实施 Cont Applicative 实例

Question

我在为 Cont 实现 Applicative 实例时遇到以下错误。

Couldn't match expected type ‘r’ with actual type ‘Cont r b’ ‘r’ is a rigid type variable bound by ...

newtype Cont r a = Cont {(>>-) :: (a -> r) -> r}

instance Functor (Cont r) where
  -- fmap :: (a -> b) -> (Cont r) a -> (Cont r) b
  fmap f (Cont cps_a) = Cont $ \cps -> cps_a (cps . f)

instance Applicative (Cont r) where
  -- pure :: a -> Cont r a
  pure x = Cont ($ x)
  -- (<*>) Cont r (a -> b) -> Cont r a -> Cont r b
  (Cont cps_f) <*> cont_cps_a = cps_f (\f -> fmap f cont_cps_a)

我正在尝试使用 fmap 来定义 (<*>)，从左侧的 cps 值中提取 f，然后对右侧的 cps 值提取 fmap f。我不确定我哪里弄错了。

Answer 1

很明显cps_f (\f -> fmap f cont_cps_a)表达式的类型是无效的。自

fmap f cont_cps_a :: Cont r b

和

\f -> fmap f cont_cps_a :: (a->b)->Cont r b

和

cps_f :: ((a->b)->r)->r

cps_f需要参数类型为(a->b)->r，但现在是(a->b)->Cont r b.

除了使用 fmap，您仍然可以实现 <*> 运算符，类似于 fmap 函数的方式

(Cont cps_f) <*> Cont cps_a = Cont $ \cps_b -> cps_f (\f -> cps_a (cps_b . f))

实施 Cont Applicative 实例

Implementing Cont Applicative instance

haskell

continuation-passing