Haskell无法与实际类型'Char'匹配预期类型'String' [英] Haskell Couldn't match expected type 'String' with actual type 'Char'

问题描述

我想知道为什么我得到这个错误。这是一个任务，我要从一个整数转换为十六进制值。当我通过 16 修改整数值时，我调用了这个帮助器转换函数。 （与递归调用中除以 16 的整数值连接）

这里是我的代码：

  changeToHex :: Integer  - >字符串
  - 主函数
 
 toHex :: Integer  - >字符串
 toHex x 
 | x == 0 ='0'
 | x == 1 ='1'
 | x == 2 ='2'
 | x == 3 ='3'
 | x == 4 ='4'
 | x == 5 ='5'
 | x == 6 ='6'
 | x == 7 ='7'
 | x == 8 ='8'
 | x == 9 ='9'
 | x == 10 ='A' 
 | x == 11 ='B'
 | x == 12 ='C'
 | x == 13 ='D'
 | x == 14 =' E'
 | x == 15 ='F'
  

解决方案

使用单引号（'F'）为您提供 Char 字面值。对于字符串，它实际上是 Char 值的列表，您应该使用双引号（<$ c由于字符串是

的别名，所以

Char 转换为字符串，请使用code> [Char] $ c>，你只能将 Char 包装在一个元素列表中。这样做的函数可能如下所示：

  stringFromChar :: Char  - >字符串
 stringFromChar x = [x] 
  

这通常是内联编写的，如（：[]），相当于 \x - > （x：[]）或 \x - >另外，您可以使用 Enum 来显着简化代码。 typeclass：

  toHexDigit :: Int  - > Char 
到HexDigit x 
 | x < 0 =错误toHex：负数字值
 | x < 10 = toEnum $ fromEnum'0'+ x 
 | x < 15 = toEnum $ fromEnum'A'+ x  -  10 
 |否则=错误toHex：数字值太大
  

更一般地说，任何时候你有一个像：

  fx 
 | x == A = ... 
 | x == B = ... 
 | x == C = ... 
 ... 
  

您可以将其转换为 case ：

  fx = case x 
 A  - > ... 
 B  - > ... 
 C  - > ... 
 ... 
  

I am wondering why I'm getting this error. It's for an assignment where i'm to convert from an integer to a hex value. I call this helper conversion function when I mod the integer value by 16. (concatenated with the integer value which I then divide by 16 in a recursive call)

Here is my code:

    changeToHex :: Integer -> String
    --main function

    toHex :: Integer -> String
    toHex x
        |x == 0         = '0'
        |x == 1         = '1'
        |x == 2         = '2'
        |x == 3         = '3'
        |x == 4         = '4'
        |x == 5         = '5'
        |x == 6         = '6'
        |x == 7         = '7'
        |x == 8         = '8'
        |x == 9         = '9'
        |x == 10        = 'A'
        |x == 11        = 'B'
        |x == 12        = 'C'
        |x == 13        = 'D'
        |x == 14        = 'E'
        |x == 15        = 'F'

解决方案

Using single quotes ('F') gives you a Char literal. For a String literal, which is in fact a list of Char values, you should use double quotes ("F").

Since String is an alias for [Char], if you want to convert from a Char to a String, you can merely wrap the Char in a one-element list. A function to do so might look like:

stringFromChar :: Char -> String
stringFromChar x = [x]

This is typically written inline, as (:[]), equivalent to \x -> (x : []) or \x -> [x].

As an aside, you can simplify your code considerably, using for example the Enum typeclass:

toHexDigit :: Int -> Char
toHexDigit x
  | x < 0 = error "toHex: negative digit value"
  | x < 10 = toEnum $ fromEnum '0' + x
  | x < 15 = toEnum $ fromEnum 'A' + x - 10
  | otherwise = error "toHex: digit value too large"

More generally, any time you have a function like:

f x
  | x == A = ...
  | x == B = ...
  | x == C = ...
  ...

You can convert that to a less repetitious, more efficient equivalent with case:

f x = case x of
  A -> ...
  B -> ...
  C -> ...
  ...

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