可遍历数据常量 a b = 常量 a 通过快速检查，但表现得很有趣

Question

haskell 这本书要我为

实现可遍历实例

newtype Constant a b = Constant { getConstant :: a }

包括所有必要的超类。下面的代码通过了 Quickcheck/Checkers，但表现得很滑稽

import Test.QuickCheck
import Test.QuickCheck.Checkers
import Test.QuickCheck.Classes

newtype Constant a b = Constant { getConstant :: a }

instance Functor (Constant a) where
  fmap f (Constant a) = Constant a

instance Foldable (Constant a) where
  foldr f z (Constant x) = z

instance Traversable (Constant a) where
  traverse f (Constant a) = pure $ Constant a    

type TI = []
main = do
  let trigger = undefined :: TI (Int, Int, [Int])
  quickBatch (traversable trigger)

当我尝试像这样使用可遍历实例时：

traverse (\x -> [x + 1]) $ Constant 5

我没有得到我希望的 Constant [5]，而是

traverse (\x -> [x + 1]) $ Constant 5
  :: (Num b, Num a) => [Constant a b]

这是什么意思？我是不是做错了什么？

Answer 1

When I try to use the traversable instance like so:
traverse (\x -> [x + 1]) $ Constant 5 
I do not get Constant [5] which I was hoping for [...]

你不会得到 Constant [5]。让我们写下 traverse 的类型：

traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)

...并将其与您的实例化对齐：

                  -- I've substituted `x` for `a` and `y` for `b` in the
                  -- first type, because otherwise my head hurts.
                  (x -> f  y) -> t            x -> f  (t            y)
(Num a, Num b) => (a -> [] a) -> (Constant b) a -> [] ((Constant b) a)

所以我们有：

t = Constant b
f = []
x = Num a => a
y = NUm b => b

请注意，traverse 的类型意味着 t 在参数和结果中是相同的。由于您使用 Constant 5 :: Num a => Constant a b 作为参数，这意味着结果中永远不会有 Constant [5] :: Num a => Constant [a] b，因为 Constant a /= Constant [a].

可遍历数据常量 a b = 常量 a 通过快速检查，但表现得很有趣

Traversable for data Constant a b = Constant a passes Quickchecks, but acts funny

haskell

traversable