haskell - 编译器是什么意思？

Question

我有以下应用表达式，我知道这是错误的:

Prelude> [Just (*2), Just (+9)] <*> [(Just 3),(Just 4), (Just 5)]

编译器提示:

<interactive>:2:2: error:
    * Couldn't match expected type `Maybe Integer -> b'
                  with actual type `Maybe (Integer -> Integer)'
    * Possible cause: `Just' is applied to too many arguments
      In the expression: Just (* 2)
      In the first argument of `(<*>)', namely `[Just (* 2), Just (+ 9)]'
      In the expression:
        [Just (* 2), Just (+ 9)] <*> [(Just 3), (Just 4), (Just 5)]
    * Relevant bindings include it :: [b] (bound at <interactive>:2:1)

编译器试图表达什么？

此错误信息:

* Couldn't match expected type `Maybe Integer -> b'
              with actual type `Maybe (Integer -> Integer)'

表示这部分[Just (*2), Just (+9)]的表达？

让我们看看函数的签名 (<*>) :

(<*>) :: f (a -> b) -> f a -> f b

将其替换为 List和 Maybe上面的类型构造函数:

(<*>) :: [] (Maybe Integer -> b) -> [] (Maybe Integer) -> [] b

b 应该是哪种类型有吗？

Answer 1

让我们更仔细地替换

(<*>) :: (Applicative f) => f (a -> b) -> f a -> f b
[Just (*2), Just (+9)] :: [Maybe (Int -> Int)]   --wlog
[Just 3, Just 4, Just 5] :: [Maybe Int]

所以 f = [] ，因此:

(<*>) :: [a -> b] -> [a] -> [b]

但是现在我们要匹配[a -> b]反对[Maybe (Int -> Int)]这是不可能的——一个函数和一个 Maybe不可能相同。您正在尝试在 两个 仿函数下进行函数应用，而不仅仅是一个。 IE。你需要一些类型的东西

f (g (a -> b)) -> f (g a) -> f (g b)

幸运的是，这很容易做到:liftA2 (<*>) .

或者，如果您想要更高级的东西，您可以改为对复合仿函数 Compose [] Maybe 进行操作.

getCompose $ Compose [Just (*2), Just (+9)] <*> Compose [Just 3, Just 4, Just 5]

(但是当 Compose 很酷时，您不会立即 getCompose 结果，而是将其用作您正在编写的任何程序中的重要抽象。)