Question

我是Haskell的新手，并且在类型系统方面遇到了麻烦。我有以下功能：

threshold price qty categorySize
    | total < categorySize = "Total: " ++ total ++ " is low"
    | total < categorySize*2 = "Total: " ++ total ++ " is medium"
    | otherwise = "Total: " ++ total ++ " is high"
    where total =  price * qty

Haskell回复：

No instance for (Num [Char])
      arising from a use of `*'
    Possible fix: add an instance declaration for (Num [Char])
    In the expression: price * qty
    In an equation for `total': total = price * qty
    In an equation for `threshold':
     ... repeats function definition

我认为问题是我需要以某种方式告诉Haskell总类型，并且可能将它与类型Show相关联，但我不知道如何实现它。谢谢你的帮助。

Answer 1

问题是您将total定义为乘法的结果，这会强制它为Num a => a，然后您将其用作带++的字符串的参数，强制它成为[Char]。

您需要将total转换为String：

threshold price qty categorySize
    | total < categorySize   = "Total: " ++ totalStr ++ " is low"
    | total < categorySize*2 = "Total: " ++ totalStr ++ " is medium"
    | otherwise              = "Total: " ++ totalStr ++ " is high"
    where total    = price * qty
          totalStr = show total

现在，这将会运行，但代码看起来有点重复。我会建议这样的事情：

threshold price qty categorySize = "Total: " ++ show total ++ " is " ++ desc
    where total = price * qty
          desc | total < categorySize   = "low"
               | total < categorySize*2 = "medium"
               | otherwise              = "high"

Answer 2

问题似乎是您需要在字符串和数字之间进行显式转换。 Haskell不会自动将字符串强制转换为数字，反之亦然。

要将数字转换为字符串显示，请使用show。

要将字符串解析为数字，请使用read。由于read实际上适用于多种类型，因此您可能需要指定结果的类型，如：

price :: Integer
price = read price_input_string

Haskell在哪里进行类型声明

2 个答案: