7

我知道这在 C# 中是可能的，它产生简单而高效的代码。--- 同一类的两个对象可以访问彼此的私有部分。

class c1
{
    private int A;

    public void test(c1 c)
    {
        c.A = 5;
    }
}

但这在 F# 中似乎是不可能的，是真的吗？

type c1()
     let A = 0
     member test (c: c1) = c.A
4

5 回答 5

6

Interesting question. It seems to work with an explicit field but not with a let binding:

// Works
type c1 =
    val private A : int
    new(a) = { A = a }
    member m.test(c : c1) = c.A

let someC1 = new c1(1)
let someMoreC1 = new c1(42);
let theAnswer = someC1.test someMoreC1

// Doesn't work
type c2() =
    let mutable A = 42
    // Compiler error: The field, constructor or member 'A' is not defined
    member m.test(c : c2) = c.A
于 2012-09-13T12:20:19.637 回答
2

是的，但是在您的示例中，它在A语义上不是 的私有成员c1，它更像是构造函数的局部变量。

@afrischke 提供了一个示例，说明如何c1使用实际的私有成员A（使用val字段）进行定义。

于 2012-09-13T12:14:46.537 回答
1

这是可能的，并且被广泛使用，例如，用于检查成员相等性：

type c1 =
    member private this.A = 0
    interface IEquatable<c1> with
        member this.Equals (that: c1) = this.A = that.A
    // of course, it can be done in a regular method as well
    member this.Equals (that: c1) = this.A = that.A
于 2012-09-13T15:24:27.860 回答
1

正如 F# 规范的第8.6.1 .3 节所述：

实例定义定义的函数和值在词法范围内（因此隐式私有）到正在定义的对象。

于 2012-09-13T14:02:47.050 回答
1

您只需a在实例方法中直接 使用

type c1()
     let A = 0
     member x.test = A

对于静态方法，这不起作用，因为 let 绑定略有不同 - 然后您需要一个类定义，例如

type c1()
     private member x.A = 0
     static member test (A:c1) = A.A
于 2012-09-13T12:04:23.940 回答