FSharp和向接口的向上转换似乎是多余的

Question

我有以下使用反应式扩展的代码段:

    let value : 't = ...

    Observable.Create<'t>(fun observer ->
        let subject = new BehaviorSubject<'t>(value)
        let d0 = subject.Subscribe(observer)
        let d1 = observable.Subscribe(subject)
        new CompositeDisposable(d0, d1) :> IDisposable
    )

这有效.但是,如果我将upcast丢弃到IDisposable,则代码无法编译,引用了模糊的重载.但是CompositeDisposable是一个IDisposable.为什么类型推理引擎无法解决此问题？注意我几乎一直在C#中使用这种模式从Observable.Create返回CompositeDisposable而不必进行upcast.

Answer 1

正如@kvb所说,函数不支持方差,因此接口和子类需要upcast.

这是一个演示子类行为的小例子:

type A() =
    member x.A = "A"

type B() =
    inherit A()
    member x.B = "B"

let f (g: _ -> A) = g()

let a = f (fun () -> A()) // works
let b = f (fun () -> B()) // fails

如果函数f是由您编写的,则添加类型约束可能有所帮助:

// This works for interface as well
let f (g: _ -> #A) = g()

let a = f (fun () -> A()) // works
let b = f (fun () -> B()) // works

否则,你必须按照你所描述的例子做一个litle upcast.