Question

--bmicalculator
bmicalculator::IO()
bmicalculator=do bmicalculator::(RealFloat a)=>a->a->String
putStrLn "Please Input your weight"
weight<-getLine
putStrLn "Please Input your height"
height<-getLine
|bmi<=17.5="You are anorexia!"
|bmi<=20.7="You are Under weight"
|bmi<=26.4="You are in normal range"
|bmi<=27.8="You are marginally overweight"
|bmi<=31.1="You are overweight"
|bmi>31.1="You are super OBESE!!"
where bmi=weight/height^2

错误发生在

weight<-getLine

我如何提示用户输入“weight”和“height”然后计算并返回字符串，就像我创建数据类型bmicalculator::(RealFloat a)=>a->a->String

一样

Answer 1

首先你的缩进是错误的（也就是说：不存在）。 do-block的内容应该缩进。

然后你给了两个与bmicalculator不相符的类型签名。第二个是在do-block里面，它显然不属于。

然后你似乎在do-block中使用模式保护，在任何模式匹配结构之外。这在语法上是无效的。你可能错过了case bmi of。您还需要将=替换为->。

最后，您不能在do - 块之后的where-block中使用do-block本地的变量。您应该在let - 块中使用do。此外weight和height是字符串，因此如果不先将它们转换为数字，就不能对它们进行算术运算。

Answer 2

另外你想要将bmi-calculator和main函数分开，因为它是一个没有任何副作用的纯函数

main :: IO ()
main = do
    putStrLn "Please Input your weight"
    w <- getLine
    let weight = read w :: Float
    putStrLn "Please Input your height"
    h <- getLine
    let height = read h :: Float
    putStrLn $ bmicalc weight height

bmicalc :: Float -> Float -> String
bmicalc weight height | bmi<=17.5 = "You are anorexic!"
                      | bmi<=20.7 = "You are underweight"
                      | bmi<=26.4 = "You are in normal range"
                      | bmi<=27.8 = "You are marginally overweight"
                      | bmi<=31.1 = "You are overweight"
                      | otherwise = "You are super OBESE!!"
                      where bmi=weight/(height*height)

haskell中的解析错误＆lt; -

2 个答案: