否有一些方法可以通过添加到Derive，或者从Vinyl记录类型编写我自己的派生，或类似的东西？如果乙烯基的好处不得不用手写出所有的序列化，那将是一种耻辱。

Answer 1

我希望在编写Binary个实例时会遇到一些问题，所有类型的技巧都在继续，但它不会更容易：

instance Binary (Rec '[]) where
  put RNil = return ()
  get = return RNil

instance (Binary t, Binary (Rec fs)) => Binary (Rec ((sy ::: t) ': fs)) where
  put ((_,x) :& xs) = put x >> put xs
  get = do
    x <- get
    xs <- get
    return ((Field, x) :& xs)

有没有办法使用Derive和Template Haskell或其他方法为黑胶唱片类型派生二进制实例

1 个答案: