Haven't been able to find much online documentation regarding begin/end in ocaml. I have two different pattern matches in the same function (which I want to be independent of each other), but vscode is parsing them to nest the second inside the first. I've tried surrounding the first pattern match in begin/end, but it's giving me syntax errors:
begin match c.r with (* first pattern match *)
| [ r1; r2; r3 ] ->
let _ = print_endline (String.make 1 r3.top) in end
match cl with (* second pattern match *)
| [] -> []
I get a red underline on end that says Syntax error after unclosed begin, expecting expr. I do not understand what this means, since I wrote end to close the begin, so why is the begin unclosed? The code compiles fine without the begin/end (except that it nests the second pattern match inside the first one). Thanks.
In OCaml begin/end is essentially identical to open/close parentheses. Inside you should have a well-formed expression (as in pretty much any programming language).
What you have inside your begin/end is not an expression, since it ends wih in. A let expression looks like let pattern = expr1 in expr2. You are missing the second expression inside the begin/end.
What you should do (I think) is put begin/end around the inner match like this:
match c.r with
| [r1; r2; r3 ] ->
let _ = print_endline (String.make 1 r3.top) in
begin
match c1 with
| [] -> []
...
end
| ...
(This code doesn't make a lot of sense but I assume it's just an example.)
As another simplification you can change let _ = a in b to a; b if a is of unit type, as it is in your code.
What Jeffrey Scofield said. Consider how this can become confusing when we nest matches. How do we read the following?
match "foo" with
| "bar" -> 42
| "foo" ->
match "baz" with
| "baz" -> 27
| _ -> 19
| _ -> 33
The indentation makes it fairly clear how this is meant, but OCaml doesn't care about your pretty indentation. It could just as easily be that you meant:
match "foo" with
| "bar" -> 42
| "foo" ->
match "baz" with
| "baz" -> 27
| _ -> 19
| _ -> 33
Or:
match "foo" with
| "bar" -> 42
| "foo" ->
match "baz" with
| "baz" -> 27
| _ -> 19
| _ -> 33
Either parentheses or begin/end disambiguate this situation.
match "foo" with
| "bar" -> 42
| "foo" ->
(match "baz" with
| "baz" -> 27
| _ -> 19)
| _ -> 33
Or:
match "foo" with
| "bar" -> 42
| "foo" ->
begin
match "baz" with
| "baz" -> 27
| _ -> 19
end
| _ -> 33
One more thing...
Nested match expressions are often unnecessary. Consider when you have a nested match if you can't more cleanly express the same as a single level match on a tuple of values.
match a with
| X -> ...
| Y ->
(match b with
| Z -> ...
| W -> ...
| _ -> ...)
| U ->
(match c with
| F -> ...
| G -> ...
| _ -> ...)
| _ -> ...
vs.
match a, b, c with
| X, _, _ -> ...
| Y, Z, _ -> ...
| Y, W, _ -> ...
| Y, _, _ -> ...
| U, _, F -> ...
| U, _, G -> ...
| U, _, _ -> ...
| _, _, _ -> ...
I write an interpreter with Ocaml but when i try :
sem(Let("Somma",Fun("x",Sum(Den "x",Eint(5))),Let("pipa",Pipe(Seq(Den "Somma",Nil)),Apply(Den "pipa",Eint(42)))),(emptyenv Unbound));;
the resault is an error : "Exception: Match_failure ("",1,41)
I think that error is in applyPipe but I don't understand where and why there is
Where did i wrong?
this is my code :
type exp =
.
.
.
| Fun of ide * exp
| Apply of exp * exp
| Letrec of ide * ide * exp * exp
| Etup of tuple (*Tupla come espressione*)
| Pipe of tuple (*Concatenazione di funzioni*)
| ManyTimes of int * exp (*Esecuzione iterata di una funzione*)
and tuple =
| Nil (*Tupla vuota*)
| Seq of exp * tuple (*Tupla di espressioni*)
;;
type eval=
| Int of int
| Bool of bool
| Unbound
| RecFunVal of ide * ide * exp * eval env
| Funval of efun
| ValTup of etuple
and efun = ide * exp * eval env
and etuple =
| Nil
| Seq of eval * etuple
;;
let rec sem ((ex: exp), (r: eval env)) = match ex with
.
.
.
| Let(i, e1, e2) -> sem(e2, bind (r, i, sem(e1, r)))
| Fun(i,a) -> Funval(i,a,r)
| Letrec(f, i, fBody, letBody) ->
let benv = bind(r, f, (RecFunVal(f, i, fBody, r)))
in sem(letBody, benv)
| Etup(tup) -> (match tup with
| Seq(ex1, tupla) ->
let evex1 = sem(ex1, r) in
let ValTup(etupl) = sem(Etup(tupla), r) in
ValTup(Seq(evex1, etupl))
| Nil -> ValTup(Nil))
| Apply(Den f, arg1) ->
(let fclosure= sem(Den f, r) in
match fclosure with
| Funval(arg, fbody, fDecEnv) ->
sem(fbody, bind(fDecEnv, arg, sem(arg1, r)))
| RecFunVal(f, arg, fbody, fDecEnv) ->
let aVal= sem(arg1, r) in
let rEnv= bind(fDecEnv, f, fclosure) in
let aEnv= bind(rEnv, arg, aVal) in
sem(fbody, aEnv)
| _ -> failwith("non functional value"))
| Apply(Pipe tup, arg) -> applyPipe tup arg r
| Apply(_,_) -> failwith("not function")
| _ -> failwith("non implementato")
and applyPipe tup argo r = (match tup with
| Seq(Den f, tupla) ->
let appf = Apply(Den f,argo) in
applyPipe tupla appf r
| Nil -> sem(argo,r)
| _ -> failwith("Not a valid Pipe"))
;;
The complete code is there : http://pastebin.com/VgpanX51
Please help me thaks
When you compile (or evaluate in a toplevel) an OCaml program, a typechecker will emit warnings about all pattern matches that are irrefutable, i.e., such patterns that may raise a Match_failure exception.
