haskell – 如何从具有功能依赖性的类型类中获取和使用依赖类型？

如何从具有功能依赖性的类型类中获取和使用依赖类型？

澄清并举例说明我最近的尝试(从我写的实际代码中最小化)：

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ghc似乎没有推断b,因为它打印出这个错误：

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,这是一个较小的示例：

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

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

这里的解决方法是将b作为类型arg添加到Graph：

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

完整的上下文：我有一个实体图表,每个实体都有一个id,每个实体都分配给1个节点.您可以按实体查找节点.我还有一个Graph’,它包含节点(可以分配一个实体),并查找一个节点,你需要提供节点的id,这是一个Int.图表在内部使用Graph’.我有一个IdMap,它将实体的id映射到Graph’中的节点ID.这是我的图表定义：

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

答：使用类型系列,请参阅Daniel Wagner’s answer.
有关完整报道,请参阅Reid Barton’s answer.

GHC抱怨你在最顶层发布的最小f值确实有点奇怪.但它似乎适用于类型系列：

{-# 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]

也许您可以将这个想法适应您的更大范例.