Haskell where类型声明

Question

我是Haskell的新手，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返回:

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 total的类型，并可能将其与类型类Show关联，但我不知道如何完成。谢谢你的帮助。

Answer 1

问题是您将total定义为乘法的结果，这迫使它成为Num a => a，然后您使用它作为带字符串的++的参数，迫使它成为[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

现在可以运行了，但是代码看起来有点重复。我建议这样写:

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

问题似乎是需要显式地在字符串和数字之间进行转换。Haskell不会自动将字符串强制转换为数字，反之亦然。

将数字转换为字符串，使用show。

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

price :: Integer
price = read price_input_string