C# Haskell:没有自己类型的实例

我正在尝试使用“call haskell to anything”包将自己的类型导出到C#，代码如下：

{-# LANGUAGE ForeignFunctionInterface #-}
{-# LANGUAGE TemplateHaskell #-}

module RectangleMover where

import Foreign.C
import FFI.Anything.TypeUncurry.Msgpack

data Point = Point Int Int
movePoint :: Point -> Int -> Int -> Point
movePoint (Point x y) dx dy = Point (x + dx) (y + dy)

data Rectangle = Rectangle { corner :: Point, width :: Int, height :: Int }
move :: Rectangle -> Int -> Int -> Rectangle
move r@(Rectangle {corner=c}) dx dy = r { corner = movePoint c dx dy }

foreign export ccall move_export :: CString -> IO CString
move_export :: CString -> IO CString
move_export = export move

当我尝试编译此文件时，会收到错误消息：

[1 of 1] Compiling RectangleMover   ( test.hs, test.o )

test.hs:19:15:
    No instance for (Data.MessagePack.Object.MessagePack Rectangle)
      arising from a use of ‘export’
    In the expression: export move
    In an equation for ‘move_export’: move_export = export move

test.hs:13:10:
    Expected a constraint, but ‘Rectangle’ has kind ‘*’
    In the instance declaration for ‘Rectangle’

所以我想我必须为“矩形”类型写一个“实例”。但正确的语法是什么？我已经尝试添加：

instance Rectangle where

但收到了以下错误消息：

[1 of 1] Compiling RectangleMover   ( test.hs, test.o )

test.hs:19:15:
    No instance for (Data.MessagePack.Object.MessagePack Rectangle)
      arising from a use of ‘export’
    In the expression: export move
    In an equation for ‘move_export’: move_export = export move

test.hs:13:10:
    Expected a constraint, but ‘Rectangle’ has kind ‘*’
    In the instance declaration for ‘Rectangle’

---

首先，您需要导入

Data.MessagePack.Object

。这将使类

MessagePack

进入范围

然后你需要说：

instance MessagePack Rectangle where
   toObject rect = 
      -- Insert your implementation here, returning an Object
   fromObject obj = 
      -- Insert your implementation here, returning a Maybe Rectangle

---

将其与的文档进行比较。

首先，您需要导入

Data.MessagePack.Object

。这将使类

MessagePack

进入范围

然后你需要说：

instance MessagePack Rectangle where
   toObject rect = 
      -- Insert your implementation here, returning an Object
   fromObject obj = 
      -- Insert your implementation here, returning a Maybe Rectangle

---

将其与的文档进行比较。

最后，我在ErikR（）的帮助下解决了这个问题。 以下是最终代码：

{-# LANGUAGE ForeignFunctionInterface #-}
{-# LANGUAGE DeriveGeneric #-}

module RectangleMover where

import Foreign.C
import FFI.Anything.TypeUncurry.Msgpack
import qualified Data.MessagePack as MP
import Data.MessagePack.Object
import Data.MessagePack.Aeson
import Data.Aeson
import GHC.Generics

instance ToJSON Point
instance FromJSON Point
data Point = Point Int Int deriving (Generic, Show)
movePoint :: Point -> Int -> Int -> Point
movePoint (Point x y) dx dy = Point (x + dx) (y + dy)

instance ToJSON Rectangle
instance FromJSON Rectangle
data Rectangle = Rectangle { corner :: Point, width :: Int, height :: Int } deriving (Generic, Show)
move :: Rectangle -> Int -> Int -> Rectangle
move r@(Rectangle {corner=c}) dx dy = r { corner = movePoint c dx dy }

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

p = Point 1 2
rect = Rectangle p 10 20
test = toMsgPack (rect)

---

最后，我在ErikR（）的帮助下解决了这个问题。 以下是最终代码：

{-# LANGUAGE ForeignFunctionInterface #-}
{-# LANGUAGE DeriveGeneric #-}

module RectangleMover where

import Foreign.C
import FFI.Anything.TypeUncurry.Msgpack
import qualified Data.MessagePack as MP
import Data.MessagePack.Object
import Data.MessagePack.Aeson
import Data.Aeson
import GHC.Generics

instance ToJSON Point
instance FromJSON Point
data Point = Point Int Int deriving (Generic, Show)
movePoint :: Point -> Int -> Int -> Point
movePoint (Point x y) dx dy = Point (x + dx) (y + dy)

instance ToJSON Rectangle
instance FromJSON Rectangle
data Rectangle = Rectangle { corner :: Point, width :: Int, height :: Int } deriving (Generic, Show)
move :: Rectangle -> Int -> Int -> Rectangle
move r@(Rectangle {corner=c}) dx dy = r { corner = movePoint c dx dy }

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

p = Point 1 2
rect = Rectangle p 10 20
test = toMsgPack (rect)

---

我不熟悉这个库，但它应该类似于

实例Data.MessagePack.Object.MessagePack矩形，其中..

，现在我得到：

test.hs:13:10:不在范围内：类型构造函数或类'Data.MessagePack.Object.MessagePack'

您可能需要

导入Data.MessagePack.Object

并使用

实例MessagePack矩形，其中

我不熟悉库，但它应该类似于

实例Data.MessagePack.Object.MessagePack矩形，其中…

。我尝试了这个，现在我得到：

test.hs:13:10:不在范围内：类型构造函数或类'Data.MessagePack.Object.MessagePack'

您可能需要

导入Data.MessagePack.Object

并使用

实例MessagePack矩形，其中

不幸的是，我对Haskell很陌生。在这种情况下，toObject和fromObject的实现是什么？请查看Data.MessagePack.Object中实例的源代码，了解一些想法。它是从文档页面链接的。不幸的是，我对Haskell很陌生。在这种情况下，toObject和fromObject的实现是什么？请查看Data.MessagePack.Object中实例的源代码，了解一些想法。其链接来自文档页面。