haskell - 围绕何时撰写以及何时使用 $ 的困惑

Question

returnGreater :: (Ord a) => a -> a -> a
returnGreater a b
  | (a > b) = a
  | otherwise = b

returnGreatest2 :: (Ord a, Num a) => a -> a -> a -> (a, a)
returnGreatest2 a b c
  | (a > b) = (a, returnGreater b c)
  | otherwise = (b, returnGreater a c)

sumOfSquares :: (Num a) => (a, a) -> a
sumOfSquares (a, b) = a^2 + b^2

鉴于上述功能，我很困惑为什么 让 x = sumOfSquares 。 returnGreatest2 返回

<interactive>:13:24: error:
    • Couldn't match type ‘a -> a -> (a, a)’ with ‘(c, c)’
      Expected type: a -> (c, c)
        Actual type: a -> a -> a -> (a, a)
    • Probable cause: ‘returnGreatest2’ is applied to too few arguments
      In the second argument of ‘(.)’, namely ‘returnGreatest2’
      In the expression: sumOfSquares . returnGreatest2
      In an equation for ‘x’: x = sumOfSquares . returnGreatest2
    • Relevant bindings include
        x :: a -> c (bound at <interactive>:13:5)

但是 sumOfSquares $ returnGreatest2 3 5 7 做对了。由于 returnGreatest2 的类型与 sumOfSquares 预期的类型相同，我认为我能够组合它们。

Answer 1

组合和柯里化(Currying)可能有点令人困惑。 平方和。 returnGreatest2与\x -> sumOfSquares (returnGreatest2 x)相同，但是returnGreatest2 x的类型是(Ord a, Num a) => a -> a -> (a, a)。在最终获得 (Ord a, Num a) => (a, a) 可接受的 类型的值之前，您需要传递所有 预期参数平方和。

另一方面，sumOfSquares $ returnGreatest2 3 5 7 的解析方式与 sumOfSquares $ (returnGreatest2 3 5 7) 相同； ($) 运算符的优先级低于函数应用程序(或任何其他运算符)。

要真正组合这两个函数，您需要多层组合:

let f = ((sumOfSquares .) . ) . returnGreatest2