如何派生 Data.Messagepack 1.0.0 的实例

Question

上一版本Data.Messagepack,0.7.2.5 supports deriving instances via Template Haskell. The current version (1.0.0), however, doesn't.

因此我想知道是否有另一种方法可以自动派生 MessagePack 1.0.0 实例，可能使用 XDeriveGeneric?

Answer 1

作为权宜之计，请查看 message-pack github 存储库的 msgpack-aeson 目录：

https://github.com/msgpack/msgpack-haskell/tree/master/msgpack-aeson

您可以从您的数据值 <-> aeson <-> 消息包。不一定高效，但方便，因为您可以使用 DeriveGeneric.

自动派生 ToJSON 和 FromJSON

示例代码：

{-# LANGUAGE DeriveGeneric, OverloadedStrings #-}

import Data.MessagePack.Aeson
import qualified Data.MessagePack as MP
import GHC.Generics
import Data.Aeson

data Foo = Foo { _a :: Int, _b :: String }
  deriving (Generic)

instance ToJSON Foo
instance FromJSON Foo

toMsgPack :: Foo  -> Maybe MP.Object
toMsgPack =   decode . encode

test = toMsgPack (Foo 3 "asd")

Answer 2

您可以编写自己的 GMessagePack class 并通过派生 Generic 来获取实例。我试着这样做来回答这个问题，但我不推荐它。 msgpack 不支持求和，并且 one sum type supported by the Haskell msgpack library, Maybe 的编码非常差。

instance MessagePack a => MessagePack (Maybe a) where
  toObject = \case
    Just a  -> toObject a
    Nothing -> ObjectNil

  fromObject = \case
    ObjectNil -> Just Nothing
    obj -> fromObject obj

Maybe的编码无法区分Nothing :: Maybe (Maybe a)和Just Nothing :: Maybe (Maybe a)，两者都会被编码为ObjectNil，解码为Nothing.如果我们要将显而易见的定律 fromObject . toObject == pure 强加给 MessagePack 个实例，MessagePack 的这个实例就会违反它。

如何派生 Data.Messagepack 1.0.0 的实例

How to derive instances of Data.Messagepack 1.0.0

haskell

msgpack