Haskell中的函数curry

Question

我有一个功能:

powerOf :: Int -> Int -> Int

示例os用法:

*Main Data.List> powerOf 100 2
2
*Main Data.List> powerOf 100 5
2

我有两个问题.首先 - 为什么它不起作用:

map (powerOf 100) [2, 5]

我想得到[2,2].

第二个问题.我试图创建pariatl函数.像这样的东西:

powerOfN :: Int -> Int
powerOfN num = powerOf num

使用它这样的方式:

let powerOf100 = powerOfN 100
powerOf100 2
powerOf100 5

但我收到了错误消息:

simplifier.hs:31:15:
    Couldn't match expected type `Int'
           against inferred type `Int -> Int'
    In the expression: powerOf num
    In the definition of `powerOfN': powerOfN num = powerOf num

这里有很多代码:

divided :: Int -> Int -> Bool
divided a b = 
  let x = fromIntegral a
      y = fromIntegral b
  in (a == truncate (x / y) * b)

listOfDividers :: Int -> [Int]
listOfDividers num =
               let n = fromIntegral num
                   maxN = truncate (sqrt n)
               in [n | n <- [1.. maxN], divided num n]


isItSimple :: Int -> Bool
isItSimple num = length(listOfDividers num) == 1

listOfSimpleDividers :: Int -> [Int]
listOfSimpleDividers num = [n | n <- listOfAllDividers, isItSimple n]
                     where listOfAllDividers = listOfDividers num

powerOfInner :: Int -> Int -> Int -> Int
powerOfInner num p power
             | divided num p = powerOfInner (quot num p) p (power + 1)
             | otherwise = power

powerOf :: Int -> Int -> Int
powerOf num p = powerOfInner num p 0


powerOfN :: Int -> Int
powerOfN num = powerOf num

powerOf返回p的最大功率p.例如:100 = 2*2*5*5,因此powerOf 100 2 = 2. 10 = 2*5,因此powerOf 10 2 = 1.

如何修复错误？谢谢.

Answer 1

使用您的代码,除了powerOfN功能.我不能重现你的问题map (powerOf 100) [2,5].

*Main> map (powerOf 100) [2,5]
[2,2]

你有任何错误吗？

关于你的第二个问题:

powerOfN :: Int -> Int
powerOfN num = powerOf num

类型签名不正确.

powerOfN取一个整数并返回一个取整数并返回整数的函数.

所以类型签名应该是

powerOfN :: Int -> (Int -> Int)

这是相同的(感谢delnan确认):

powerOfN :: Int -> Int -> Int