Types 编译器接受的OCaml中的幻象类型显然无效
我试图回答这个问题: 使用幻影类型。所以我打算提出这个准则:Types 编译器接受的OCaml中的幻象类型显然无效,types,ocaml,Types,Ocaml,我试图回答这个问题: 使用幻影类型。所以我打算提出这个准则: type colour = Red | Blue | Yellow type shape = Rectangle | Square module ColouredShape : sig (* Type parameterized by 'a, just for
type colour = Red | Blue | Yellow
type shape = Rectangle | Square
module ColouredShape : sig
(* Type parameterized by 'a, just for the type system. 'a does not appear in the
right hand side *)
type 'a t = shape * colour
(* Dummy types, used as labels in the phantom type *)
type red
type yellow
val make_red : shape -> red t
val make_yellow : shape -> yellow t
val make_rectangle : unit -> red t
val make_square : unit -> yellow t
val f : 'a t -> colour
val g : red t -> colour
val h : yellow t -> colour
end
=
struct
type 'a t = shape * colour
type red
type yellow
let make_red s = (s, Red)
let make_yellow s = (s, Yellow)
let make_rectangle () = make_red Rectangle
let make_square () = make_yellow Square
let f x = snd x
let g x = snd x
let h x = snd x
end
open ColouredShape
open Printf
let _ =
let rectangle = make_rectangle () in
let square = make_square () in
let c = f square in
printf "%b\n" (c = Red);
let c = f rectangle in
printf "%b\n" (c = Red);
let c = g square in
printf "%b\n" (c = Red);
let c = g rectangle in
printf "%b\n" (c = Red);
let c = h square in
printf "%b\n" (c = Red);
let c = h rectangle in
printf "%b\n" (c = Red)
我希望编译器在第二行拒绝代码
let c = g square in
因为g
是red t->color
类型,square
是yellow t
类型。但是一切都经过编译,程序也可以执行
我错过了什么?这是编译器的预期行为吗?因为您在
coloredshape
的签名中公开了couluredshape.t
的结构,所以类型检查器知道红色t=shape*color
和黄色t=shape*color
,然后就知道红色t=yellow t
但是,如果将coloredshape.t
抽象化,则这些类型等式在coloredshape
之外是未知的,因此您将得到相应的错误:
let c = g square
^^^^^^
Error: This expression has type ColouredShape.yellow ColouredShape.t
but an expression was expected of type
ColouredShape.red ColouredShape.t
Type ColouredShape.yellow is not compatible with type
ColouredShape.red
由于您在
coloredshape
的签名中公开了CoulouredShape.t
的结构,因此类型检查器知道red t=shape*color
和yellow t=shape*color
,随后是red t=yellow t
但是,如果将coloredshape.t
抽象化,则这些类型等式在coloredshape
之外是未知的,因此您将得到相应的错误:
let c = g square
^^^^^^
Error: This expression has type ColouredShape.yellow ColouredShape.t
but an expression was expected of type
ColouredShape.red ColouredShape.t
Type ColouredShape.yellow is not compatible with type
ColouredShape.red
一种解决方案是使类型抽象,即使模块接口仅公开以下内容:
(* abstract *)
type 'a t
而不是
(* concrete *)
type 'a t = shape * colour
使用最新版本的OCaml的中间解决方案是将类型声明为私有:
type 'a t = private (shape * colour)
为了模式匹配的目的公开类型的结构,同时强制用户通过调用模块的函数来创建格式良好的对象,这通常非常有用
使用private
的一个简单示例是创建唯一ID:
module ID : sig
type t = private int
val create : unit -> t
end = struct
type t = int (* note: no 'private' *)
let counter = ref 0
let create () =
let res = !counter in
if res < 0 then
failwith "ID.create: int overflow";
incr counter;
res
end
模块ID:sig
类型t=专用int
val创建:单位->t
end=struct
类型t=int(*注:无“专用”*)
设计数器=ref 0
让我们创建()=
让res=!抵挡
如果res<0,则
带有“ID.create:int overflow”的故障;
递增计数器;
物件
终止
一种解决方案是将类型抽象化,即让模块接口仅公开以下内容:
(* abstract *)
type 'a t
而不是
(* concrete *)
type 'a t = shape * colour
使用最新版本的OCaml的中间解决方案是将类型声明为私有:
type 'a t = private (shape * colour)
为了模式匹配的目的公开类型的结构,同时强制用户通过调用模块的函数来创建格式良好的对象,这通常非常有用
使用private
的一个简单示例是创建唯一ID:
module ID : sig
type t = private int
val create : unit -> t
end = struct
type t = int (* note: no 'private' *)
let counter = ref 0
let create () =
let res = !counter in
if res < 0 then
failwith "ID.create: int overflow";
incr counter;
res
end
模块ID:sig
类型t=专用int
val创建:单位->t
end=struct
类型t=int(*注:无“专用”*)
设计数器=ref 0
让我们创建()=
让res=!抵挡
如果res<0,则
带有“ID.create:int overflow”的故障;
递增计数器;
物件
终止