How to call the constructor of a module - ocaml

Assuming I have the following module signature and implementation in the same file named monoids.ml:
module type MonoidADT = sig
type 'a monoid
(* . . . *)
end
module Monoid : MonoidADT = struct
type 'a monoid = Monoid of ('a list) * ('a -> 'a -> 'a) * ('a)
(* . . . *)
end
How am I supposed to use a constructor, of the type defined in the implementation, in an other file (module)? Is it directly accesible or should I create something like a factory function with 'a monoid as return type?
What I tried to do is simply open the module and call its constructor as I would do inside the module, but of course it doesn't work and gives me Unbound constructor Monoid error:
open Monoids.Monoid;;
let boolean_monoid = Monoid ([true; false], ( || ), false);;

Signature constraints remove information. Moreover, if you remove too much information you can perfectly end up with unusable module.
Typically, when you write
module Monoid : MonoidADT = struct
...
end
you are asking the compiler to restrict your monoid to the signature shared by every monoids. And if your monoid module type was the standard one:
module type MonoidADT = sig
type t
val e: t
val ( * ): t -> t -> t
end
this would mean that you are restricting yourself to writing code that is compatible with the one element monoid:
module One = struct
type t = unit
let e = ()
let ( * ) () () = ()
end
This is unlikely to be what you want.
If you only wish to check that your module is a subtype of some module type you can write:
module _ : MonoidADT = Monoid

Apparently defining the type in both signature and implementation resolve the error:
module type MonoidADT = sig
type 'a monoid = Monoid of ('a list) * ('a -> 'a -> 'a) * ('a)
(* . . . *)
end
module Monoid : MonoidADT = struct
type 'a monoid = Monoid of ('a list) * ('a -> 'a -> 'a) * ('a)
(* . . . *)
end
But what if I'd like to give a different implementation? The non-abstract signature limits me in the definition of the concrete module.

Related

Parametrically polymorphic modules

Here is a simple OCaml module type for a monad:
module type Monad = sig
type 'a t
val return : 'a -> 'a t
val bind : 'a t -> ('a -> 'b t) -> 'b t
end
I can instantiate this with any particular monad, such as the reader monad for some type r:
module Reader_monad : Monad = struct
type 'a t = r -> 'a
let return a = fun _ -> a
let bind o f = fun x -> f (o x) x
end
And I can parametrize it over the type r by using a functor:
module type Readable = sig type r end
module Reader (R : Readable) : Monad = struct
type 'a t = R.r -> 'a
let return a = fun _ -> a
let bind o f = fun x -> f (o x) x
end
However, the latter approach requires that I instantiate different instances of the functor for different types r. Is there any way to define a "parametrically polymorphic" module of type Monad that would give parametrically polymorphic functions like return : 'a -> ('r -> 'a)?
I think can get more or less what I want with a separate module type for "families of monads":
module type Monad_family = sig
type ('c, 'a) t
val return : 'a -> ('c, 'a) t
val bind : ('c, 'a) t -> ('a -> ('c, 'b) t) -> ('c, 'b) t
end
module Reader_family : Monad_family = struct
type ('c, 'a) t = 'c -> 'a
let return a = fun _ -> a
let bind o f = fun x -> f (o x) x
end
But if I have a substantial library of general facts about monads, this would require modifying it everywhere manually to use families. And then some monads are parametrized by a pair of types (although I suppose that could be encoded by a product type), etc. So I would rather avoid having to do it this way.
If this isn't directly possible, is there at least a way to instantiate the module Reader locally inside a parametrically polymorphic function? I thought I might be able to do this with first-class modules, but my naive attempt
let module M = Reader(val (module struct type r = int end) : Readable) in M.return "hello";;
produces the error message
Error: This expression has type string M.t
but an expression was expected of type 'a
The type constructor M.t would escape its scope
which I don't understand. Isn't the type M.t equal to int -> string?
I think this is the same issue as The type constructor "..." would escape its scope when using first class modules, where the module M doesn't live long enough. If you instead wrote
# module M = Reader(struct type r = int end);;
# M.return "hello";;
- : string M.t = <fun>
then this would work fine.
Separately, the Reader functor loses some type equalities that you might want. You can restore them by defining it as such:
module Reader (R : Readable) : Monad with type 'a t = R.r -> 'a = struct
type 'a t = R.r -> 'a
let return a = fun _ -> a
let bind o f = fun x -> f (o x) x
end

How to achieve the type-class of sml?

