F#计算表达式中展开嵌套元组

Question

我有一个计算表达式，想要返回一个扁平化的元组作为第一个元素，以及一个 int 作为第二个元素。 我试图使用方法重载来实现这一点。 现在编译器报错说它找不到唯一的重载。我不知道如何帮助编译器弄清楚。对我来说，这似乎是确定性的。

type IntBuilder () =

    member inline this.Yield (i:int) =
        i

    member inline this.For(source:seq<'a>, body:'a -> seq<'b * int>) =
        source
        |> Seq.collect (fun x -> body x |> Seq.map (fun (idx, i) -> (x, idx), i))

    member inline this.For(source:seq<'a>, body:'a -> int) =
        source |> Seq.map (fun x -> x, body x)

    member inline this.Run(source:seq<('a * ('b * ('c * 'd))) * 'v>) =
        source 
        |> Seq.map (fun ((x, (y, (z, a))), d) -> (x, y, z, a), d)

    member inline this.Run(source:seq<('a * ('b * 'c)) * 'v>) =
        source 
        |> Seq.map (fun ((x, (y, z)), d) -> (x, y, z), d)

    member inline this.Run(source:seq<('a * 'b) * 'v>) =
        source 
        |> Seq.map (fun ((x, y), d) -> (x, y), d)

    member inline this.Run(source:seq<'a * 'v>) =
        source 
        |> Seq.map (fun (x, d) -> x, d)

let intBuilder = IntBuilder ()
let c = 
    intBuilder {
        for i in 1..2 do
            for j in 1..2 do
                for k in 1..2 do
                    for l in 1..2 -> 
                         i + j + k + l
    }

// What I get
c : seq<(int * (int * (int * int))) * int>

// What I want
c : seq<(int * int * int * int) * int>

在这种情况下，c 的类型是 seq<(int * (int * (int * int))) * int>。我希望 IntBuilder 计算返回 seq<(int * int * int * int), int>。我该如何做到这一点？

- Matthew Crews

Answer 1

这可以通过内联和重载解决，但一般来说（据我所知），不可能有一个函数能够根据它所接收到的任意类型的输入返回类型为int * int * ...的值。也许其他人可以就此发表意见。

但我们可以通过运行时类型测试实现扁平化。

let flatten tuple =
    let rec fold (tuple: obj) acc = 
        match tuple with
        | :? int as v -> v::acc
        | :? ITuple as rest ->  
            let next, value = rest.[0], rest.[1] :?> int
            fold next (value::acc)
        | _ -> failwith "Unexpected type"

    fold tuple []

> flatten ((1, 2), 3);;
val it : int list = [1; 2; 3]

- Asti

Answer 2

看起来唯一让这个生效的方法是将它们全部包装在具体类型中：

type T1<'a> = | T1 of seq<'a * int>
type T2<'a,'b> = | T2 of seq<('a * 'b) * int>
type T3<'a,'b,'c> = | T3 of seq<('a * ('b * 'c)) * int>
type T4<'a,'b,'c,'d> = | T4 of seq<('a * ('b * ('c * 'd))) * int>
type T5<'a,'b,'c,'d,'e> = | T5 of seq<('a * ('b * ('c * ('d * 'e)))) * int>

type IntBuilder () =
    member this.Yield (i:int) =
        i

    member this.For(source:seq<'a>, body:'a -> int) =
        source |> Seq.map (fun x -> x, body x)
        |> T1.T1

     member this.For(source:seq<'a>, body:'a -> T1<'b>) =
        source
        |> Seq.collect (fun x -> 
            body x 
            |> fun (T1.T1 x) -> x
            |> Seq.map (fun (idx, i) -> (x, idx), i))
        |> T2.T2

    member this.For(source:seq<'a>, body:'a -> T2<'b,'c>) =
       source
       |> Seq.collect (fun x -> 
           body x 
           |> fun (T2.T2 x) -> x
           |> Seq.map (fun (idx, i) -> (x, idx), i))
       |> T3.T3

    member this.For(source:seq<'a>, body:'a -> T3<'b,'c,'d>) =
       source
       |> Seq.collect (fun x -> 
           body x 
           |> fun (T3.T3 x) -> x
           |> Seq.map (fun (idx, i) -> (x, idx), i))
       |> T4.T4

    member this.For(source:seq<'a>, body:'a -> T4<'b,'c,'d,'e>) =
        source
        |> Seq.collect (fun x -> 
            body x 
            |> fun (T4.T4 x) -> x
            |> Seq.map (fun (idx, i) -> (x, idx), i))
        |> T5.T5

    member inline this.Run(T1.T1 source) =
        source 
        |> Seq.map (fun (x, d) -> x, d)

    member inline this.Run(T2.T2 source) =
        source 
        |> Seq.map (fun ((x, y), d) -> (x, y), d)

    member inline this.Run(T3.T3 source) =
        source 
        |> Seq.map (fun ((x, (y, z)), d) -> (x, y, z), d)

    member inline this.Run(T4.T4 source) =
        source 
        |> Seq.map (fun ((x, (y, (z, a))), d) -> (x, y, z, a), d)

let intBuilder = IntBuilder ()


let c = 
    intBuilder {
        for i in 1..2 do
            for j in 1..2 do
                for k in 1..2 do
                    for l in 1..2 -> 
                         i + j + k + l
    }

FSI输出：

val c : seq<(int * int * int * int) * int>

> c;;
val it : seq<(int * int * int * int) * int> =
  seq
    [((1, 1, 1, 1), 4); ((1, 1, 1, 2), 5); ((1, 1, 2, 1), 5);
     ((1, 1, 2, 2), 6); ...]

- Cody Johnson

Answer 3

回答编辑后的问题：

编译器无法解析正确的重载，因为

seq<('a * ('b* ('c * 'd))) * 'v>) 

是以下类型的子类型：

seq<'u * 'v> where 'u = 'a * ('b* ('c * 'd)))

对于其他类型也是如此。因此，将某些类型限制为具体类型。限制最内层元组类型应该足够。

例如：

member  this.Run(source:seq<('a * ('b* ('c* int))) * 'v>) =
    source 
    |> Seq.map (fun ((x, (y, (z, a))), d) -> (x, y, z, a), d)

member  this.Run(source:seq<('a * ('b* int)) * 'v>) =
    source 
    |> Seq.map (fun ((x, (y, z)), d) -> (x, y, z), d)

member  this.Run(source:seq<('a * int) * 'v>) =
    source 
    |> Seq.map (fun ((x, y), d) -> (x, y), d)

member  this.Run(source:seq<int * 'v>) =
    source 
    |> Seq.map (fun (x, d) -> x, d)

- Asti