使用OCaml中的GADT的简单lambda演算DSL

时间：2014-03-27 02:11:27

标签： ocaml dsl lambda-calculus gadt

如何使用GADT在OCaml中定义一个简单的lambda演算类DSL？具体来说，我无法弄清楚如何正确定义类型检查器以从无类型AST转换为类型化AST，也无法找出上下文和环境的正确类型。

以下是使用OCaml中传统方法的简单lambda演算语言的一些代码

(* Here's a traditional implementation of a lambda calculus like language *)

type typ =
| Boolean
| Integer
| Arrow of typ*typ

type exp =
| Add of exp*exp
| And of exp*exp
| App of exp*exp
| Lam of string*typ*exp
| Var of string
| Int of int
| Bol of bool

let e1=Add(Int 1,Add(Int 2,Int 3))
let e2=Add(Int 1,Add(Int 2,Bol false)) (* Type error *)
let e3=App(Lam("x",Integer,Add(Var "x",Var "x")),Int 4)

let rec typecheck con e =
    match e with
    | Add(e1,e2) ->
        let t1=typecheck con e1 in
        let t2=typecheck con e2 in
        begin match (t1,t2) with 
        | (Integer,Integer) -> Integer
        | _ -> failwith "Tried to add with something other than Integers"
        end
    | And(e1,e2) ->
        let t1=typecheck con e1 in
        let t2=typecheck con e2 in
        begin match (t1,t2) with 
        | (Boolean,Boolean) -> Boolean 
        | _ -> failwith "Tried to and with something other than Booleans"
        end
    | App(e1,e2) ->
        let t1=typecheck con e1 in
        let t2=typecheck con e2 in
        begin match t1 with 
        | Arrow(t11,t12) ->
            if t11 <> t2 then
                failwith "Mismatch of types on a function application"
            else
                t12
        | _ -> failwith "Tried to apply a non-arrow type" 
        end
    | Lam(x,t,e) ->
        Arrow (t,typecheck ((x,t)::con) e)
    | Var x  ->
        let (y,t) = List.find (fun (y,t)->y=x) con in
        t
    | Int _ -> Integer
    | Bol _ -> Boolean

let t1 = typecheck [] e1
(* let t2 = typecheck [] e2 *)
let t3 = typecheck [] e3

type value = 
| VBoolean of bool
| VInteger of int
| VArrow of ((string*value) list -> value -> value)

let rec eval env e = 
    match e with
    | Add(e1,e2) ->
        let v1=eval env e1 in
        let v2=eval env e2 in
        begin match (v1,v2) with 
        | (VInteger i1,VInteger i2) -> VInteger (i1+i2) 
        | _ -> failwith "Tried to add with something other than Integers"
        end
    | And(e1,e2) ->
        let v1=eval env e1 in
        let v2=eval env e2 in
        begin match (v1,v2) with 
        | (VBoolean b1,VBoolean b2) -> VBoolean (b1 && b2) 
        | _ -> failwith "Tried to and with something other than Booleans"
        end
    | App(e1,e2) ->
        let v1=eval env e1 in
        let v2=eval env e2 in
        begin match v1 with 
        | VArrow a1 -> a1  env v2 
        | _ -> failwith "Tried to apply a non-arrow type" 
        end
    | Lam(x,t,e) ->
        VArrow (fun env' v' -> eval ((x,v')::env') e) 
    | Var x  ->
        let (y,v) = List.find (fun (y,t)->y=x) env in
        v 
    | Int i -> VInteger i 
    | Bol b -> VBoolean b

let v1 = eval [] e1
let v3 = eval [] e3

现在，我正试图将其翻译成使用GADT的东西。这是我的开始

(* Now, we try to GADT the process *) 

type exp =
| Add of exp*exp
| And of exp*exp
| App of exp*exp
| Lam of string*typ*exp
| Var of string
| Int of int
| Bol of bool

let e1=Add(Int 1,Add(Int 2,Int 3))
let e2=Add(Int 1,Add(Int 2,Bol false))
let e3=App(Lam("x",Integer,Add(Var "x",Var "x")),Int 4)

type _ texp =
| TAdd : int texp * int texp -> int texp
| TAnd : bool texp * bool texp -> bool texp
| TApp : ('a -> 'b) texp * 'a texp -> 'b texp
| TLam : string*'b texp -> ('a -> 'b) texp
| TVar : string -> 'a texp
| TInt : int -> int texp
| TBol : bool -> bool texp

let te1 = TAdd(TInt 1,TAdd(TInt 2,TInt 3))

let rec typecheck : type a. exp -> a texp = fun e ->
    match e with
    | Add(e1,e2) ->
        let te1 = typecheck e1 in
        let te2 = typecheck e2 in
        TAdd (te1,te2)
    | _ -> failwith "todo"