I want to write a comparable set as below.
signature COMPARABLE_SET=
sig
type 'a set
val empty: 'a set
val insert: 'a * 'a set -> 'a set
val member: 'a * 'a set -> bool
end
I need to limit the element in 'a set type to be comparable:(there is a function with type:'a * 'a -> order).
How to achieve it?
If you want to do it in OCaml, this is simply a functor case :
First, you need to define the type of your elements :
module type OrderedType = sig
type t
val compare : t -> t -> int
end
And then you'll define a functor on this type :
module MakeComparableSet (Ord : OrderedType) :
sig
type elt = Ord.t
type t
val empty : t
val insert : elt -> t -> t
val member : elt -> t -> bool
end = struct
type elt = Ord.t
type t
let empty = failwith "TODO"
let insert = failwith "TODO"
let member = failwith "TODO"
end
Which is exactly what is made here.
You can see a functor as a function on module that will create new modules. Here, the functor ComparableSet takes a module of signature OrderedType and returns a module that is a set.

Dynamic binding using a ref cell

I understand that you can't do this, but want to understand precisely why.
module M : sig
type 'a t
val call : 'a t -> 'a option
end = struct
type 'a t
let state : ('a t -> 'a option) ref = ref (fun _ -> None)
let call : ('a t -> 'a option) = fun x -> !state x
end
Results in:
Error: Signature mismatch:
Modules do not match:
sig
type 'a t
val state : ('_a t -> '_a option) ref
val call : '_a t -> '_a option
end
is not included in
sig
type 'a t
val call : 'a t -> 'a option
end
Values do not match:
val call : '_a t -> '_a option
is not included in
val call : 'a t -> 'a option
Why are the abstract types not compatible here?
My gut tells me it has everything to do with early vs late binding, but I'm looking for an exact description of what the type system is doing here.
One way to look at it is that your field state can't have the polymorphic value you ascribe to it, because mutable values can't be polymorphic. References are at most monomorphic (as indicated by the '_a notation for the type variable).
If you just try to declare a similar reference in the toplevel, you'll see the same effect:
# let lfr: ('a list -> 'a option) ref = ref (fun x -> None);;
val lfr : ('_a list -> '_a option) ref = {contents = <fun>}
The type variable '_a indicates some single type that hasn't yet been determined.
The reason that references can't be polymorphic is that it's unsound. If you allow references to be generalized (polymorphic) it's easy to produce programs that go horribly wrong. (In practice this usually means a crash and core dump.)
The issue of soundness is discussed near the beginning of this paper: Jacques Garrigue, Relaxing the Value Restriction (which I refer to periodically when I forget how things work).
Update
What I think you want is "rank 2 polymorphism". I.e., you want a field whose type is polymorphic. You can actually get this in OCaml as long as you declare the type. The usual method is to use a record type:
# type lfrec = { mutable f: 'a. 'a list -> 'a option };;
type lfrec = { mutable f : 'a. 'a list -> 'a option; }
# let x = { f = fun x -> None };;
val x : lfrec = {f = <fun>}
# x.f ;;
- : 'a list -> 'a option = <fun>
The following code compiles for me using lfrec instead of a reference:
module M : sig
type 'a t
val call : 'a t -> 'a option
end = struct
type 'a t
type lfrec = { mutable f: 'a. 'a t -> 'a option }
let state: lfrec = { f = fun _ -> None }
let call : ('a t -> 'a option) = fun x -> state.f x
end

OCaml variance (+'a, -'a) and invariance

After writing this piece of code
module type TS = sig
type +'a t
end
module T : TS = struct
type 'a t = {info : 'a list}
end
I realised I needed info to be mutable.
I wrote, then :
module type TS = sig
type +'a t
end
module T : TS = struct
type 'a t = {mutable info : 'a list}
end
But, surprise,
Type declarations do not match:
type 'a t = { mutable info : 'a list; }
is not included in
type +'a t
Their variances do not agree.
Oh, I remember hearing about variance. It was something about covariance and contravariance. I'm a brave person, I'll find about my problem alone!
I found these two interesting articles (here and here) and I understood!
I can write
module type TS = sig
type (-'a, +'b) t
end
module T : TS = struct
type ('a, 'b) t = 'a -> 'b
end
But then I wondered. How come that mutable datatypes are invariant and not just covariant?
I mean, I understand that an 'A list can be considered as a subtype of an ('A | 'B) list because my list can't change. Same thing for a function, if I have a function of type 'A | 'B -> 'C it can be considered as a subtype of a function of type 'A -> 'C | 'D because if my function can handle 'A and 'B's it can handle only 'A's and if I only return 'C's I can for sure expect 'C or 'D's (but I'll only get 'C's).
But for an array? If I have an 'A array I can't consider it as a an ('A | 'B) array because if I modify an element in the array putting a 'B then my array type is wrong because it truly is an ('A | 'B) array and not an 'A array anymore. But what about a ('A | 'B) array as an 'A array. Yes, it would be strange because my array can contain 'B but strangely I thought it was the same as a function. Maybe, in the end, I didn't understand everything but I wanted to put my thoughts on it here because it took me long to understand it.
TL;DR :
persistent : +'a
functions : -'a
mutable : invariant ('a) ? Why can't I force it to be -'a ?
I think that the easiest explanation is that a mutable value has two intrinsic operations: getter and setter, that are expressed using field access and field set syntaxes:
type 'a t = {mutable data : 'a}
let x = {data = 42}
(* getter *)
x.data
(* setter *)
x.data <- 56
Getter has a type 'a t -> 'a, where 'a type variable occurs on the right-hand side (so it imposes a covariance constraint), and the setter has type 'a t -> 'a -> unit where the type variable occurs to the left of the arrow, that imposes a contravariant constraint. So, we have a type that is both covariant and contravariant, that means that type variable 'a is invariant.
Your type t basically allows two operations: getting and setting. Informally, getting has type 'a t -> 'a list and setting has type 'a t -> 'a list -> unit. Combined, 'a occurs both in a positive and in a negative position.
[EDIT: The following is a (hopefully) clearer version of what I wrote in the first place. I consider it superior, so I deleted the previous version.]
I will try to make it more explicit. Suppose sub is a proper subtype of super and witness is some value of type super which is not a value of type sub. Now let f : sub -> unit be some function which fails on the value witness. Type safety is there to ensure that witness is never passed to f. I will show by example that type safety fails if one is allowed to either treat sub t as a subtype of super t, or the other way around.
let v_super = ({ info = [witness]; } : super t) in
let v_sub = ( v_super : sub t ) in (* Suppose this was allowed. *)
List.map f v_sub.info (* Equivalent to f witness. Woops. *)
So treating super t as a subtype of sub t cannot be allowed. This shows covariance, which you already knew. Now for contravariance.
let v_sub = ({ info = []; } : sub t) in
let v_super = ( v_sub : super t ) in (* Suppose this was allowed. *)
v_super.info <- [witness];
(* As v_sub and v_super are the same thing,
we have v_sub.info=[witness] once more. *)
List.map f v_sub.info (* Woops again. *)
So, treating sub t as a subtype of super t cannot be allowed either, showing contravariance. Together, 'a t is invariant.

