如何解决具有静态解析类型参数的递归映射中的奇怪类型错误？

Question

type CudaInnerExpr<'t> = CudaInnerExpr of expr: string with 
    member t.Expr = t |> fun (CudaInnerExpr expr) -> expr 

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 CudaAr2D<'t> = CudaAr2D of CudaScalar<int> * CudaScalar<int> * name: string with 
    member t.Name = t |> fun (CudaAr2D (_, _, name)) -> name 

type ArgsPrinter = ArgsPrinter with 
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar<float32>) = sprintf "float %s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaScalar<int>) = sprintf "int %s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr1D<float32>) = sprintf "float *%s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr1D<int>) = sprintf "int *%s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr2D<float32>) = sprintf "float *%s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr2D<int>) = sprintf "int *%s" t.Name 

    static member inline PrintArg(_: ArgsPrinter, (x1, x2)) = 
     let inline print_arg x = 
      let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x) 
      call ArgsPrinter 
     [|print_arg x1;print_arg x2|] |> String.concat ", " 
    static member inline PrintArg(_: ArgsPrinter, (x1, x2, x3)) = 
     let inline print_arg x = 
      let inline call (tok : ^T) = ((^T or ^in_) : (static member PrintArg: ArgsPrinter * ^in_ -> string) tok, x) 
      call ArgsPrinter 
     [|print_arg x1;print_arg x2;print_arg x3|] |> String.concat ", " 

在行static member inline PrintArg(_: ArgsPrinter, (x1, x2, x3)) =，表达(x1, x2, x3)给了我以下错误：

Script1.fsx(26,52): error FS0001: This expression was expected to have type 
    'in_  
but here has type 
    'a * 'b * 'c 

任何想法是在这里做，使这项工作？

Answer 1

想要做这样的事情在我看来：

... 

    static member inline PrintArg(_: ArgsPrinter, t: CudaAr2D<float32>) = sprintf "float *%s" t.Name 
    static member inline PrintArg(_: ArgsPrinter, t: CudaAr2D<int>) = sprintf "int *%s" t.Name 

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 ", " 

您在类型中定义的通用功能，因为你会使用它的最后两个重载，这将成为一种'递归重载'。

请注意，这种技术目前在FSharpPlus使用，实际上是一种简化的技术。

最后请注意，你的解决方案对我来说似乎也是正确的（尽管更详细），但由于某种原因，F＃编译器感到困惑，我无法向你解释为什么，但遇到了像这样的许多情况，我所能做的就是找到一个最小的repro，一个解决方法并将它报告给F＃的人。约束求解器中还有很多事情需要解决。