What you should do, is to go through all warnings and fix them.
There are quite a few irrefutable matches in your code, e.g., the sem function final match Apply(_,_) -> failwith("not function") will only catch Apply terms, but will not catch all others, adding something like _ -> failwith "unimplemented" will fix it.
QA
the error is in the try-code or in my interpreter?
It is in the interpreter, you didn't include all possible cases in your pattern-matching code.
.I do extend the typechecker
You don't need to. The typechecker verifies whether you anticipated all possible cases, for example, let's take the simple example:
type abc = A | B | C
let string_of_abc abc = match abc with
| A -> "A"
| B -> "B"
When you will try to compile (or interpret) the above code the typechecker will tell you:
Warning 8: this pattern-matching is not exhaustive.
Here is an example of a value that is not matched:
C
type abc = A | B | C
So, it gives you a hint, that you forgot to match with the C constructor, so the expression string_of_abc C will terminate with a Match_failure exception.
You can follow the hints and add the cases one-by-one. Given your example, the pattern matching in the sema function is incomplete, and the type checker hits you with the following:
Warning 8: this pattern-matching is not exhaustive.
Here is an example of a value that is not matched:
(Pipe _|ManyTimes (_, _))
And indeed, you missed the case with Pipe, so when the interpreter sees
Pipe(...)
It can't find a match, as according to your code you're expecting the Pipe constructor only as a first argument to Apply, when in your example, you're actually passing it as a second argument to Let.
How do you define a simple lambda calculus-like DSL in OCaml using GADTs? Specifically, I can't figure out how to properly define the type checker to translate from an untyped AST to a typed AST nor can I figure out the correct type for the context and environment.
Here's some code for a simple lambda calculus-like language using the traditional approach in OCaml
(* 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
Now, I'm trying to translate this into something that uses GADTs. Here's my start
(* 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"
Here's the problem. First, I'm not sure how to define the correct type for TLam and TVar in the type texp. Generally, I would provide the type with the variable name, but I'm not sure how to do that in this context. Second, I don't know the correct type for the context in the function typecheck. Before, I used some kind of list, but now I'm sure sure of the type of the list. Third, after leaving out the context, the typecheck function doesn't type check itself. It fails with the message
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
which makes complete sense. This is more of an issue of that I'm not sure what the correct type for typecheck should be.
In any case, how do you go about fixing these functions?
Edit 1
Here's a possible type for the context or environment
type _ ctx =
| Empty : unit ctx
| Item : string * 'a * 'b ctx -> ('a*'b) ctx
Edit 2
The trick with the environment is to make sure that the type of the environment is embedded into the type of the expression. Otherwise, there's not enough information in order to make things type safe. Here's a completed interpreter. At the moment, I do not have a valid type checker to move from untyped expressions to typed expressions.
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
Then, we have
# eval () te1;;
- : int = 6
# eval () te3;;
- : int = 8
# eval () te5;;
- : '_a -> '_b -> '_a = <fun>
# eval () te4;;
- : int = 9
If you want the term representation to enforce well-typedness, you need to change the way type environments (and variables) are represented: you cannot finely type a mapping from strings to value (type to represent mapping are homogeneous). The classic solution is to move to a representation of variables using De Bruijn indices (strongly-typed numbers) instead of variable names. It may help you to perform that conversion in the untyped world first, and then only care about typing in the untyped -> GADT pass.
Here is, rouhgly sketched, a GADT declaration for strongly typed variables:
type (_, _) var =
| Z : ('a, 'a * 'g) var
| S : ('a, 'g) var -> ('a, 'b * 'g) var
A value at type ('a, 'g) var should be understood as a description of a way to extract a value of type 'a out of an environment of type 'g. The environment is represented by a cascade of right-nested tuples. The Z case corresponds to picking the first variable in the environment, while the S case ignores the topmost variables and looks deeper in the environment.
Shayan Najd has a (Haskell) implementation of this idea on github. Feel free to have a look at the GADT representation or the type-checking/translating code.
Alright, so I finally worked things out. Since I may not be the only one who finds this interesting, here's a complete set of code that does both type checking and evaluation:
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
Here are the pieces I had trouble with and how I managed to resolve them
In order to correctly type the typed expressions texp, the type of the environment needed to be built into the type of texp. This implies, as gasche correctly noted, that we needed some sort of De Bruijin notation. The easiest was just Var0 and VarS. In order to use variable names, we'd just have to preprocess the AST.
The type of the expression, typ, needed to include both variant types to match on as well as the type we use in the typed expression. In other words, that also needed to be a GADT.
We require three proofs in order to ferret out the correct types in the type checker. These are is_equal, is_integer, and is_bool. The code for is_equal is actually in the OCaml manual under Advanced examples. Specifically, look at the definition of eq_type.
The type exp, for the untyped AST, actually needs to be a GADT also. The lambda abstraction needs access to typ, which is a GADT.
The type checker returns an existential type of both a typed expression as well as the type. We need both to get the program to check type. Also, we need the existential because the untyped expression may or may not have a type.
The existential type, exists_texp, exposes the type of the environment/context, but not the type. We need this type exposed in order to type check properly.
Once everything is setup, the evaluator follows the type rules exactly.
The result of combining the type checker with the evaluator must be another existential type. A priori, we don't know the resulting type, so we have to hide it in an existential package.