OCaml 中的可变数据

发布于 2024-08-09 05:01:52 字数 777 浏览 12 评论 0原文

我在 OCaml 中创建了一个可变数据结构，但是当我去访问它时，它给出了一个奇怪的错误，

这是我的代码

type vector = {a:float;b:float};;
type vec_store = {mutable seq:vector array;mutable size:int};;

let max_seq_length = ref 200;;

exception Out_of_bounds;;
exception Vec_store_full;;

let vec_mag {a=c;b=d} = sqrt( c**2.0 +. d**2.0);;


let make_vec_store() = 
    let vecarr = ref ((Array.create (!max_seq_length)) {a=0.0;b=0.0}) in
         {seq= !vecarr;size=0};;

当我在 ocaml 顶级中执行此操作

let x = make _ vec _store;;

然后尝试执行 x.size< /code> 我收到此错误，

Error: This expression has type unit -> vec_store
       but an expression was expected of type vec_store

这似乎是什么问题？我不明白为什么这行不通。

谢谢，费萨尔

原文

I've created a mutable data structure in OCaml, however when I go to access it, it gives a weird error,

Here is my code

type vector = {a:float;b:float};;
type vec_store = {mutable seq:vector array;mutable size:int};;

let max_seq_length = ref 200;;

exception Out_of_bounds;;
exception Vec_store_full;;

let vec_mag {a=c;b=d} = sqrt( c**2.0 +. d**2.0);;


let make_vec_store() = 
    let vecarr = ref ((Array.create (!max_seq_length)) {a=0.0;b=0.0}) in
         {seq= !vecarr;size=0};;

When I do this in ocaml top-level

let x = make _ vec _store;;

and then try to do x.size I get this error

Error: This expression has type unit -> vec_store
       but an expression was expected of type vec_store

Whats seems to be the problem? I cant see why this would not work.

Thanks,
Faisal

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回眸一笑 2024-08-16 05:01:53

尝试 x = make_vec_store()

回复收藏 0 原文

提赋 2024-08-16 05:01:53

作为对所提供的优秀答案的跟进。您可以知道您的示例行：

# let x = make_vec_store;;
val x : unit -> vec_store = <fun>

返回一个函数，因为 repl 会告诉您这一点。您可以从输出中看到 x 的类型为，不带参数 unit 并返回类型 vec_store。

声明进行对比。

# let x = 1;;
val x : int = 1

将此与告诉您 x 的类型为 int 且值为 1 的

As a follow up to the excellent answere provided. You can tell that your example line:

# let x = make_vec_store;;
val x : unit -> vec_store = <fun>

returns a function as the repl will tell you this. You can see from the output that x is of type <fun> that takes no parameters unit and returns a type vec_store.

Contrast this to the declaration

# let x = 1;;
val x : int = 1

which tells you that x is of type int and value 1.

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旧竹 2024-08-16 05:01:52

make_vec_store 是一个函数。当您说 let x = make_vec_store 时，您将 x 设置为该函数，就像您编写 let x = 1 一样，这将使 x 成为数字 1您想要是调用该函数的结果。根据 make_vec_store 的定义，它以 () （也称为“unit”）作为参数，因此您可以编写 let x = make_vec_store ()。