编写我的第一个Haskell函数的问题

Question

我是Haskell的新手,我的代码不会编译.

multipleSum :: Int -> Int
multipleSum x = let recSum 0 b = b
                    recSum a b | a mod 3 == 0     = recSum a-1 b+a
                               | a mod 5 == 0     = recSum a-1 b+a
                               | otherwise        = recSum a-1 b
                in recSum x 0 

这是我得到的两个错误,第一个出现在第3行,第二个出现在第6行.我做错了什么？(该函数应该是n下3和5的所有倍数之和)

1.
Occurs check: cannot construct the infinite type: a ~ a -> a -> a
Expected type: (a -> a -> a) -> a -> a
  Actual type: ((a -> a -> a) -> a -> a) -> (a -> a -> a) -> a -> a
Relevant bindings include
  b :: (a -> a -> a) -> a -> a
    (bound at src\Main.hs:5:30)
  a :: (a -> a -> a) -> a -> a
    (bound at src\Main.hs:5:28)
  recSum :: ((a -> a -> a) -> a -> a)
            -> ((a -> a -> a) -> a -> a) -> (a -> a -> a) -> a -> a
    (bound at src\Main.hs:4:21)
In the first argument of `(-)', namely `recSum a'
In the first argument of `(+)', namely `recSum a - 1 b'


2.Couldn't match expected type `(a0 -> a0 -> a0) -> a0 -> a0'
            with actual type `Int'
In the first argument of `recSum', namely `x'
In the expression: recSum x 0

Couldn't match expected type `Int'
            with actual type `(a0 -> a0 -> a0) -> a0 -> a0'
Probable cause: `recSum' is applied to too few arguments
In the expression: recSum x 0
In the expression:
  let
    recSum 0 b = b
    recSum a b
      | a mod 3 == 0 = recSum a - 1 b + a
      | a mod 5 == 0 = recSum a - 1 b + a
      | otherwise = recSum a - 1 b
  in recSum x 0

Answer 1

您有两个与语法相关的问题.第一个与函数和运算符优先级有关.函数应用程序在Haskell中具有最高优先级,因此recSum a-1 b+a被视为相同(recSum a)-(1 b)+a.相反,你需要写recSum (a-1) (b+a).

第二个问题是,a mod 3是函数a调用的参数mod和3.要mod用作中缀运算符,请将其写为

a `mod` 3

把这两个变化放在一起我们就有了

multipleSum :: Int -> Int
multipleSum x = let recSum 0 b = b
                    recSum a b | a `mod` 3 == 0  = recSum (a-1) (b+a)
                               | a `mod` 5 == 0  = recSum (a-1) (b+a)
                               | otherwise       = recSum (a-1) b
                in recSum x 0 

Answer 2

首先,您可以获得的类型签名越多,调试就越容易,所以我将其重写为

multipleSum :: Int -> Int
multipleSum x = recSum x 0

recSum :: Int -> Int -> Int
recSum 0 b = b
recSum a b | a mod 3 == 0     = recSum a-1 b+a
           | a mod 5 == 0     = recSum a-1 b+a
           | otherwise        = recSum a-1 b

并用ghci或拥抱来解雇它.

这样我就得到了一个错误a mod 3.

好吧,我必须用反引号编写中缀函数,所以应该这样

recSum :: Int -> Int -> Int
recSum 0 b = b
recSum a b | a `mod` 3 == 0     = recSum a-1 b+a
           | a `mod` 5 == 0     = recSum a-1 b+a
           | otherwise        = recSum a-1 b

现在我收到有关参数数量的错误recSum a-1 b+a.那是因为应该只有两个,所以如果我要传递比单个变量更复杂的东西,我需要括号,所以我应该写

recSum :: Int -> Int -> Int
recSum 0 b = b
recSum a b | a `mod` 3 == 0     = recSum (a-1) (b+a)
           | a `mod` 5 == 0     = recSum (a-1) (b+a)
           | otherwise        = recSum (a-1) b -- don't need brackets for b on its own

现在它已经编译好了,现在是时候用各种输入来测试它,看看它是否能达到预期效果.