On Tue, Jun 30, 2020 at 12:26 PM Sam Kuper <sampablokuper@posteo.net> wrote:
Dear list,

Forgive me for asking a very basic question, but I have not so far been
able to find an answer in any of the OCaml books to which I have access,
nor in the OCaml documentation or mailing list archive.

How does one define a type whose values are restricted to one of some
specified chars?

E.g. suppose I want to define a type `ab` whose values can only be
either 'a' or 'b'.  I imagine that should work something like this:

    # type ab = Ab of char 'a' | Ab of char 'b' ;;
    type ab = Ab of char 'a' | Ab of char 'b'

and thereby give the following functionality:

    # Ab 'a';;
    - : ab = Ab 'a'
    # Ab 'b';;
    - : ab = Ab 'b'
    # Ab 'c';;
    Error: <some error>

The definition above is essentially pseudo-code to illustrate what I
would like to achieve with real, valid OCaml code.  (If I knew how to
write valid OCaml to achieve this, then I would not be posting this
question on the mailing list.)

Here are several of my failed attempts at writing OCaml code for what I
want to achieve:

    # type ab = 'a' | 'b';;
    Error: Syntax error

    # type ab = char 'a' | char 'b';;
    Error: Syntax error

    # type ab = Ab of char 'a' | Ab of char 'b';;
    Error: Syntax error

    # type 'a ab = 'a constraint 'a = 'a' | 'b';;
    Error: Syntax error

    # type 'a ab = 'a constraint 'a = 'a' | 'a = 'b';;
    Error: Syntax error

How can I actually achieve it?

Thank you in advance,

Sam

--
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Hi Sam.

As a small variation of Yawar's excellent reply, consider the following (the file is named `sam.ml`):

    module AB : sig
      type t = private char

      val a : t
      val b : t
      val char : t -> char
    end = struct
      type t = char

      let a = 'a'
      let b = 'b'
      let char t = t
    end

    let () =
      let x = AB.a in
      let y = AB.b in
      assert (AB.char x = 'a') ;
      assert (AB.char y = 'b')

I like this approach because it statically guarantees that a value of `Ab.t` is always either 'a' or 'b', and attempting to do otherwise results in a compilation error.

Best,
--
Jesse