这是问题所在。首先，我不确定如何在texp类型中为TLam和TVar定义正确的类型。通常，我会提供带有变量名称的类型，但我不确定如何在此上下文中执行此操作。其次，我不知道函数类型检查中上下文的正确类型。以前，我使用了某种列表，但现在我确定列表的类型。第三，在省略上下文后，类型检查功能不会自行检查。它失败并显示消息

File "test03.ml", line 32, characters 8-22:
Error: This expression has type int texp
       but an expression was expected of type a texp
       Type int is not compatible with type a

这完全有道理。这更像是一个问题，我不确定类型检查的正确类型是什么。

无论如何，你如何修复这些功能？

编辑1

这是上下文或环境的可能类型

type _ ctx =
| Empty : unit ctx
| Item :  string * 'a * 'b ctx -> ('a*'b) ctx

编辑2

环境的诀窍是确保环境的类型嵌入到表达式的类型中。否则，没有足够的信息来使事情类型安全。这是一个完整的翻译。目前，我没有有效的类型检查器来从无类型表达式转换为类型化表达式。

type (_,_) texp =
| TAdd : ('e,int) texp * ('e,int) texp -> ('e,int) texp
| TAnd : ('e,bool) texp * ('e,bool) texp -> ('e,bool) texp
| TApp : ('e,('a -> 'b)) texp * ('e,'a) texp -> ('e,'b) texp
| TLam : (('a*'e),'b) texp -> ('e,('a -> 'b)) texp
| TVar0 : (('a*'e),'a) texp
| TVarS : ('e,'a) texp -> (('b*'e),'a) texp
| TInt : int -> ('e,int) texp
| TBol : bool -> ('e,bool) texp

let te1 = TAdd(TInt 1,TAdd(TInt 2,TInt 3))
(*let te2 = TAdd(TInt 1,TAdd(TInt 2,TBol false))*)
let te3 = TApp(TLam(TAdd(TVar0,TVar0)),TInt 4)
let te4 = TApp(TApp(TLam(TLam(TAdd(TVar0,TVarS(TVar0)))),TInt 4),TInt 5)
let te5 = TLam(TLam(TVarS(TVar0)))

let rec eval : type e t. e -> (e,t) texp -> t = fun env e -> 
    match e with
    | TAdd (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 + v2
    | TAnd (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 && v2
    | TApp (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 v2
    | TLam e ->
        fun x -> eval (x,env) e 
    | TVar0 ->
        let (v,vs)=env in
        v
    | TVarS e ->
        let (v,vs)=env in
        eval vs e 
    | TInt i -> i
    | TBol b -> b

然后，我们有

# eval () te1;;
- : int = 6
# eval () te3;;
- : int = 8
# eval () te5;;
- : '_a -> '_b -> '_a = <fun>
# eval () te4;;
- : int = 9

2 个答案:

答案 0 :(得分：7)

如果您希望术语表示强制执行良好类型，则需要更改表示类型环境（和变量）的方式：您不能精确地键入从字符串到值的映射（表示映射的类型是同类的）。经典的解决方案是使用De Bruijn indices（强类型数字）而不是变量名称来转换为变量的表示。它可以帮助您首先在无类型世界中执行转换，然后只关心输入无类型 - ＆gt; GADT传球。

这里是粗略描绘的强类型变量的GADT声明：

type (_, _) var =
  | Z : ('a, 'a * 'g) var
  | S : ('a, 'g) var -> ('a, 'b * 'g) var

类型('a, 'g) var的值应理解为从'a类型的环境中提取类型'g的值的方法的说明。环境由一系列右嵌套元组表示。 Z情况对应于选择环境中的第一个变量，而S情况忽略了最顶层的变量并且在环境中看得更深。

Shayan Najd（Haskell）实现了这个想法on github。请随意查看the GADT representation或the type-checking/translating code。

