f#中的重载取消引用(!)和赋值(:=)运算符

时间:2016-02-09 05:17:59

标签: f# overloading operator-keyword

我正在尝试重新引用解除引用(!)和赋值(:=)运算符,但不是全局运算符。我仍然希望保持通常的ref op重载。这里有一些代码来说明问题:

type MyVar<'a>(init:'a) =
    let mutable _value = init
    member __.Get() = _value
    member __.Set x = _value <- x
    //static member (!) (s:MyVar<'a>) = s.Get()      // compiles, doesn't work
    //static member (:=) (d:MyVar<'a>, s) = d.Set(s) // warning, doesn't work

//let inline (!) (x :MyVar<'a>) = x.Get()           // overrides !ref
//let inline (:=) (x :MyVar<'a>) (v :'a) = x.Set(v) // overrides ref := v 
let inline (!!) (x :MyVar<'a>) = x.Get()            // works but ugly
let inline (.=) (x :MyVar<'a>) (v :'a) = x.Set(v)   // works ... meh

let test_myvar() =
    let mv = new MyVar<_>("wee")
    let r = ref 100
    let x = !mv
    let y = !!mv
    let z = !r
    mv .= "haaa"
    r := 42

解决方案:

@ Carsten的解决方案正是我所寻求的&amp;作品。然而,事实证明我正在使用Websharper,它使用Quotations进行编译,而@Carstens解决方案变得有点more complex。由于Websharper.UI.Next包含该解决方案,我所需要的只是包含在我的项目中,并且它可以工作!

1 个答案:

答案 0 :(得分:5)

您可以通过重复onCreate()(!)运算符来尝试使用static constraints

(:=)

我删除了您的访问者,因为我只需要与type MyVar<'a>(init:'a) = let mutable _value = init member __.Value with get () = _value and set v = _value <- v let inline (!) a = (^a : (member Value : ^b) a) let inline (:=) a v = (^a : (member Value : ^b with set) (a, v)) 相同(但您可以重新添加)

示范

这是一个F#-interactive会话,用你的价值证明这一点:

Ref<'a>

<强>备注

我不确定我是否真的会这样做 - 你只是重新发明val mv : MyVar<string> val r : int ref = {contents = 100;} > !mv;; val it : string = "wee" > !r;; val it : int = 100 > mv := "It works";; val it : unit = () > !mv;; val it : string = "It works" > r := 50;; val it : unit = () > !r;; val it : int = 50 - 细胞(作为一个班级)并且什么也得不到,当然它可能很难为其他人阅读 - 所以请对待在意。