对于此＆＃x27的正确Haskell语法是什么条款？

Question

该语法的适当校正是什么？这是一个问题吗？我复制了 SO 上的答案。

我敢肯定，有更好的写作方法，但是我在免费代码方面确实没有任何好处。我是新手级别，并试图掌握非常基本的语法事物，这些内容继续通过进行大量的编码器练习来使我陷入困境。

import Data.List (findIndices)

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z
    where op x = head $ findIndices (== x) "+-*/" :: Char -> Int
    operations = [(+), (-), (*), (div)]  :: [(Int -> Int -> Int)]

当我将这样写为一个函数时，引用了另外两个功能。

import Data.List (findIndices)

operations :: [(Int -> Int -> Int)]
operations = [(+), (-), (*), (div)]

op :: Char -> Int
op x = head $ findIndices (== x) "+-*/" 

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z

带有Where子句返回以下错误的代码：

codewars.hs:103:34: error:
    • Couldn't match expected type ‘Int’ with actual type ‘Char -> Int’
    • Probable cause: ‘op’ is applied to too few arguments
      In the second argument of ‘(!!)’, namely ‘(op x)’
      In the expression: (operations !! (op x)) y z
      In an equation for ‘basicOp’:
          basicOp x y z
            = (operations !! (op x)) y z
            where
                op x = head $ findIndices (== x) "+-*/" :: Char -> Int
                operations = [(+), ....] :: [(Int -> Int -> Int)]
    |
103 | basicOp x y z = (operations  !! (op x)) y z
    |                                  ^^^^

codewars.hs:104:18: error:
    • Couldn't match expected type ‘Char -> Int’ with actual type ‘Int’
    • Possible cause: ‘($)’ is applied to too many arguments
      In the expression: head $ findIndices (== x) "+-*/" :: Char -> Int
      In an equation for ‘op’:
          op x = head $ findIndices (== x) "+-*/" :: Char -> Int
      In an equation for ‘basicOp’:
          basicOp x y z
            = (operations !! (op x)) y z
            where
                op x = head $ findIndices (== x) "+-*/" :: Char -> Int
                operations = [(+), ....] :: [(Int -> Int -> Int)]
    |
104 |     where op x = head $ findIndices (== x) "+-*/" :: Char -> Int
    |                  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

我的原始尝试可能更容易阅读，但这只是尝试使用我尚不熟悉的更多Haskell构造的练习。

basicOp :: Char -> Int -> Int -> Int
basicOp oper x y
    | oper == '+'  = (+)   x y 
    | oper == '-'  = (-)   x y
    | oper == '*'  = (*)   x y
    | oper == '/'  = (div) x y

从@DanielWagner的评论中，这得到了改进到：

basicOp c = case c of
    '+' -> (+)
    '-' -> (-)
    '*' -> (*)
    '/' -> div

Answer 1

op x表达式的主体类型是int，而不是char-＆gt; int。您还应将操作定位在与op的同一

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z
    where op x = head $ findIndices (== x) "+-*/" :: Int
          operations = [(+), (-), (*), (div)]  :: [(Int -> Int -> Int)]

中

basicOp :: Char -> Int -> Int -> Int
basicOp x = (operations  !! op x)
    where op x = head $ findIndices (== x) "+-*/"
          operations = [(+), (-), (*), (div)]

列 ：//hackage.haskell.org/package/base-4.16.1.0/docs/prelude.html#v：lookup“ rel =“ nofollow noreferrer”> lookup :: eq a =＆gt; a - ＆gt; [（a，b）] - ＆gt;也许b ：

basicOp :: Char -> Int -> Int -> Int
basicOp x | Just y <- lookup x operations = y
          | otherwise = …
    where operations = [('+', (+)), ('-', (-)), ('*', (*)), ('/', div)]

其中…是一个表达式

Answer 2

来提供该函数本身的类型。

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z
    where
        op :: Char -> Int
        op x = head $ findIndices (== x) "+-*/"
        operations = [(+), (-), (*), (div)]  :: [(Int -> Int -> Int)]

请注意，您 can 在单独的行上或通过提供显式lambda表达式作为op op的值

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z
    where op = (\x -> head $ findIndices (== x) "+-*/") :: Char -> Int
          operations = [(+), (-), (*), (div)]  :: [(Int -> Int -> Int)]

Answer 3

哦，删除op和操作的类型声明工作。

我认为有时将类型声明“内联”放入类型声明是可以的。

basicOp :: Char -> Int -> Int -> Int
basicOp x y z = (operations  !! (op x)) y z
    where 
        op x = head $ findIndices (== x) "+-*/" 
        operations = [(+), (-), (*), (div)]