Haskell类型变量不解决

Question

我有以下代码段，

class LatticeElement a where
    next :: a -> a -- next element     

class (Ord a, LatticeElement a) => LatticeDate a where
    prev :: a -> a -- prev date

data LatticeSlice d v = forall d v. (LatticeDate d, LatticeElement v) => LatticeSlice{date :: d, slice :: v}

map :: (v -> w) -> LatticeSlice d v -> LatticeSlice d w
map f LatticeSlice{date = d, slice = sl} = LatticeSlice{date = d, slice = f sl}

该代码会导致编译错误。现在，据我了解代码，首先，我将latticeElement作为Typeclass设置，然后将latticedate作为latticeelement typecleass，typecleass 要求类型d是ord的实例。

现在，我创建latticeslice，并在声明中读到“ latticeslice是类型上的数据类型d和v ，仅适用于所有类型d实例化latticedate和v实例化latticeElement

MAP的实例化：

src/Lattice.hs:9:77: error:
    • Couldn't match expected type ‘v’ with actual type ‘v1’
      ‘v1’ is a rigid type variable bound by
        a pattern with constructor:
          LatticeSlice :: forall d1 v1 d2 v2.
                          (LatticeDate d2, LatticeElement v2) =>
                          d2 -> v2 -> LatticeSlice d1 v1,
        in an equation for ‘map’
        at src/Lattice.hs:9:7-40
      ‘v’ is a rigid type variable bound by
        the type signature for:
          Lattice.map :: forall v w d.
                         (v -> w) -> LatticeSlice d v -> LatticeSlice d w
        at src/Lattice.hs:8:1-55
    • In the first argument of ‘f’, namely ‘sl’
      In the ‘slice’ field of a record
      In the expression: LatticeSlice {date = d, slice = f sl}
    • Relevant bindings include
        sl :: v1 (bound at src/Lattice.hs:9:38)
        f :: v -> w (bound at src/Lattice.hs:9:5)
        map :: (v -> w) -> LatticeSlice d v -> LatticeSlice d w
          (bound at src/Lattice.hs:9:1)

我认为编译器告诉我，它在（v - ＆gt; w）中没有解决v 与Latticeslice D V中的V相同。但是我不知道为什么要这样做，或者更重要的是，该怎么办！

Answer 1

首先，forall dv。阴影latticeslice的参数。

                  d1           d2                d2                                           d2
                  |            |                 |                                            |
                  v            v                 v                                            v
data LatticeSlice d v = forall d v. (LatticeDate d, LatticeElement v) => LatticeSlice{date :: d, slice :: v}
                    ^            ^                                 ^                                      ^
                    |            |                                 |                                      |
                    v1           v2                                v2                                     v2

latticeslice的类型演示了

>> :set -fprint-explicit-foralls
>> :t +v LatticeSlice
LatticeSlice
  :: forall d1 v1 d2 v2.
     (LatticeDate d2, LatticeElement v2) =>
     d2 -> v2 -> LatticeSlice d1 v1

值构造函数的参数类型d2和v2与类型构造函数的参数无关。 > d1 和v1：

instance LatticeDate    ()
instance LatticeElement ()

ls :: LatticeSlice d v
ls = LatticeSlice () ()

如果删除forall量化器量化您的数据类型携带的约束，在这种情况下，我再次建议它！它等同于此GADT：

data LatticeSlice d v where
  LatticeSlice :: (LatticeDate d, LatticeElement v) => {date :: d, slice :: v} -> LatticeSlice d v

约束您的定义，MAP在w参数中无法参数。仅仅因为v是latticeElement并不意味着w是。因此，我们必须见证f sl :: w是latticelement w：

map :: LatticeElement w => (v -> w) -> LatticeSlice d v -> LatticeSlice d w
map f LatticeSlice{date = d, slice = sl} = LatticeSlice{date = d, slice = f sl}

这使其与functor（>）的基本基础结构不相容（）适用的，monad ..）和bifunctor（biapplicative，..）。

这是最好的版本，如果需要晶格约束，则将它们添加到在latticeslice上操作的功能中，

data LatticeSlice d v = LatticeSlice {date :: d, slice :: v}
  deriving stock Functor

instance Bifunctor LatticeSlice where
  bimap :: (d -> d') -> (s -> s') -> (LatticeSlice d s -> LatticeSlice d' s')
  bimap f g as = bipure f g <<*>> as

instance Biapplicative LatticeSlice where
  bipure :: d -> s -> LatticeSlice d s
  bipure = LatticeSlice

  biliftA2 :: (d1 -> d2 -> d3)
           -> (s1 -> s2 -> s3)
           -> (LatticeSlice d1 s1 -> LatticeSlice d2 s2 -> LatticeSlice d3 s3)
  biliftA2 (·) (×) (LatticeSlice d1 s1) (LatticeSlice d2 s2) =
    LatticeSlice (d1 · d2) (s1 × s2)

请观看Edward Kmett的类型类与世界有关更多信息。