令人困惑的类型声明？

发布于 2024-11-06 09:52:25 字数 294 浏览 0 评论 0原文

我已经有一段时间没有使用 SML 了，我遇到了这行代码：

type memory = string -> int;

这是否将“内存”定义为一个接受字符串 a 返回 int 的函数，或者完全是其他东西？我搜索过类似的声明，但我似乎找不到一个或弄清楚它的作用。

当我将其放入 SML/NJ 时，我得到的是：

- type memory = string -> int;
type memory = string -> int

原文

I haven't worked with SML in awhile and I came across this line of code:

type memory = string -> int;

Does this define 'memory' to be a function which takes a string a returns an int, or something else entirely? I've searched for a similar declaration but I can't seem to find one or figure out what it does.

When I put it into SML/NJ, I just get this:

- type memory = string -> int;
type memory = string -> int

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dawn曙光 2024-11-13 09:52:25

memory 不是一个函数，它只是一个类型的缩写，该类型是一个以字符串作为输入并返回 int 的函数。
因此，每当您想编写 string->int 类型的内容时，您都可以编写它的 memory 类型。

例如，

- fun foo(f : string->int, s) = f s;
val foo = fn : (string -> int) * string -> int

您可以编写：

- fun foo( f: memory, s) = f s;
val foo = fn : memory * string -> int

这样的 type 声明可以使您的代码更具可读性（例如，不要编写一对 x 的类型为 int* int 与 (x: int*int) 类似，您只需创建一个缩写 typepair = int*int 然后您就可以编写 x 是 pair 类型，就像这样 (x:pair))。

memory is not a function , it's just an abbreviation for a type that is a function that takes as input a string and returns an int.
So whenever you would like to write that something is of type string->int you can just write it's of type memory.

For example instead of writing:

- fun foo(f : string->int, s) = f s;
val foo = fn : (string -> int) * string -> int

you could write:

- fun foo( f: memory, s) = f s;
val foo = fn : memory * string -> int

Such type declarations can make your code more readable (e.g. instead of writing that a pair x is of type int*int like (x: int*int), you can just create an abbreviation type pair = int*int and then you can write that x is of type pair like this (x: pair)).