答案 1 :(得分：3)

好吧，所以我终于解决了问题。由于我可能不是唯一一个发现这个有趣的人，这里有一套完整的代码，可以进行类型检查和评估：

type (_,_) texp =
| TAdd : ('gamma,int) texp * ('gamma,int) texp -> ('gamma,int) texp
| TAnd : ('gamma,bool) texp * ('gamma,bool) texp -> ('gamma,bool) texp
| TApp : ('gamma,('t1 -> 't2)) texp * ('gamma,'t1) texp -> ('gamma,'t2) texp
| TLam : (('gamma*'t1),'t2) texp -> ('gamma,('t1 -> 't2)) texp
| TVar0 : (('gamma*'t),'t) texp
| TVarS : ('gamma,'t1) texp -> (('gamma*'t2),'t1) texp
| TInt : int -> ('gamma,int) texp
| TBol : bool -> ('gamma,bool) texp

type _ typ =
| Integer : int typ
| Boolean : bool typ
| Arrow : 'a typ * 'b typ -> ('a -> 'b) typ

type (_,_) iseq = IsEqual : ('a,'a) iseq
let rec is_equal : type a b. a typ -> b typ -> (a,b) iseq option = fun a b ->
    match a, b with
    | Integer, Integer -> Some IsEqual
    | Boolean, Boolean -> Some IsEqual
    | Arrow(t1,t2), Arrow(u1,u2) ->
        begin match is_equal t1 u1, is_equal t2 u2 with
        | Some IsEqual, Some IsEqual -> Some IsEqual
        | _ -> None
        end
    | _ -> None

type _ isint = IsInt : int isint
let is_integer : type a. a typ -> a isint option = fun a -> 
    match a with
    | Integer -> Some IsInt
    | _ -> None

type _ isbool = IsBool : bool isbool
let is_boolean : type a. a typ -> a isbool option = fun a -> 
    match a with
    | Boolean -> Some IsBool 
    | _ -> None

type _ context =
| CEmpty : unit context 
| CVar : 'a context * 't typ -> ('a*'t) context 

type exp =
| Add of exp*exp
| And of exp*exp
| App of exp*exp
| Lam : 'a typ * exp -> exp
| Var0
| VarS of exp
| Int of int
| Bol of bool

type _ exists_texp =
| Exists : ('gamma,'t) texp * 't typ -> 'gamma exists_texp

let rec typecheck
    : type gamma t. gamma context -> exp -> gamma exists_texp =
fun ctx e ->
    match e with
    | Int i -> Exists ((TInt i) , Integer)
    | Bol b -> Exists ((TBol b) , Boolean)
    | Var0 ->
        begin match ctx with
        | CEmpty -> failwith "Tried to grab a nonexistent variable"
        | CVar(ctx,t) -> Exists (TVar0 , t)
        end
    | VarS e ->
        begin match ctx with
        | CEmpty -> failwith "Tried to grab a nonexistent variable"
        | CVar(ctx,_) ->
            let tet = typecheck ctx e in
            begin match tet with
            | Exists (te,t) -> Exists ((TVarS te) , t)
            end
        end
    | Lam(t1,e) ->
        let tet2 = typecheck (CVar (ctx,t1)) e in
        begin match tet2 with
        | Exists (te,t2) -> Exists ((TLam te) , (Arrow(t1,t2)))
        end
    | App(e1,e2) ->
        let te1t1 = typecheck ctx e1 in
        let te2t2 = typecheck ctx e2 in
        begin match te1t1,te2t2 with
        | Exists (te1,t1),Exists (te2,t2) ->
            begin match t1 with
            | Arrow(t11,t12) ->
                let p = is_equal t11 t2 in
                begin match p with
                | Some IsEqual -> 
                    Exists ((TApp (te1,te2)) , t12)
                | None -> 
                    failwith "Mismatch of types on a function application"
                end
            | _ -> failwith "Tried to apply a non-arrow type" 
            end
        end
    | Add(e1,e2) ->
        let te1t1 = typecheck ctx e1 in
        let te2t2 = typecheck ctx e2 in
        begin match te1t1,te2t2 with
        | Exists (te1,t1),Exists (te2,t2) ->
            let p = is_equal t1 t2 in
            let q = is_integer t1 in
            begin match p,q with
            | Some IsEqual, Some IsInt ->
                Exists ((TAdd (te1,te2)) , t1)
            | _ ->
                failwith "Tried to add with something other than Integers"
            end
        end
    | And(e1,e2) ->
        let te1t1 = typecheck ctx e1 in
        let te2t2 = typecheck ctx e2 in
        begin match te1t1,te2t2 with
        | Exists (te1,t1),Exists (te2,t2) ->
            let p = is_equal t1 t2 in
            let q = is_boolean t1 in
            begin match p,q with
            | Some IsEqual, Some IsBool ->
                Exists ((TAnd (te1,te2)) , t1)
            | _ ->
                failwith "Tried to and with something other than Booleans"
            end
        end

let e1 = Add(Int 1,Add(Int 2,Int 3))
let e2 = Add(Int 1,Add(Int 2,Bol false))
let e3 = App(Lam(Integer,Add(Var0,Var0)),Int 4)
let e4 = App(App(Lam(Integer,Lam(Integer,Add(Var0,VarS(Var0)))),Int 4),Int 5)
let e5 = Lam(Integer,Lam(Integer,VarS(Var0)))
let e6 = App(Lam(Integer,Var0),Int 1)
let e7 = App(Lam(Integer,Lam(Integer,Var0)),Int 1)
let e8 = Lam(Integer,Var0)
let e9 = Lam(Integer,Lam(Integer,Var0))

let tet1 = typecheck CEmpty e1
(*let tet2 = typecheck CEmpty e2*)
let tet3 = typecheck CEmpty e3
let tet4 = typecheck CEmpty e4
let tet5 = typecheck CEmpty e5
let tet6 = typecheck CEmpty e6
let tet7 = typecheck CEmpty e7
let tet8 = typecheck CEmpty e8
let tet9 = typecheck CEmpty e9

let rec eval : type gamma t. gamma -> (gamma,t) texp -> t = fun env e -> 
    match e with
    | TAdd (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 + v2
    | TAnd (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 && v2
    | TApp (e1,e2) ->
        let v1 = eval env e1 in
        let v2 = eval env e2 in
        v1 v2
    | TLam e ->
        fun x -> eval (env,x) e 
    | TVar0 ->
        let (env,x)=env in
        x
    | TVarS e ->
        let (env,x)=env in
        eval env e 
    | TInt i -> i
    | TBol b -> b

type exists_v =
| ExistsV : 't -> exists_v

let typecheck_eval e =
    let tet = typecheck CEmpty e in
    match tet with
    | Exists (te,t) -> ExistsV (eval () te)

let v1 = typecheck_eval e1
let v3 = typecheck_eval e3
let v4 = typecheck_eval e4
let v5 = typecheck_eval e5
let v6 = typecheck_eval e6
let v7 = typecheck_eval e7
let v8 = typecheck_eval e8
let v9 = typecheck_eval e9

以下是我遇到麻烦的部分以及我如何设法解决这些问题

  1. 为了正确键入类型化表达式texp，需要将内置类型构建到texp类型中。正如加油管正确指出的那样，这意味着我们需要某种De Bruijin符号。最简单的只是Var0和VarS。为了使用变量名，我们只需要预处理AST。
  2. 表达式的类型，typ，需要包括要匹配的变体类型以及我们在类型化表达式中使用的类型。换句话说，这也需要是一个GADT。
  3. 我们需要三个样张才能找出类型检查器中的正确类型。这些是is_equal，is_integer和is_bool。 is_equal的代码实际上位于Advanced examples下的OCaml手册中。具体来说，请查看eq_type的定义。
  4. 类型exp，对于无类型的AST，实际上也需要是GADT。 lambda抽象需要访问typ，这是一个GADT。
  5. 类型检查器返回类型化表达式和类型的存在类型。我们需要两个让程序检查类型。此外，我们需要存在主义因为无类型表达式可能有也可能没有类型。
  6. 存在类型exists_texp公开环境/上下文的类型，但不公开类型。我们需要暴露这种类型才能正确地进行检查。
  7. 一旦设置好所有内容，评估者就会完全遵循类型规则。
  8. 将类型检查器与求值程序组合的结果必须是另一种存在类型。先验地，我们不知道结果类型，因此我们必须将其隐藏在存在的包中。
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