抽象类类型签名中的可选参数

Question

班级:

type NotAbstract () = 
    member this.WithOptionalParameters (x, ?y) = 
        let y = defaultArg y 10
        x + y

具有以下类型签名:

type NotAbstract =
  class
    new : unit -> NotAbstract
    member WithOptionalParameters : x:int * ?y:int -> int
  end

但是,这不起作用:

[<AbstractClass>]
type AbstractExample () = 
    abstract WithOptionalParameters: int * ?int -> int /// Ouch...

type NotAbstract () = 
    inherit AbstractExample ()
    override this.WithOptionalParameters (x, ?y) = 
        let y = defaultArg y 10
        x + y

如何在带有可选参数的函数的抽象定义中编写正确的类型签名？我没有在这里找到任何暗示.

PS:我知道(类似的)结果可以通过多态实现

Answer 1

将参数声明为Option类型并不会使参数成为可选参数.

NotAbstract().WithOptionalParameters(2)
// This expression was expected to have type
//     int * Option<int>    
// but here has type
//     int    

该规范§8.13.6有它:

在签名中,可选参数如下所示: static member OneNormalTwoOptional : arg1:int * ?arg2:int * ?arg3:int -> int

因此,在抽象成员签名中命名可选参数

[<AbstractClass>]
type AbstractExample () = 
    abstract WithOptionalParameters: int * ?y:int -> int      

type NotAbstract () = 
    inherit AbstractExample ()
    override this.WithOptionalParameters (x, ?y) = 
        let y = defaultArg y 10
        x + y

NotAbstract().WithOptionalParameters(42)  // val it : int = 52