Syntax F中项属性的类型#

考虑接口：

type IVector = 
    abstract Item : int -> float

现在，让我们定义类：

type DenseVector(size : int) = 
    let mutable data = Array.zeroCreate size

    interface IVector with 
        member this.Item with get n = data.[n]

提供一种方法来改变稠密向量的第n个条目怎么样？然后，最好将上述代码修改为：

type DenseVector(size : int) = 
    let mutable data = Array.zeroCreate size

    interface IVector with 
        member this.Item with get n = data.[n]
                          and set n value = data.[n] <- value

类型DenseVector（大小：int）=
让可变数据=Array.zeroCreate size
接口IVector与
使用get n=data成员此.Item。[n]
并设置n value=data。[n]您可以在不更改原始接口的情况下实现DenseVector，同时提供如下设置器：

type IVector = 
    abstract Item: int -> float with get

type DenseVector(size : int) = 
    let data = Array.zeroCreate size
    interface IVector with 
        member this.Item with get i = data.[i]
    member this.Item 
        with get i = (this :> IVector).[i]
        and set i value = data.[i] <- value

类型IVector=
抽象项：int->float with get
类型DenseVector（大小：int）=
让data=Array.zeroCreate size
接口IVector与
使用get i=data来成员此.Item。[i]
成员：本项目
使用get i=（this:>IVector）。[i]
设置i值=数据。[i]非常好。非常感谢你@毛里西奥，是的。也就是说，如果您希望通过接口调用setter，则是这样。如果没有，并且接口中只有getter，那么在这里取消对setter的注释仍然是非法的：“public class V:IV{double IV.this[int x]{get{return 0.0；}/*set{}*/}”（回想一下，F#中的所有接口都是显式的），您必须与接口分开实现该属性。
type IVector = 
    abstract Item: int -> float with get

type DenseVector(size : int) = 
    let data = Array.zeroCreate size
    interface IVector with 
        member this.Item with get i = data.[i]
    member this.Item 
        with get i = (this :> IVector).[i]
        and set i value = data.[i] <- value