F#计算表达式和返回语句

在Try F#网站上，他们给出了一个计算表达式的示例：

type Age =
| PossiblyAlive of int
| NotAlive

type AgeBuilder() =
    member this.Bind(x, f) =
        match x with
        | PossiblyAlive(x) when x >= 0 && x <= 120 -> f(x)
        | _ -> NotAlive
    member this.Delay(f) = f()
    member this.Return(x) = PossiblyAlive x

let age = new AgeBuilder()

let willBeThere (a:int) (y:int) =
  age { 
    let! current = PossiblyAlive a
    let! future = PossiblyAlive (current + y)

    return future
  }

let! current = age { return a }
let! future = age { return (current + y) }

将是：

let! current = return a
let! future = return (current + y)

但是它不起作用。我最接近的是：

let! current = age.Return a
let! future = age.Return (current + y)

---

但这看起来很脏。是否有任何方法可以使用

return

，而不显式使用计算生成器函数

您可以创建嵌套表达式：

type Age =
| PossiblyAlive of int
| NotAlive

type AgeBuilder() =
    member this.Bind(x, f) =
        match x with
        | PossiblyAlive(x) when x >= 0 && x <= 120 -> f(x)
        | _ -> NotAlive
    member this.Delay(f) = f()
    member this.Return(x) = PossiblyAlive x

let age = new AgeBuilder()

let willBeThere (a:int) (y:int) =
  age { 
    let! current = PossiblyAlive a
    let! future = PossiblyAlive (current + y)

    return future
  }

let! current = age { return a }
let! future = age { return (current + y) }

尽管您可以使用

来代替：

let current = a
let future = current + y

请注意，此生成器自

return 150>>=return
与
return 150
我已经更详细地研究了这个问题，并且我认为我已经找到了一个合理的替代方法，可以替代Lee在回答中显示的
age{return}
语法

我对这种语法的主要不满是我们已经处于
age
monad中，因此身体中的任何
return
语句都应该自动解析为
age.return
。然而，对于F#团队来说，解决这个问题的优先级可能很低，因为解决方法非常简单

我的替代方法是使用一个函数重载
Bind
方法，该函数获取一个值，然后将其提升；然后将该提升值发送给另一个
Bind
函数：

type Age =
| PossiblyAlive of int
| NotAlive

type AgeBuilder() =
    let reasonableAge (x:int) = x >= 0 && x <= 120

    member __.Return x = 
        if reasonableAge x then PossiblyAlive x else NotAlive

    member __.Bind(x:Age, f) =
        match x with
        | PossiblyAlive x when reasonableAge x -> f x
        | _ -> NotAlive

    member this.Bind(x:int, f) =
        this.Bind(this.Return(x), f)

let age = new AgeBuilder()

let willBeThere (a:int) (y:int) =
    age { 
        let! current = a
        let! future = (current + y)
        return future
    }

类型年龄=
|int的可能性
|不活着
类型AgeBuilder（）=
设reasonalAge（x:int）=x>=0&&xfx
|现场直播
成员this.Bind（x:int，f）=
this.Bind（this.Return（x），f）
let age=new AgeBuilder（）
让willBeThere（a:int）（y:int）=
年龄{
让！电流=a
让！future=（当前+y）
回归未来
}
你的代码对我有用-你得到了什么错误？没有错误-关于计算表达式语法和“return”语句。完美。。。回答了我的问题，也证实了我的怀疑，即“Return”定义没有进行边界检查，这有点“不确定”。