F#记录会员评价

发布于 2024-09-02 02:58:09 字数 166 浏览 4 评论 0原文

为什么每次调用都要评估 tb ?有什么办法让它只评估一次吗?

type test =
  { a: float }
  member x.b =
    printfn "oh no"
    x.a * 2.

let t = { a = 1. }
t.b
t.b

Why is t.b evaluated on every call? And is there any way how to make it evaluate only once?

type test =
  { a: float }
  member x.b =
    printfn "oh no"
    x.a * 2.

let t = { a = 1. }
t.b
t.b

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评论(4

沦落红尘 2024-09-09 02:58:09

Brian 答案的替代版本最多评估一次 b,但如果从未使用过 B 则根本不会评估它

type Test(a:float) =
    // constructor
    let b = lazy
                 printfn "oh no"
                 a * 2.
    // properties
    member this.A = a
    member this.B = b.Value

An alternative version of Brian's answer that will evaluate b at most once, but won't evaluate it at all if B is never used

type Test(a:float) =
    // constructor
    let b = lazy
                 printfn "oh no"
                 a * 2.
    // properties
    member this.A = a
    member this.B = b.Value
不交电费瞎发啥光 2024-09-09 02:58:09

它是一种财产;您基本上是在调用 get_b() 成员。

如果您希望构造函数只发生一次效果,您可以使用一个类:

type Test(a:float) =
    // constructor
    let b =   // compute it once, store it in a field in the class
        printfn "oh no"
        a * 2.
    // properties
    member this.A = a
    member this.B = b

It's a property; you're basically calling the get_b() member.

If you want the effect to happen once with the constructor, you could use a class:

type Test(a:float) =
    // constructor
    let b =   // compute it once, store it in a field in the class
        printfn "oh no"
        a * 2.
    // properties
    member this.A = a
    member this.B = b
哀由 2024-09-09 02:58:09

为了回应您在 Brian 帖子中的评论,您可以使用可选/命名参数来伪造复制和更新记录表达式。例如:

type Person(?person:Person, ?name, ?age) =

    let getExplicitOrCopiedArg arg argName copy =
        match arg, person with
        | Some(value), _ -> value
        | None, Some(p) -> copy(p)
        | None, None -> nullArg argName

    let name = getExplicitOrCopiedArg name "name" (fun p -> p.Name)
    let age = getExplicitOrCopiedArg age "age" (fun p -> p.Age)

    member x.Name = name
    member x.Age = age

let bill = new Person(name = "Bill", age = 20)
let olderBill = new Person(bill, age = 25)

printfn "Name: %s, Age: %d" bill.Name bill.Age
printfn "Name: %s, Age: %d" olderBill.Name olderBill.Age

In response to your comments in Brian's post, you can fake copy-and-update record expressions using optional/named args. For example:

type Person(?person:Person, ?name, ?age) =

    let getExplicitOrCopiedArg arg argName copy =
        match arg, person with
        | Some(value), _ -> value
        | None, Some(p) -> copy(p)
        | None, None -> nullArg argName

    let name = getExplicitOrCopiedArg name "name" (fun p -> p.Name)
    let age = getExplicitOrCopiedArg age "age" (fun p -> p.Age)

    member x.Name = name
    member x.Age = age

let bill = new Person(name = "Bill", age = 20)
let olderBill = new Person(bill, age = 25)

printfn "Name: %s, Age: %d" bill.Name bill.Age
printfn "Name: %s, Age: %d" olderBill.Name olderBill.Age
不必了 2024-09-09 02:58:09

之前的回复建议切换到类,而不是使用记录。如果您想保留记录(因为其简单的语法和不变性),您可以采用这种方法:

type test =
    { a : float
      b : float }
    static member initialize (t: test) =
        { t with b = t.a * 2. }

如果 test 的实例是由另一个库(例如来自 Web 的数据提供者)创建的,这非常有用服务或数据库)。使用这种方法时,您必须记住先通过初始化函数传递从该 API 接收到的任何 test 实例,然后再在代码中使用它。

The previous responses suggest switching to a class, instead of using a record. If you want to stay with records (for their simple syntax and immutability), you can take this approach:

type test =
    { a : float
      b : float }
    static member initialize (t: test) =
        { t with b = t.a * 2. }

This is useful if the instance of test is created by another library (like a data provider from a web service or database). With this approach, you must remember to pass any instance of test that you receive from that API through the initialize function before using it in your code.

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