Question

我想在变量参数函数上定义fmap：

type family VarArg (args :: [*]) e where
  VarArg '[] e = e
  VarArg (a ': as) e = a -> VarArg as e

mapVarArg :: forall args e e'
           . (e -> e') -> VarArg args e -> VarArg args e'
mapVarArg f = _

这是我找到的最接近的解决方案：

mapVarArg :: forall args e e' . VarArgIso args
          => (e -> e') -> VarArg args e -> VarArg args e'
mapVarArg f = (^. Lens.from varArgIso) . fmap f . (^. varArgIso @args)

data VarArgD (args :: [*]) e where
  DNil  :: e -> VarArgD '[] e
  DCons :: (a -> VarArgD as e) -> VarArgD (a ': as) e

class VarArgIso (args :: [*]) where
  varArgIso :: Iso' (VarArg args e) (VarArgD args e)

instance VarArgIso '[] where
  varArgIso = iso DNil (\(DNil x) -> x)

instance VarArgIso as => VarArgIso (a ': as) where
  varArgIso = iso (\f -> DCons ((^. varArgIso) . f)) (\(DCons f) -> ((^. Lens.from varArgIso) . f))

instance Functor (VarArgD args) where
  fmap f (DNil a)  = DNil (f a)
  fmap f (DCons g) = DCons (fmap f . g)

是否有更简单的解决方案，或没有额外VarArgIso约束的任何解决方案？

Answer 1

我认为没有额外的类约束，非模板解决方案是不可能的。有一个简单且可能有效的实现与重叠实例：

class VarArg a b c d where
  mapVarArg :: (a -> b) -> c -> d

instance (a ~ c, b ~ d) => VarArg a b c d where
  mapVarArg = id

instance {-# overlapping #-} (VarArg a b c2 d2, c1 ~ d1) =>
  VarArg a b (c1 -> c2) (d1 -> d2) where
  mapVarArg f g = mapVarArg f . g

如果将overlapping替换为incoherent，它通常也可以使用多态/约束函数作为参数：

> mapVarArg (+100) (+) 0 0
100

但是，对于incoherent实例，部分应用mapVarArg - s往往具有无法使用的推断类型。

> let foo = mapVarArg (+100) (+)
> :t foo
foo :: (Num (a -> a -> a), Num a) => a -> a -> a

fmap over variable argument function

1 个答案: