如何从具有函数依赖项的类型类中获取和使用依赖项类型?

本文介绍了如何从具有函数依赖项的类型类中获取和使用依赖项类型?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

如何从具有功能依赖项的类型类中获取和使用依赖项类型?

How do you get and use the dependent type from a type class with functional dependencies?

澄清并举例说明我的最新尝试(与我编写的实际代码相比有所减少):

To clarify and give an example of my latest attempt (minimised from actual code I was writing):

class Identifiable a b | a -> b where -- if you know a, you know b idOf :: a -> b instance Identifiable Int Int where idOf a = a f :: Identifiable Int b => Int -> [b] -- Does ghc infer b from the functional dependency used in Identifiable, and the instance? f a = [5 :: Int]

但是ghc似乎无法推断b，因为它会显示此错误:

But ghc does not infer b, it seems, as it prints this error:

Couldn't match expected type ‘b’ with actual type ‘Int’ ‘b’ is a rigid type variable bound by the type signature for f :: Identifiable Int b => Int -> [b] at src/main.hs:57:6 Relevant bindings include f :: Int -> [b] (bound at src/main.hs:58:1) In the expression: 5 :: Int In the expression: [5 :: Int] In an equation for ‘f’: f a = [5 :: Int]

对于上下文，这是一个最小化的示例:

For context, here's a less minimised example:

data Graph a where Graph :: (Identifiable a b) => GraphImpl b -> Graph a getImpl :: Identifiable a b => Graph a -> GraphImpl b getImpl (Graph impl) = impl

此处的解决方法是将b作为arg类型添加到Graph:

The workaround here would be to add b as type arg to Graph:

data Graph a b | a -> b where Graph :: (Identifiable a b) => GraphImpl b -> Graph a

完整上下文:我有一个Graph实体，每个实体都有一个ID，每个实体分配给1个节点.您可以按实体查找节点.我还有一个Graph'，它由节点组成(可以分配一个实体)，要查找节点，您需要提供节点的ID(即Int). Graph在内部使用Graph'.我有一个IdMap，它将实体的ID映射到Graph'中的节点的ID.这是我的Graph定义:

The full context: I have a Graph of entities that each have an id, each entity is assigned to 1 node. You can look up a node by entity. I also have a Graph'which consists of nodes (which can be assigned an entity), and to lookup a node you need to provide the node's id, which is an Int. Graph uses Graph' internally. I have an IdMap which maps ids of entities to ids of nodes in Graph'. This is my Graph definition:

data Graph a where Graph :: (Identifiable a b) => { _idMap :: IdMap b, _nextVertexId :: Int, _graph :: Graph' a } -> Graph a

答案:使用类型族，请参见 Daniel Wagner的答案. 有关完整的故事，请参见里德·巴顿的答案.

Answer: Use type families, see Daniel Wagner's answer. For the full story, see Reid Barton's answer.

推荐答案

GHC抱怨您在最顶部发布的最小f确实有点奇怪.但这对于类型家族似乎还可以:

It does indeed seem a bit odd that GHC complains about the minimal f you posted at the very top. But it seems to work okay with type families:

{-# LANGUAGE TypeFamilies #-} class Identifiable a where type IdOf a idOf :: a -> IdOf a instance Identifiable Int where type IdOf Int = Int idOf a = a f :: a -> [IdOf Int] f _ = [5 :: Int]

也许您可以将此想法改编为更大的示例.

Perhaps you can adapt this idea to your larger example.