Question

我怎么能在这种类型中调用send（a：'T []）？

type Test() =
  member o.send(value:'T) = 4
  member o.send(values:'T []) = 5

let test = Test()
let b = test.send [|4|]

当我这样做时，我获得了

A unique overload for method 'Send' could not be determined based on type information 
prior to this program point. The available overloads are shown below...

关键是MPI.NET正好有这个名为Send的方法，我无法向其中发送数组。

谢谢你，
Oldrich

Answer 1

好吧，F＃无法确定你的T是int还是int数组，所以我能看到解决问题的唯一方法是这样的

let b = test.send<int> [|4|]

或者

let b = test.send<int array> [|4|]

Answer 2

type Test() =
  member o.send(a:'T) = 4
  member o.send(a:'T []) = 5

let test = Test()
let b = test.send<int>([|4|] : _[]) // 5

Answer 3

您可以通过在类型上使用模式匹配来完成此操作：

type MyTest() =
    member this.Send (a : obj)  = 
        match a with
        | :? System.Collections.IEnumerable -> 5
        | _ -> 4

这将匹配F＃（array，list，seq）中的任何Seq类型。一旦你有正确的匹配，通常更容易实现特定类型的私有函数。

F＃独特的过载方法

3 个答案: