haskell - 不在范围 : data constructor `Movement'

Question

我想做一个函数，根据传递的字母改变特定的值。 (这些基本上给出了方向:东，西......)

代码是:

data Movement  = N Int | S Int | E Int | W Int deriving (Eq, Show)

step :: Movement -> (Int, Int) -> (Int, Int)
step (Movement x h) (y, z) 
    | x == N = (y, z+h)
    | x == S = (y, z-h)
    | x == W = (y-h, z)
    | x == E = (y+h, z)

一个例子:

step (N 1) (239, 578) == (239, 579)
step (S 1) (240, 578) == (240, 577)
step (W 1) (239, 578) == (238, 578)
step (E 1) (239, 577) == (240, 577)
step (N 61) (239, 578) == (239,639)
step (N 2) (-4, 0) == (-4, 2)
step (E 1) (-4, 0) == (-3, 0)
step (S (-61)) (239, 578) == (239,639)

我不断得到

Not in scope: data constructor `Movement'

错误信息。

Answer 1

Movement 是一种类型，而不是一个值。你不能在模式中使用它。

此外，N 和其他构造函数都是函数，您不能使用== 函数。

你需要改用模式匹配，忘记守卫。

step :: Movement -> (Int, Int) -> (Int, Int)
step (N h) (y,z) = ...
step (S h) (y,z) = ...
step (W h) (y,z) = ...
step (E h) (y,z) = ...

或者，重构您的类型:

data Direction = N | S | E | W deriving (Eq, Show)
data Movement  = Movement Direction Int deriving (Eq, Show)

step :: Movement -> (Int, Int) -> (Int, Int)
step (Movement x h) (y,z) 
    | x == N = (y, z+h)
    | x == S = (y, z-h)
    | x == W = (y-h, z)
    | x == E = (y+h, z)

现在您的代码可以运行了，因为 Movement 也是一个数据构造函数，并且 N 和 friends 不再是函数。不过，我还是宁愿避开 guard ，而使用

step :: Movement -> (Int, Int) -> (Int, Int)
step (Movement N h) (y,z) = (y, z+h)
step (Movement S h) (y,z) = (y, z-h)
step (Movement W h) (y,z) = (y-h, z)
step (Movement E h) (y,z) = (y+h, z)