与可区分联合相比，在 F# 中使用记录类型时更加详细

发布于 2024-12-01 10:02:54 字数 396 浏览 1 评论 0原文

let Method = { Name:string } //oversimplification

let method_parser =
  spaces >>. many1Satisfy isLetter .>> spaces
  |>> (fun name -> { Name=name })

如果我选择使用方法区分联合，事情会更简洁一些：

let method_parser =
  spaces >>. many1Satisfy isLetter .>> spaces
  |>> Method

我相信在 F# 中使用记录类型时没有办法避免这种冗长的情况。我说得对吗？

原文

let Method = { Name:string } //oversimplification

let method_parser =
  spaces >>. many1Satisfy isLetter .>> spaces
  |>> (fun name -> { Name=name })

Had I chosen to use a Method discriminated union, instead, and things would be a bit more succinct:

let method_parser =
  spaces >>. many1Satisfy isLetter .>> spaces
  |>> Method

I believe there is no way to avoid this kind of verbosity when using record types in F#. Am I right?

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苦行僧 2024-12-08 10:02:54

记录与只有一个案例的受歧视联盟非常相似。在某些情况下，我也更喜欢联合，因为它更容易使用。然而，记录有两个主要优点：

它们为字段命名，这在一定程度上导致了冗长，但使代码更加不言自明
如果字段数量较少，可以使用元组或单例联合。
它们允许您使用 { info with Name = "Tomas" } 语法来克隆记录
如果您不需要这个，您可以使用标准 F# 类（具有更多 .NET 风格）

如果你想获得记录的好处，但仍然使用简单的语法来创建，那么你可以定义一个静态成员来构造你的记录：

type Info = 
  { Name : string 
    Count : int }
  static member Create(name:string, count:int) = 
    { Name = name; Count = count }

然后你可以写：

// Simple example using the above type:
let res = ("hello", 5) |> Info.Create

// I expect this would work for your type like this:
let method_parser = 
   spaces >>. many1Satisfy isLetter .>> spaces 
   |>> Method.Create

A record is very similar to a discriminated union with just a single case. In some cases, I would also prefer a union, because it is easier to use. However, records have two main advantages:

They give names to the fields, which partly causes the verbosity, but makes code more self-explanatory
If you have small number of fields, you can use tuple or single-case union.
They allow you to use the { info with Name = "Tomas" } syntax for cloning records
If you don't need this, you can use standard F# classes (which have more .NET feel)

If you want to get the benefits of records, but still use simple syntax for creating then, then you can define a static member for constructing your record:

type Info = 
  { Name : string 
    Count : int }
  static member Create(name:string, count:int) = 
    { Name = name; Count = count }

Then you can write:

// Simple example using the above type:
let res = ("hello", 5) |> Info.Create

// I expect this would work for your type like this:
let method_parser = 
   spaces >>. many1Satisfy isLetter .>> spaces 
   |>> Method.Create

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