Question

为什么这是可以接受的：

type SomeClass<'T> =
    val mutable id : int
    val mutable result : 'T

但这不是：

type SomeIface = 
    abstract id : int

type SomeClass<'T> = 
    interface SomeIface with
        val mutable id : int
        val mutable result : 'T

编译器抱怨我使用'val'告诉我使用'member'但是我不能使用mutable。

Answer 1

移动接口实现上方的字段

type SomeIface = 
    abstract id : int

type SomeClass<'T>() = 
    [<DefaultValue>]
    val mutable id : int
    [<DefaultValue>]
    val mutable result : 'T
    interface SomeIface with
        member this.id = this.id

let x = SomeClass<int>(id = 10)
let y : SomeIface = upcast x
printfn "%d" y.id

Answer 2

Desco的回答是正确的。至于为什么你的方法不起作用，关键是在F＃中，当你有类似的东西时

interface Iface with
   [indented lines here]

缩进行只能包含接口成员的实现。它们不应包含您定义的类型的其他字段或成员（例如您的两个可变字段）。因此，desco的答案有效，如下所示：

type SomeIface = 
    abstract id : int

type SomeClass<'T> = 
    interface SomeIface with
        member this.id = this.id
    val mutable id : int
    val mutable result : 'T

带有和不带接口的F＃类定义

2 个答案: