F# 在计算表达式中展平嵌套元组

Question

我有一个计算表达式，我想返回一个扁平元组作为第一个元素和一个 int作为第二个。我正在尝试使用方法重载来完成此操作。现在编译器抛出一个错误，说它找不到唯一的重载。我不知道如何帮助编译器解决这个问题。这对我来说似乎是确定性的。

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类型为 seq<(int * (int * (int * int))) * int> .我想要 IntBuilder计算返回 seq<(int * int * int * int), int> .我该如何做到这一点？

Answer 1

这可以通过内联和重载决议来解决，但在一般意义上，(afaik)不可能有一个可以返回类型 int * int * ... 的函数基于给定的任意类型的输入。也许其他人可以权衡这一点。

不过，我们可以通过运行时类型测试来实现扁平化。

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]