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我正在为 Zoho REST API 编写一个客户端库，并且有一堆具有所有Maybe a字段的不同记录类型，即：

data Approval = Approval
  { apDelegate :: Maybe Bool
  , apApprove :: Maybe Bool
  , apReject :: Maybe Bool
  , apResubmit :: Maybe Bool
  } deriving (Eq, Show, Generic)

data ContactSpecialFields = ContactSpecialFields
  { csfCurrencySymbol :: Maybe Text -- $currency_symbol
  , csfState :: Maybe Text -- $state
  , csfProcessFlow :: Maybe Bool -- $process_flow
  , csfApproved :: Maybe Bool -- $approved
  , csfApproval :: Approval  -- $approval
  , csfEditable :: Maybe Bool -- $editable
  } deriving (Eq, Show)

-- and so on

我需要一种方法来定义此类类型的“空”记录，例如：

emptyApproval :: Approval
emptyApproval = Approval
  { apDelegate = Nothing
  , apApprove = Nothing
  , apReject = Nothing
  , apResubmit = Nothing
  }

因此，我伸出手GHC.Generics并得到了一些工作（这是错误的！）：

-- These parts seem logically correct to me...

class GEmptyZohoStructure f where
  gEmptyZohoStructure :: f p

instance (GEmptyZohoStructure f, GEmptyZohoStructure g) => GEmptyZohoStructure (f :*: g) where
  gEmptyZohoStructure = (gEmptyZohoStructure :: f p) :*: (gEmptyZohoStructure :: g p)

instance GEmptyZohoStructure Maybe where
  gEmptyZohoStructure = Nothing

class EmptyZohoStructure a where
  emptyZohoStructure :: a

  default emptyZohoStructure :: (Generic a, (GEmptyZohoStructure (Rep a))) => a
  emptyZohoStructure = GHC.Generics.to gEmptyZohoStructure

-- Whereas these parts are random type-class instances that I've written, just 
-- to get the code to compile.

instance (GEmptyZohoStructure f) => GEmptyZohoStructure (M1 i t f) where
  gEmptyZohoStructure = (gEmptyZohoStructure :: f p)

instance (GEmptyZohoStructure f) => GEmptyZohoStructure (K1 i (f p)) where
  gEmptyZohoStructure = gEmptyZohoStructure

instance EmptyZohoStructure Approval

在代码编译时，以下（可以理解）导致运行时堆栈溢出：

ghci> emptyZohoStructure  :: Approval
*** Exception: stack overflow

我正在关注https://www.stackage.org/haddock/lts-12.1/base-4.11.1.0/GHC-Generics.html#g:12encode给出的教程，由于参数被传递给函数，它允许人们有机会解开/构造函数并构建一些有意义的递归层次结构。如何为我的用例编写泛型（泛型函数实际上没有任何参数）？encodeM1K1M1K1

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1 回答 1

3

GEmptyZohoStructure Maybe在类型类中为泛型 定义是没有意义的。

class G f where 
  gempty' :: f p
instance (G f, G g) => G ( f :*: g) where 
  gempty' = gempty' :*: gempty'

instance G c => G (D1 x c) where
  gempty' = M1 gempty'
instance G s => G (C1 x s) where
  gempty' = M1 gempty'    
instance E t => G (S1 m (Rec0 t)) where -- the key instance
  gempty' = M1 (K1 gempty)

class E a where
  gempty :: a
  default gempty :: (Generic a, G (Rep a)) => a
  gempty = to gempty'    
instance E (Maybe a) where 
  gempty = Nothing

之后，您可以定义由 Maybe 值组成的任何产品类型。

于 2019-12-29T13:01:39.237 回答