Creating GADT expression in OCaml

问题:

There is my toy GADT expression:

type _ expr =
  | Num : int -> int expr
  | Add : int expr * int expr -> int expr
  | Sub : int expr * int expr -> int expr
  | Mul : int expr * int expr -> int expr
  | Div : int expr * int expr -> int expr
  | Lt  : int expr * int expr -> bool expr
  | Gt  : int expr * int expr -> bool expr
  | And : bool expr * bool expr -> bool expr
  | Or  : bool expr * bool expr -> bool expr

Evaluation function:

let rec eval : type a. a expr -> a = function
  | Num n -> n
  | Add (a, b) -> (eval a) + (eval b)
  | Sub (a, b) -> (eval a) - (eval b)
  | Mul (a, b) -> (eval a) * (eval b)
  | Div (a, b) -> (eval a) / (eval b)
  | Lt  (a, b) -> (eval a) < (eval b)
  | Gt  (a, b) -> (eval a) > (eval b)
  | And (a, b) -> (eval a) && (eval b)
  | Or  (a, b) -> (eval a) || (eval b)

Creating expression is trivial when we limited to int expr:

let create_expr op a b =
  match op with
  | '+' -> Add (a, b)
  | '-' -> Sub (a, b)
  | '*' -> Mul (a, b)
  | '/' -> Div (a, b)
  | _ -> assert false

The question is how to support both int expr and bool expr in create_expr function.

My try:

type expr' = Int_expr of int expr | Bool_expr of bool expr

let concrete : type a. a expr -> expr' = function
  | Num _ as expr -> Int_expr expr
  | Add _ as expr -> Int_expr expr
  | Sub _ as expr -> Int_expr expr
  | Mul _ as expr -> Int_expr expr
  | Div _ as expr -> Int_expr expr
  | Lt  _ as expr -> Bool_expr expr
  | Gt  _ as expr -> Bool_expr expr
  | And _ as expr -> Bool_expr expr
  | Or  _ as expr -> Bool_expr expr

let create_expr (type a) op (a:a expr) (b:a expr) : a expr =
  match op, concrete a, concrete b with
  | '+', Int_expr  a, Int_expr  b -> Add (a, b)
  | '-', Int_expr  a, Int_expr  b -> Sub (a, b)
  | '*', Int_expr  a, Int_expr  b -> Mul (a, b)
  | '/', Int_expr  a, Int_expr  b -> Div (a, b)
  | '<', Int_expr  a, Int_expr  b -> Lt  (a, b)
  | '>', Int_expr  a, Int_expr  b -> Gt  (a, b)
  | '&', Bool_expr a, Bool_expr b -> And (a, b)
  | '|', Bool_expr a, Bool_expr b -> Or  (a, b)
  | _ -> assert false

It still can't return value of generalized type.

Error: This expression has type int expr but an expression was expected of type a expr Type int is not compatible with type a

UPDATE

Thanks to @gsg, I was able to implement type safe evaluator. Two tricks are crucial there:

• existential wrapper Any
• type tagging (TyInt and TyBool) that lets us to pattern match Any type

see

type _ ty =
  | TyInt : int ty
  | TyBool : bool ty

type any_expr = Any : 'a ty * 'a expr -> any_expr

let create_expr op a b =
  match op, a, b with
  | '+', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyInt,  Add (a, b))
  | '-', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyInt,  Sub (a, b))
  | '*', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyInt,  Mul (a, b))
  | '/', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyInt,  Div (a, b))
  | '<', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyBool, Lt  (a, b))
  | '>', Any (TyInt,  a), Any (TyInt,  b) -> Any (TyBool, Gt  (a, b))
  | '&', Any (TyBool, a), Any (TyBool, b) -> Any (TyBool, And (a, b))
  | '|', Any (TyBool, a), Any (TyBool, b) -> Any (TyBool, Or  (a, b))
  | _, _, _ -> assert false

let eval_any : any_expr -> [> `Int of int | `Bool of bool] = function
  | Any (TyInt, expr) -> `Int (eval expr)
  | Any (TyBool, expr) -> `Bool (eval expr)

回答1:

As you've found, this approach doesn't type check. It also has a more fundamental problem: GADTs can be recursive, in which case it is flatly impossible to enumerate their cases.

Instead you can reify types as a GADT and pass them around. Here's a cut-down example:

type _ expr =
  | Num : int -> int expr
  | Add : int expr * int expr -> int expr
  | Lt  : int expr * int expr -> bool expr
  | And : bool expr * bool expr -> bool expr

type _ ty =
  | TyInt : int ty
  | TyBool : bool ty

let bin_op (type a) (type b) op (l : a expr) (r : a expr) (arg_ty : a ty) (ret_ty : b ty) : b expr =
  match op, arg_ty, ret_ty with
  | '+', TyInt, TyInt -> Add (l, r)
  | '<', TyInt, TyBool -> Lt (l, r)
  | '&', TyBool, TyBool -> And (l, r)
  | _, _, _ -> assert false

At some point you are going to want to have a value that can be 'any expression'. Introducing an existential wrapper allows this. Cheesy example: generating random expression trees:

type any_expr = Any : _ expr -> any_expr

let rec random_int_expr () =
  if Random.bool () then Num (Random.int max_int)
  else Add (random_int_expr (), random_int_expr ())

let rec random_bool_expr () =
  if Random.bool () then Lt (Num (Random.int max_int), Num (Random.int max_int))
  else And (random_bool_expr (), random_bool_expr ())

let random_expr () =
  if Random.bool () then Any (random_int_expr ())
  else Any (random_bool_expr ())

回答2:

Your stated type for create_expr is char -> 'a expr -> 'a expr -> 'a expr. But the type for the '>' case would be char -> int expr -> int expr -> bool expr. So it seems there are problems with the basic plan.

In essence you want the type of the result to depend on the value of the character. I'm not absolutely positive, but I suspect this isn't possible in OCaml. Seems like it would require a more powerful type system.