我仍然非常想进入哈斯克尔，但我注意到一些让我烦恼的东西 .
在书中"Learn You a Haskell for Great Good!"这一部分显示了在模式匹配中使用警卫，在本书的情况下它是一个计算人的bmi的小函数，它有点像这样（部分略有改变，没有侵犯版权或其他）：
bmiCalc :: (RealFloat a) => a -> a -> String
bmiCalc weight height
| bmi <= 18.5 = "skinny"
| bmi <= 25.0 = "normal"
| bmi <= 30.0 = "fat"
| otherwise = "obese"
where bmi = weight / height ^ 2
这一切都很好，花花公子的代码就像宣传的那样，但我想，如果它还显示了它计算的bmi与文本一起怎么办？
所以我重新编写了代码：
bmiCalc :: (RealFloat a) => a -> a -> String
bmiCalc weight height
| bmi <= 18.5 = "skinny, " ++ show bmi
| bmi <= 25.0 = "normal, " ++ show bmi
| bmi <= 30.0 = "fat, " ++ show bmi
| otherwise = "obese, " ++ show bmi
where bmi = weight / height ^ 2
期望"show"像.toString一样工作在java和c＃中
男孩，我错了 .
ghci给了我这个非常讨厌的错误信息：
Could not deduce (Show a) arising from a use of `show'
from the context (RealFloat a)
bound by the type signature for
bmiCalc :: RealFloat a => a -> a -> String
at file.hs:1:16-48
Possible fix:
add (Show a) to the context of
the type signature for bmiCalc :: RealFloat a => a -> a -> String
In the second argument of `(++)', namely `show bmi'
In the expression: "skinny, " ++ show bmi
In an equation for `bmiCalc':
bmiCalc weight height
| bmi <= 18.5 = "skinny, " ++ show bmi
| bmi <= 25.0 = "normal, " ++ show bmi
| bmi <= 30.0 = "fat, " ++ show bmi
| otherwise = "obese, " ++ show bmi
where
bmi = weight / height ^ 2
Failed, modules loaded: none.
这是为什么？为什么't it allow me to append what appears to return a string, to a string? I mean as far as I' ve了解 "skinny, " ++ show bmi 是一个字符串...这正是类型签名所说的我必须返回
那我在这里做错了什么？
2 回答
将类型签名更改为：
因为要使用成员函数
show，来自Show类型类;但是你还没有在函数约束中指定，并且ghci没有办法推断它是正确的 .RealFloat不是一个可显示的类型 . 你必须添加一个show约束 .