Question

The following (simplified) snippet is taken from an application I’m implementing which consistently uses Type Parameters resolved statically.

type A< ^B when ^B : (static member MyMember : Unit -> Unit)> = {
  Field : unit
}
type TestA = {
  AField : A< BTy >
}
and BTy = {
  BField : Unit
} with
  static member MyMember () = ()

IntelliSense gives me the following error when I define the type of field AField (AField : A< BTy >) which is: The type ‘BTy’ does not support any operators named ‘MyMember’.

EDITED.
Declaring them separately works, but if I have a mutual reference and I cannot declare a third type to put in the top which contains the common Infos of the two types. What should I do to avoid this problem? Anyway, if I define below the definitions let pluto = ("" :> obj) :?> A< BTy > it works, I imagine because both types are visible from the let binding.

Answer 1

Your implementation @Tomas satisfies the question but there is some hesitation over the style of your solution because it does not respected the functional programming paradigm. I thought to a solution arising from the Type Classes implementation of Haskell.
Interfaces, abstract classes, etc. have been implemented in order to allow the F# environment to “interact” with the .Net environment, for a reason of uniformity of style the use of code which implements interfaces, abstract classes etc. in a context where no need of interact with the .Net libraries, it is a “Overhead” in my opinion (although F# is a multi-paradigm language). This is the reason why I found the implementation of the Type Classes of Haskell very elegant. Below I implemented through F# the “Haskell Type Class” code to solve my problem.

type Operations< ^a> = 
  {  
    MyMember : unit -> unit
  }

type A< ^a> = {
  Operations : Operations< ^a>
}

and TestA = { 
  AField : A< BTy > 
}

and BTy = { 
  BField : Unit 
}

let it = 
  let BTy_operations : Operations< BTy > = { MyMember = fun () -> () }
  let A_of_BTy = { Operations = BTy_operations }
  { AField = A_of_BTy }