如何让记录实现接口？

Question

如果我有一个接口：

type IData = 
  abstract member firstName: string
  abstract member lastName: string

如何定义符合此接口的记录类型。

我试过如下：

> type Data = { firstName: string; lastName: string } interface IData ;;


Snippet.js(43,63): error FS0366: No implementation was given for 'abstract member IData.firstName : string'. Note that all interface members must be implemented
and listed under an appropriate 'interface' declaration, e.g. 'interface ... with member ...'.

来自Records的官方参考：

记录字段与类的不同之处在于它们自动公开为属性

我的第一个问题是：如果属性是“自动公开的”，那么为什么我需要“做某事”来实现它们。

由于错误消息要求我提供接口的实现，因此我尝试了以下操作：

> type Data = { firstName: string; lastName: string; } interface IData with
-   member this.firstName with get () = this.firstName
-   member this.lastName with get () = this.lastName 

type Data =
  {firstName: string;
   lastName: string;}
  with
    interface IData
  end 

到目前为止一切顺利，但是现在当我尝试使用它时，我遇到了问题：

> let d: IData = { firstName = "john"; lastName = "doe" } ;;

error FS0001: This expression was expected to have type
    'IData'
but here has type
    'Data'

另一种尝试：

> let d = { firstName = "john"; lastName = "doe" }
- ;;
val d : Data = {firstName = "john";
                lastName = "doe";}

> let d2: IData = d ;;


C:\Users\loref\Workspace\source-nly10r\Untitled-1(25,17): error FS0001: This expression was expected to have type
    'IData'
but here has type
    'Data'

所以，我的第二个问题是，如果Data实现，IData那么为什么我不能将Datatype的值分配给type 的变量IData？

Answer 1

正如Gustavo所指出的，隐式接口实现正在由 F# 实现者讨论，目前尚不可用。

写的。我的第二个问题，需要显式转换：

> let d2: IData = d :> IData ;;
val d2 : IData = {firstName = "john";
                  lastName = "doe";}