Question

我尝试了两种变体，它们都有同样的问题。

type CudaScalar<'t> = CudaScalar of name: string with
    member t.Name = t |> fun (CudaScalar name) -> name

type TypePrinter<'t>() = class end

let inline print_type x =
    ((^T or ^in_) : (static member PrintType: TypePrinter< ^in_> -> string) x)

type TypePrinter with
    static member inline PrintType(_: TypePrinter<float32>) = "float32"
    static member inline PrintType(_: TypePrinter<int>) = "int"

type ArgsPrinter = ArgsPrinter
let inline print_arg x =
    let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x)
    call ArgsPrinter       

type ArgsPrinter with
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar< ^t>) = 
        [|print_type (TypePrinter< ^t>()); t.Name|] |> String.concat " "

type CudaScalar<'t> = CudaScalar of name: string with
    member t.Name = t |> fun (CudaScalar name) -> name

type TypePrinter = TypePrinter

let inline print_type x =
    let call (tok: ^T) = ((^T or ^in_) : (static member PrintType: TypePrinter * ^in_ -> string) tok, x)
    call TypePrinter

type TypePrinter with
    static member inline PrintType(_: TypePrinter,_: float32) = "float32"
    static member inline PrintType(_: TypePrinter,_ : int) = "int"

type ArgsPrinter = ArgsPrinter
let inline print_arg x =
    let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x)
    call ArgsPrinter       

type ArgsPrinter with
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar< ^t>) = 
        [|print_type Unchecked.defaultof< ^t>; t.Name|] |> String.concat " "

起初我得到了与previous question完全相同的错误，但在我删除之后：

static member inline PrintArg(_: ArgsPrinter, (x1, x2)) = 
    [|print_arg x1;print_arg x2|] |> String.concat ", "
static member inline PrintArg(_: ArgsPrinter, (x1, x2, x3)) = 
    [|print_arg x1;print_arg x2;print_arg x3|] |> String.concat ", "

我在最后3行中得到了这个（对于两种变体）：

Script1.fsx(16,34): error FS0193: Type constraint mismatch. The type 
    'in_    
is not compatible with type
    FSI_0002.CudaScalar<'a>    
The type ''in_' does not match the type 'FSI_0002.CudaScalar<'a>'

这让我感到惊讶，因为错误没有出现在Intellisense中，只有当我尝试在F＃Interactive中运行它时才会出现。我不确定再次在这做什么。我可能会过多地滥用类型系统，但我决心以无标记的方式完成这个Cuda编译器，无论如何。我需要这样才能传播后期阶段的类型信息。

编辑：Gustavo的修复工作正常，但是当我再往前走时它会再次破坏。

type CudaScalar<'t> = CudaScalar of name: string with
    member t.Name = t |> fun (CudaScalar name) -> name
type CudaAr1D<'t> = CudaAr1D of CudaScalar<int> * name: string with
    member t.Name = t |> fun (CudaAr1D (_, name)) -> name

type TypePrinter<'t>() = class end

let inline print_type x =
    ((^T or ^in_) : (static member PrintType: TypePrinter< ^in_> -> string) x)

type TypePrinter with
    static member inline PrintType(_: TypePrinter<float32>) = "float32"
    static member inline PrintType(_: TypePrinter<int>) = "int"

type ArgsPrinter = ArgsPrinter with
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar< ^t>) = 
        [|print_type (TypePrinter< ^t>()); t.Name|] |> String.concat " "
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr1D< ^t>) = 
        [|print_type (TypePrinter< ^t>()); "*"; t.Name|] |> String.concat " "

let inline print_arg x =
    let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x)
    call ArgsPrinter       

type ArgsPrinter with
    static member inline PrintArg(_: ArgsPrinter, (x1, x2)) = 
        [|print_arg x1;print_arg x2|] |> String.concat ", "
    static member inline PrintArg(_: ArgsPrinter, (x1, x2, x3)) = 
        [|print_arg x1;print_arg x2;print_arg x3|] |> String.concat ", "

再次，在最后两行中，我表现出与pt完全相同的错误。其中一个。我在这里要做的是成为嵌入式Cuda DSL的一部分。之所以我这样做而不是使用DU，是因为使用DUs，我必须将接口作为一个单独的部分。如果我这样做，我将能够编写cuda_map (fun x -> x*x)并编译它，因为类型信息将存在。特别是上面的片段应该打印出内核方法的参数。

如果这是Haskell，我会在Ocaml GADT中使用类型类+ HKTs +无标记样式来实现它，但在F＃中，对我来说唯一真实的选择是使用静态解析类型参数来模仿finally tagless style，但是我没有指望存在编译器错误。

我打算将这个帖子传递给我打开的问题。

Answer 1

我不确定我是否理解你的问题/你想要实现的目标。

我不知道为什么你在这个类型的中间插入了这个函数，在你之前的问题中它是必要的但是在这里我不这么认为。

如果你按自然顺序编写它，它会编译：

type CudaScalar<'t> = CudaScalar of name: string with
    member t.Name = t |> fun (CudaScalar name) -> name

type TypePrinter<'t>() = class end

let inline print_type x =
    ((^T or ^in_) : (static member PrintType: TypePrinter< ^in_> -> string) x)

type TypePrinter with
    static member inline PrintType(_: TypePrinter<float32>) = "float32"
    static member inline PrintType(_: TypePrinter<int>) = "int"

type ArgsPrinter = ArgsPrinter with
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar< ^t>) = 
        [|print_type (TypePrinter< ^t>()); t.Name|] |> String.concat " "

let inline print_arg x =
    let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x)
    call ArgsPrinter

如果我错过了什么，请告诉我。

修改

尝试添加更多重载后，请尝试按以下方式添加：

type ArgsPrinter with static member inline PrintArg(_: ArgsPrinter, (x1, x2)) = [|print_arg x1;print_arg x2|] |> String.concat ", " type ArgsPrinter with static member inline PrintArg(_: ArgsPrinter, (x1, x2, x3)) = [|print_arg x1;print_arg x2;print_arg x3|] |> String.concat ", "

我的意思是'延伸'两次。

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