Question

我与对象和课程有一些混乱。我试图从课堂上的班级小伙子那里调用一种方法，在家里随时添加一个孩子，但我得到一个错误，我不会安静地理解。

class virtual person name lastName sex =
    object
        val name = (name:string)
        val mutable lastName = (lastName:string)
        val sex = (sex:char)
    end;;

class man name lastName=
    object
        inherit person name lastName 'm'
        method printNameAndLastName = (name ^ " " ^ lastName)
        method changeLastName newLastName = lastName <- newLastName
        method printLastName = lastName
end;;

class woman name lastName =
    object
        inherit person name lastName 'w'
        method printNameAndLastName = (name ^ " " ^ lastName)
        method changeLastName newLastName = lastName <- newLastName
        method printLastName = lastName
end;;

class child name sex=
    object
        inherit person name "" 'm'
        method printNameAndLastName = (name ^ " " ^ lastName)
        method changeLastName newLastName = lastName <- newLastName
        method printLastName = lastName
end;;

class family familyName man woman=
    object
        val man = man
        val woman = woman
        val mutable kids = ([] :string list)
        initializer woman#changeLastName (man#printLastName)
        method addKid kid = kids <- (kid#printLastName)::kids
        method printFamily = (man#printNameAndLastNAme ^ ", " ^ woman#printNameAndLastName)
end;;



Error: Some type variables are unbound in this type:
         class family :
           'a ->
           (< printLastName : 'c; printNameAndLastNAme : string; .. > as 'b) ->
           (< changeLastName : 'c -> unit; printNameAndLastName : string;
              .. >
            as 'd) ->
           object
             val mutable kids : string list
             val man : 'b
             val woman : 'd
             method addKid : < printLastName : string; .. > -> unit
             method printFamily : string
           end
       The method addKid has type
         (< printLastName : string; .. > as 'e) -> unit
       where 'e is unbound

添加方法addKid时发生错误。也是最后一个方法printFamily，我想在男女名字后加上孩子名字的列表，看起来像这样：

（string * string）* string list =（（＆＃34; man＆＃34;，＆＃34; woman＆＃34;），[＆＃34; kid1＆＃34 ;; ＆＃34; KID2＆＃34;。]）

我一直在试验这个但是我无法将字符串与列表合并，得到预期的错误（字符串*字符串）*字符串。

谢谢你的时间：）

Answer 1

类中所有值的类型应绑定到其定义。例如：

# class ident x = object
    method x = x
  end;;
    Characters 6-11:
  class ident x = object
        ^^^^^
Error: Some type variables are unbound in this type:
         class ident : 'a -> object method x : 'a end
       The method x has type 'a where 'a is unbound

这意味着，您需要将其绑定到具体类型：

class ident x = object
    method x : int = x
  end;;
class ident : int -> object method x : int end

或者将您的类多态和绑定值赋予新引入的类型变量：

# class ['a] ident x = object
    method x : 'a = x
  end;;
class ['a] ident : 'a -> object method x : 'a end

在您的情况下，man，woman和kid不受约束。因为，每次定义一个类时，都会创建同名的类型，你可以将它们绑定到man，woman和child（并修复方法中的拼写错误）

class family familyName man woman =
    object
        val man : man  = man
        val woman : woman = woman
        val mutable kids = ([] :string list)
        initializer woman#changeLastName (man#printLastName)
        method addKid (kid : child) = kids <- (kid#printLastName)::kids
        method printFamily = (man#printNameAndLastName ^ ", " ^ woman#printNameAndLastName)
end;;

Ocaml如何从另一个类

1 个答案: