Question

我是Haskell的新手，我正在尝试为家庭作业实施计算器。我被困在一个需要对两个值进行除法的地方，我认为问题在于它们的类型无法推断或需要声明/转换。我正在努力学习如何解决这个问题，但是在此过程中的任何见解都会有所帮助。

以下是代码：

data Value e = OK e | Error String deriving (Eq)

-- assuming we know how to type e can be shown, i.e. Show e, then
-- we know how to show a Value e type
instance (Show e) => Show (Value e) where
    show (OK x) = (show x)
    show (Error s) = "ERROR: " ++ s

type Token = String
type Result = Value Int
type Intermediate = [ (Value Int) ]

-- an algebra is a things that knows about plus and times
class Algebra a where
    plus :: a -> a -> a
    times :: a -> a -> a
    subtraction :: a -> a -> a
    division :: a -> a-> a

-- assuming that we know how to + and * things of type e, (i.e.
-- we have Num e, then we have algebra's over Value e 
instance (Num e) => Algebra (Value e) where
    plus (OK x) (OK y) = (OK (x+y))
    times (OK x) (OK y) = (OK (x*y))
    subtraction (OK x) (OK y) = (OK (x-y))
    division (OK x) (OK 0) = (Error "div by 0")
    division (OK x) (OK y) = (OK (x `div` y))   <-- this is line 44 that it complains about

当我尝试通过ghci test.hs

运行程序时出现错误

test.hs:44:34:  
    Could not deduce (Integral e)  
      from the context (Algebra (Value e), Num e)  
      arising from a use of `div' at test.hs:44:34-42  
    Possible fix:  
      add (Integral e) to the context of the instance declaration  
    In the first argument of `OK', namely `(x `div` y)'  
    In the expression: (OK (x `div` y))  
    In the definition of `division':  
        division (OK x) (OK y) = (OK (x `div` y))

还有更多的代码，我想我会把它留下来以保持清晰，但如果不清楚我可以随时编辑它。

Answer 1

div :: (Integral a) => a -> a -> a
(/) :: (Fractional a) => a -> a -> a

Num a并不意味着Integral a或Fractional a（当然，反过来也适用）。如果您想使用div，则必须提供至少与Integral a上下文一样严格的内容。

Answer 2

div仅对Integral个值（标准前奏中仅Int或Integer）起作用。 Float和Double支持/运算符的除法，但问题的根源是Num类型类实际上并不强制执行任何类型的除法运算。< / p>

这是有道理的 - 我们可能想要制作Num实例的大量数字集合，其中乘法是不可逆的 - 除法运算本质上是没有意义的。

haskell分裂类型不匹配？

2 个答案: