haskell分裂类型不匹配？

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.

这是代码:

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))  

有更多的代码,我想我会把它留下来清楚,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并不暗示任何一种Integral a也不是Fractional a(虽然反过来也适用).如果你想使用div,你必须提供至少与Integral a上下文一样限制的东西.