具有类型约束的GADT的Functor实例

问题描述

今天，我想调查是否有可能以这种方式构造数据类型，它不会存储类型签名类型的数据，而是存储它的另一种表示形式。因此，我尝试了一个GADT，它的类型构造函数类型为 a ，但是类型为 ByteString 的数据构造函数>。

  { - ＃LANGUAGE GADTs＃ - } 
 import Data.ByteString.Char8 
 import Data 。序列化
 
数据序列化a where 
 MkSerialized ::（Serialize a）=> ByteString  - >序列化
  

现在我可以定义一个 decode'函数按以下方式：

  decode'::（Serialize a）=>序列化的a  - > 
解码'（MkSerialized bs）=让右r =（解码bs）in r 
  

它的工作原理：

$ pre $ let $ s = MkSerialized（encodetest）:: Serialized String
print $ decode 's - printstest

现在我的问题是我想要序列化为 Functor 的实例。

  
 
 
 
 
 
 
 $ f $ / code> 

但是我得到的错误（序列化b）无法推断。我如何限制Functor实例，以便在 fmap 中实施 Serialize ？

解决方案

您可以使用 CoYoneda函子完成此操作。 很简单：有一个额外的功能字段，可以累积 fmap ing函数。当你解码你的值，然后应用该函数。

以下是代码：

  { - ＃LANGUAGE GADTs＃ - } 
 import Data.ByteString.Char8 
 import Data.Serialize 
 
 data序列化a其中
 MkSerialized 
 ::（序列化a）
 => ByteString  - > （a  - > b） - >序列化b 
 
解码'::序列化a  - > a 
 decode'（MkSerialized bs f）= let Right r =解码bs in fr 
 
实例Functor序列化其中
 fmap f（MkSerialized bs g）= MkSerialized bs（f。 g）
  

这也有自动融合多个 fmap s而不是重复的解码和编码，就像你的情况一样。

Today I wanted to investigate if it is possible to construct a data type in such a way, that it does not store the data of the type of its type signature, but another representation of it. So, here is my attempt of an GADT which has a type constructor of type a, but a data constructor of type ByteString.

{-# LANGUAGE GADTs #-}
import Data.ByteString.Char8
import Data.Serialize

data Serialized a where
    MkSerialized :: (Serialize a) => ByteString -> Serialized a

Now I can define a decode' function in the following way:

decode' :: (Serialize a) => Serialized a -> a
decode' (MkSerialized bs) = let Right r = (decode bs) in r

And it works:

let s = MkSerialized (encode "test") :: Serialized String
print $ decode' s     -- prints "test"

My problem is now that I'd like Serialized to be an instance of Functor.

instance Functor Serialized where
    fmap f (MkSerialized bs) = MkSerialized (encode (f (right (decode bs))))
                               where right (Right r) = r

But I get the error (Serialize b) can not be deduced. How can I constraint the Functor instance so that Serialize is enforced in the fmap?

解决方案

You can do this using a CoYoneda functor.

The idea is simple: have an additional functional field where you accumulate your fmaping functions. When you decode your value, then apply that function.

Here's the code:

{-# LANGUAGE GADTs #-}
import Data.ByteString.Char8
import Data.Serialize

data Serialized a where
    MkSerialized
      :: (Serialize a)
      => ByteString -> (a -> b) -> Serialized b

decode' :: Serialized a -> a
decode' (MkSerialized bs f) = let Right r = decode bs in f r

instance Functor Serialized where
    fmap f (MkSerialized bs g) = MkSerialized bs (f . g)

This also has the benefit of automatically fusing multiple fmaps instead of repeated decodings and encodings, as would be in your case.

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