Question

在我为函数式编程课程所做的练习中，系统要求我找到x mod a = b的最低x，并给出一系列(a, b)对。

在给定三对（元组）的情况下，我正在使用以下代码：

solveModularEq :: [(Integer,Integer)] -> Integer
solveModularEq [(a),(b),(c)] = lowestModThree(fst(a) snd(a) fst(b) snd(b) fst(c) snd(c) 1)

lowestModThree :: Integer -> Integer -> Integer -> Integer -> Integer -> 
Integer -> Integer -> Integer 
lowestModThree a b c aa bb cc k
  | k `mod` a == aa && k `mod` b == bb && k `mod` c == cc = k
  | k > (aa * bb * cc) = aa * bb * cc
  | otherwise = lowestModThree a b c aa bb cc (k+1)

如果没有这样的x，请返回模的乘积。

我遇到的错误很奇怪，因为看来我没有匹配任何类型。

modEq.hs:3:32:
   Couldn't match expected type ‘Integer’
              with actual type ‘Integer
                                 -> Integer -> Integer -> Integer -> Integer -> Integer -> Integer’
   Probable cause: ‘lowestModThree’ is applied to too few arguments
  In the expression:
     lowestModThree (fst (a) snd (a) fst (b) snd (b) fst (c) snd (c) 1)
   In an equation for ‘solveModularEq’:
      solveModularEq [(a), (b), (c)]
         = lowestModThree
             (fst (a) snd (a) fst (b) snd (b) fst (c) snd (c) 1)

modEq.hs:3:51:
  Couldn't match type ‘Integer’
                  with ‘((a0, b0) -> b0)
                       -> (Integer, Integer)
                        -> ((a1, b1) -> a1)
                        -> (Integer, Integer)
                        -> ((a2, b2) -> b2)
                        -> (Integer, Integer)
                        -> ((a3, b3) -> a3)
                        -> (Integer, Integer)
                        -> ((a4, b4) -> b4)
                        -> (Integer, Integer)
                        -> Integer
                        -> Integer’
   Expected type: (((a0, b0) -> b0)
                   -> (Integer, Integer)
                   -> ((a1, b1) -> a1)
                   -> (Integer, Integer)
                   -> ((a2, b2) -> b2)
                   -> (Integer, Integer)
                   -> ((a3, b3) -> a3)
                   -> (Integer, Integer)
                   -> ((a4, b4) -> b4)
                   -> (Integer, Integer)
                   -> Integer
                   -> Integer,
                   Integer)
     Actual type: (Integer, Integer)
   In the first argument of ‘fst’, namely ‘(a)’
   In the first argument of ‘lowestModThree’, namely
     ‘(fst (a) snd (a) fst (b) snd (b) fst (c) snd (c) 1)’

在我的递归素性测试实现中，同样发生这种情况。

isPrimeRec :: Int -> Bool
isPrimeRec n = isPrimeRec'(isqrt(n) n)

isPrimeRec' :: Int -> Int -> Bool
isPrimeRec' divisor n
  | mod n divisor == 0 = isPrimeRec' (divisor-1) n
  | mod n divisor /= 0 = False
  | divisor < 2 = True

此错误是

palPrimes.hs:10:16:
    Couldn't match expected type ‘Bool’ with actual type ‘Int -> Bool’
    Probable cause: ‘isPrimeRec'’ is applied to too few arguments
    In the expression: isPrimeRec' (isqrt (n) n)
    In an equation for ‘isPrimeRec’:
        isPrimeRec n = isPrimeRec' (isqrt (n) n)

palPrimes.hs:10:28:
    Couldn't match expected type ‘Int -> Int’ with actual type ‘Int’
    The function ‘isqrt’ is applied to two arguments,
    but its type ‘Int -> Int’ has only one
    In the first argument of ‘isPrimeRec'’, namely ‘(isqrt (n) n)’
    In the expression: isPrimeRec' (isqrt (n) n)

Answer 1

将函数f应用于参数x的语法是f x，而不是f(x)。应用程序向左关联，因此f x y的意思是(f x) y，即将f应用于x，并将结果函数应用于y。如果x本身是一个包含函数应用程序的复杂表达式，则可以使用多余的括号来消除歧义；因此，例如：

a b c d将函数a应用于参数b，c和d
a (b c) d将函数a应用于参数b c和d
a b (c d)将函数a应用于参数b和c d
a (b c d)将函数a应用于单个参数b c d

在Haskell中无法匹配Integer类型

1 个答案: