我有一个计算表达式，我想返回一个扁平化的元组作为第一个元素，并int作为第二个元素。我正在尝试使用方法重载来完成此任务。现在，编译器抛出错误，表明找不到唯一的重载。我不知道如何帮助编译器解决问题。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是类型seq<(int * (int * (int * int))) * int>。我希望IntBuilder计算返回seq<(int * int * int * int), int>。我如何做到这一点？
看起来唯一可行的方法是将它们全部包装为具体类型：
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); ...]
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我来说两句