Why can't I add type constraints when implementing a module type?

I was trying (just out of interest) to do this:
module type CAT = sig
type ('a, 'b) t
val id : ('a, 'a) t
val (#) : ('b, 'c) t -> ('a, 'b) t -> ('a, 'c) t
end
module Lst = struct
type ('a, 'b) t = 'a list constraint 'a = 'b
let id = []
let (#) = (#)
end
module L : CAT = Lst (* (error) *)
But I get:
Type declarations do not match:
type ('b, 'a) t = 'b list constraint 'a = 'b
is not included in
type ('a, 'b) t
Why isn't this safe? Everything that can see the concrete type can also see the constraint, so I don't think you could make something with a wrong type (e.g. call # with a (string, int) t argument).
Update: to those saying that my module doesn't implement the signature because it requires the types to be the same, consider that the following (which just wraps the lists in a List variant) is accepted despite having the same behaviour:
module Lst = struct
type ('a, 'b) t =
List : 'a list -> ('a, 'a) t
let id = List []
let (#) (type a) (type b) (type c) (a:(b, c) t) (b:(a, b) t) : (a, c) t =
match a, b with
| List a, List b -> List (a # b)
end
The example can be reduced to the type definition alone:
module type S =
sig
type ('a, 'b) t
end
module M =
struct
type ('a, 'b) t = 'a list constraint 'a = 'b
end
As Jeffrey already pointed out, M is not of type S, because it allows fewer applications of t: according to signature S, the type (int, string) t would be perfectly legal (it is well-formed), but M does not allow this type ((int, string) M.t is not a legal type, because it violates the explicit constraint).
All that is completely independent from the question whether the type is actually inhabited, i.e., whether you can construct values of the type. In your second example, the module makes the respective type well-formed, though it is uninhabited. Uninhabited types are legal, however, and sometimes even useful (see e.g. the concept of phantom types).
The type signature CAT is more general than the type of the Lst module. You need to put the type constraint on the abstract type too, i.e. type ('a, 'b) t constraint 'a = 'b.
This gives us the following:
module type CAT = sig
type ('a, 'b) t constraint 'a = 'b
val id : ('a, 'a) t
val (#) : ('b, 'c) t -> ('a, 'b) t -> ('a, 'c) t
end
which is printed as follows by the toplevel, showing a single type variable in the signature of (#):
module type CAT =
sig
type ('b, 'a) t constraint 'a = 'b
val id : ('a, 'a) t
val ( # ) : ('c, 'c) t -> ('c, 'c) t -> ('c, 'c) t
end
Error messages of the form "type x is not included in type y" refer to types or module types as specifications of sets of possible values, hence the use of the term "included".
In the case of a module implementation (Lst), we have a module type for it. Applying a signature (module type CAT) to a module is only allowed if that signature is as specialized (equal set) or more specialized (strict subset) than the original signature of the module.
One can write module X : sig val f : unit -> unit end = struct let f x = x end
but not module X : sig val f : 'a -> 'a end = struct let f () = () end. The latter gives the following error:
Error: Signature mismatch:
Modules do not match:
sig val f : unit -> unit end
is not included in
sig val f : 'a -> 'a end
Values do not match:
val f : unit -> unit
is not included in
val f : 'a -> 'a
This is different than placing type constraints on certain expressions, in which case the constraint is a mask to be applied (a set to intersect with) rather than a subset. For example it is fine to write let f : unit -> 'a = fun x -> x even though f's signature ends up being unit -> unit, a strict subset - or subtype - of unit -> 'a.
Your Lst module doesn't seem to me to have the type CAT. CAT allows the two types 'a and 'b to be independent. The Lst module requires them to be the same. If the L module were of type CAT then it should allow me to make something of type (string, int) t but it doesn't.
The error message is a little confusing, at least to me.