我目前正在尝试使用通过OCaml语言实现的静态类型系统来实现类型分析器。
我正在使用的算法是先生成类型方程,然后使用统一算法求解这些方程。除了递归letrec-exp
绑定类型之外,我已经能够很好地实现代码。
这是完整代码:
type program = exp
and exp =
| NUM of int
| VAR of var
| SUM of exp * exp
| DIFF of exp * exp
| MULT of exp * exp
| DIV of exp * exp
| ZERO of exp
| IF of exp * exp * exp
| LET of var * exp * exp
| LETREC of var * var * exp * exp
| FUNC of var * exp
| CALL of exp * exp
and var = string
;;
type typ = TypeInt | TypeBool | TypeFun of typ * typ | TypeVar of tyvar
and tyvar = string
;;
type type_equation = (typ * typ) list;;
module TypeEnv = struct
type t = var -> typ
let empty
= fun _ -> raise (Failure "Type Env is empty")
let extend (x, t) tenv
= fun y -> if x = y then t else (tenv y)
let find tenv x = tenv x
end
;;
let typevar_num = ref 0;;
let new_typevar () = (typevar_num := !typevar_num + 1; (TypeVar ("t" ^ string_of_int !typevar_num)));;
let rec generate_eqns : TypeEnv.t -> exp -> typ -> type_equation
= fun tenv e ty ->
match e with
| NUM n -> [(ty, TypeInt)]
| VAR x -> [(ty, TypeEnv.find tenv x)]
| SUM (e1, e2) -> let eq1 = [(ty, TypeInt)] in
let eq2 = generate_eqns tenv e1 TypeInt in
let eq3 = generate_eqns tenv e2 TypeInt in
eq1 @ eq2 @ eq3
| DIFF (e1, e2) -> let eq1 = [(ty, TypeInt)] in
let eq2 = generate_eqns tenv e1 TypeInt in
let eq3 = generate_eqns tenv e2 TypeInt in
eq1 @ eq2 @ eq3
| DIV (e1, e2) -> let eq1 = [(ty, TypeInt)] in
let eq2 = generate_eqns tenv e1 TypeInt in
let eq3 = generate_eqns tenv e2 TypeInt in
eq1 @ eq2 @ eq3
| MULT (e1, e2) -> let eq1 = [(ty, TypeInt)] in
let eq2 = generate_eqns tenv e1 TypeInt in
let eq3 = generate_eqns tenv e2 TypeInt in
eq1 @ eq2 @ eq3
| ISZERO e -> let eq1 = [(ty, TypeBool)] in
let eq2 = generate_eqns tenv e TypeInt in
eq1 @ eq2
| IF (e1, e2, e3) -> let eq1 = generate_eqns tenv e1 TypeBool in
let eq2 = generate_eqns tenv e2 ty in
let eq3 = gen_equations tenv e3 ty in
eq1 @ eq2 @ eq3
| LET (x, e1, e2) -> let t1 = new_typevar () in
let eq1 = generate_eqns tenv e1 t1 in
let eq2 = generate_eqns (TypeEnv.extend (x, t1) tenv) e2 ty in
eq1 @ eq2
| LETREC (f, x, e1, e2) -> (* let t1 = new_typevar () in
let new_env = TypeEnv.extend (x, t1) tenv in
let eq1 = generate_eqns new_env f *)
| FUNC (x, e) -> let t1 = new_typevar () in
let t2 = new_typevar () in
let eq1 = [(ty, TypeFun (t1, t2))] in
let eq2 = generate_eqns (TypeEnv.extend (x, t1) tenv) e t2 in
eq1 @ eq2
| CALL (e1, e2) -> let t1 = new_typevar () in
let eq1 = generate_eqns tenv e1 (TypeFun (t1, ty)) in
let eq2 = generate_eqns tenv e2 t1 in
eq1 @ eq2
;;
执行类型方程式生成的主要功能是generate_eqns
。它以空类型环境,表达式和初始类型作为参数,并被称为:generate_eqns TypeEnv.empty (NUM 3) (new_typevar ())
。
我无法实现LETREC
递归调用。我一直在尝试在线查找资料,但它们似乎对我的问题没有太大帮助。
尤其是,我一直在尝试从 Progrmaming Languages(3e)-Friedman&Wand 的基本语言分析此键入规则:
有人愿意给我一些指导或建议吗?
谢谢。
答案 0 :(得分:2)
所以我浏览了您的代码,未经测试的代码,等等。但是乍一看,我想应该是沿着这些思路写的东西,
| LETREC (f, x, e1, e2) -> let tx = new_typevar () in (** type of x **) let tfx = new_typevar () in (** type of f x **) let tf = TypeFun (tx, tfx) in (** type of f **) let tenvf = TypeEnv.extend (f, tf) tenv in (** f in env **) let tenvxf = TypeEnv.extend (x, tx) tenvf in (** x and f in env **) let eq1 = generate_eqns tenvxf e1 tfx in (** type e1 = type (f x) **) let eq2 = generate_eqns tenvf e2 ty in (** type e2 = typ **) eq1 @ eq2