Question

我试图为我的ZipList编写一个Applicative实例，而且我得到了一些令人困惑的结果。

data List a =
    Nil
  | Cons a (List a)
  deriving (Eq, Show)

newtype ZipList' a =
  ZipList' (List a)
  deriving (Eq, Show)

instance Applicative ZipList' where
  pure = ZipList' . flip Cons Nil
  (<*>) (ZipList' Nil) _ = ZipList' Nil
  (<*>) _ (ZipList' Nil) = ZipList' Nil
  (<*>) (ZipList' (Cons f fs)) (ZipList' (Cons x xs)) =
    ZipList' $ Cons (f x) (fs <*> xs)

对于长度为1或2的ZipLists，它按预期工作：

> ZipList' (Cons (*2) (Cons (+9) Nil)) <*> ZipList' (Cons 5 (Cons 9 Nil))
ZipList' (Cons 10 (Cons 18 Nil))

但是当我去3+时，我会得到一些奇怪的结果：

> ZipList' (Cons (*2) (Cons (+99) (Cons (+4) Nil))) <*> ZipList' (Cons 5 (Cons 9 (Cons 1 Nil)))
ZipList' (Cons 10 (Cons 108 (Cons 100 (Cons 13 (Cons 5 Nil)))))

结果应该是一个10,108,5的ZipList - 但不知何故100和13正在使派对崩溃。

所以我尝试将我的函数从实例中拉出来，以便我可以检查Haskell推断的类型：

(<**>) (ZipList' Nil) _ = ZipList' Nil
(<**>) _ (ZipList' Nil) = ZipList' Nil
(<**>) (ZipList' (Cons f fs)) (ZipList' (Cons x xs)) =
  ZipList' $ Cons (f x) (fs <**> xs)

但它不会编译！

17-applicative/list.hs:94:26: error:
    • Couldn't match expected type ‘ZipList' (a0 -> b0)’
                  with actual type ‘List (a -> b)’
    • In the first argument of ‘(<**>)’, namely ‘fs’
      In the second argument of ‘Cons’, namely ‘(fs <**> xs)’
      In the second argument of ‘($)’, namely ‘Cons (f x) (fs <**> xs)’
    • Relevant bindings include
        xs :: List a (bound at 17-applicative/list.hs:93:49)
        x :: a (bound at 17-applicative/list.hs:93:47)
        fs :: List (a -> b) (bound at 17-applicative/list.hs:93:26)
        f :: a -> b (bound at 17-applicative/list.hs:93:24)
        (<**>) :: ZipList' (a -> b) -> ZipList' a -> ZipList' b
          (bound at 17-applicative/list.hs:91:1)

该错误告诉我，我正在尝试传递一个预期有ZipList的List，我可以看到。但是，我的Applicative实例如何编译呢？

Answer 1

问题是<*>中的ZipList' $ Cons (f x) (fs <*> xs)。

它不是ZipList'＆＃39; <*>，List。

试试ZipList' $ Cons (f x) (case ZipList' fs <*> ZipList' xs of ZipList ys -> ys) `

Haskell ZipList Applicative

1 个答案: