不在范围内：数据构造函数`Movement'

Question

我想做一个函数，根据传递的字母改变某些值。（这些基本上是给定的方向：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

不在范围内：数据构造函数`Movement'

错误信息。

Answer 1

Movement是一种类型，而不是值。你不能在模式中使用它。

此外，N其他构造函数是函数，您不能使用==函数。

您需要改用模式匹配，而忘记了守卫。

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)

现在你的代码可以工作了，因为Movement它也是一个数据构造函数，并且N和朋友不再是函数。不过，我仍然更愿意避开守卫，并使用

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)