Work with class in OCaml - ocaml

I want create function with my type test_type in class t. My code:
type test_type = [`t1|`t2]
let get_types =
function
| `t1 -> "t1"
| `t2 -> "t2";;
class type class_types =
object
method t_types : test_type
method test : (string -> string -> test_type -> unit) -> unit
end;;
class t : class_types =
object
method test par1 ?(par2="no par2") ?(par3=`t1) () =
print_endline("--->"^par1);
print_endline("--->"^par2);
print_endline("--->"^get_types par3)
end;;
let t_run = new t;;
t_run # test "parametr1" ~par3:`t2 ();;
is return error
The class type is not matched by the class type class_types
The first class type has no method t_types
How to do this?

Simply your implementation of t lacks the method t_types, which is defined in the class type class_types.
In addition to this, method test's type is string -> ?par2: string -> ?par3: test_type -> unit -> unit, which is incompatible with one at the class type.

Related

How to apply [##deriving show] to a type from module parameter of my functor?

I have a functor that takes a Set type like:
module type MySet = functor (S : Set.S) -> sig
val my_method : S.t -> S.elt -> S.elt list option
end
module MySet_Make : MySet = functor (S : Set.S) -> struct
let my_method set el = Some [el] (* whatever *)
end
module IntSet = Set.Make(Int)
module MyIntSet = MySet_Make(IntSet)
S.elt is the type of elements of the set
I want to apply [##deriving show] (from https://github.com/ocaml-ppx/ppx_deriving#plugin-show) to S.elt within my functor somehow, so that in one of my methods I can rely on having a show : S.elt -> string function available.
I feel like it must be possible but I can't work out the right syntax.
Alternatively - if there's a way to specify in the signature that the Set type S was made having elements of a "showable" type.
e.g. I can define:
module type Showable = sig
type t [##deriving show]
end
...but I can't work out how to specify that as a type constraint to elements of (S : Set.S)
You can construct new signatures that specify the exact function show you need:
module MySet_Make(S : sig
include Set.S
val show : elt -> string
end) = struct
let my_method _set el =
print_endline (S.show el);
Some [el]
end
Then you can build the actual module instance by constructing the module with the needed function:
module IntSet = struct
include Set.Make(Int)
(* For other types, this function could be created by just using [##deriving show] *)
let show = string_of_int
end
module MyIntSet = MySet_Make(IntSet)
Ok, after a couple of hours more fumbling around in the dark I found a recipe that does everything I wanted...
First we define a "showable" type, representing a module type that has had [##deriving show] (from https://github.com/ocaml-ppx/ppx_deriving#plugin-show) applied to it:
module type Showable = sig
type t
val pp : Format.formatter -> t -> unit
val show : t -> string
end
(I don't know if there's some way to get this directly from ppx_deriving.show without defining it manually?)
Then we re-define and extend the Set and Set.OrderedType (i.e. element) types to require that the elements are "showable":
module type OrderedShowable = sig
include Set.OrderedType
include Showable with type t := t
end
module ShowableSet = struct
include Set
module type S = sig
include Set.S
end
module Make (Ord : OrderedShowable) = struct
include Set.Make(Ord)
end
end
I think with the original code in my question I had got confused and used some kind of higher-order functor syntax (?) ...I don't know how it seemed to work at all, but at some point I realised my MySet_Make was returning a functor rather than a module. So we'll fix that now and just use a normal functor.
The other thing we can fix is to make MySet a further extension of ShowableSet ... so MySet_Make will take the element type as a parameter instead of another Set type. This makes the eventual code all simpler too:
module type MySet = sig
include ShowableSet.S
val my_method : t -> elt -> elt list option
val show_el : elt -> string
end
module AdjacencySet_Make (El : OrderedShowable) : AdjacencySet
with type elt = El.t
= struct
include ShowableSet.Make(El)
let my_method set el = Some [el] (* whatever *)
let show_el el = El.show el (* we can use the "showable" elements! *)
end
Then we just need an OrderedShowable version of Int as the element type. Int is already ordered so we just have to extend it by deriving "show" and then we can make a concrete MySet:
module Int' = struct
include Int
type t = int [##deriving show]
end
module MyIntSet = MySet_Make(Int')
And we can use it like:
# let myset = MyIntSet.of_list [3; 2; 8];;
# print_endline (MyIntSet.show_el 3);;
"3"

Use abstract module as part of type definition separate from module

I'm trying to use the module type Partial_information which is constructed via the functor Make_partial_information as the type of the field contents in the type Cell.t. However, I'm getting the error Unbound module Partial_information.
open Core
(* module which is used as argument to functor *)
module type Partial_type = sig
type t
val merge : old:t -> new_:t -> t
end
(* type of result from functor *)
module type Partial_information = sig
type a
type t = a option
val merge : old:t -> new_:t -> t
val is_nothing : t -> bool
end
(* The functor *)
module Make_partial_information(Wrapping : Partial_type):
(Partial_information with type a = Wrapping.t)
= struct
type a = Wrapping.t
type t = a option
let merge ~(old : t) ~(new_ : t) =
match (old, new_) with
| (None, None) -> None
| (None, Some a) -> Some a
| (Some a, None) -> Some a
| (Some a, Some b) -> (Wrapping.merge ~old:a ~new_:b) |> Some
let is_nothing (it: t) : bool = (is_none it)
end
(* Checking to make sure understanding of functor is correct *)
module Int_partial_type = struct
type t = int
let merge ~old ~new_ = new_ [##warning "-27"]
end
module Int_partial_information = Make_partial_information(Int_partial_type)
(* Trying to use _any_ type which might have been created by the functor as a field in the record *)
module Cell = struct
type id = { name : string ; modifier : int }
type t = {
(* Error is here stating `Unbound module Partial_information` *)
contents : Partial_information.t ;
id : id
}
end
Module types are specifications for modules. They do not define types by themselves. They are also not constructed by functors in any way.
Consequently, it is hard to tell what you are trying to do.
As far I can see, you can simply define your cell type with a functor:
module Cell(P : Partial_information) = struct
type id = { name : string ; modifier : int }
type partial
type t = {
contents : P.t;
id : id
}
end
Or it might be even simpler to make the cell type polymorphic:
type 'a cell = {
contents : 'a;
id : id
}
since the type in itself is not particularly interesting nor really dependent upon
the type of contents.
P.S:
It is possible to use first class modules and GADTs to existentially quantify over a specific implementation of a module type. But it is unclear if it is worthwhile to explode your complexity budget here:
type 'a partial_information = (module Partial_information with type a = 'a)
module Cell = struct
type id = { name : string ; modifier : int }
type t = E: {
contents : 'a ;
partial_information_implementation: 'a partial_information;
id : id
} -> t
end

Is it possible to create a polymorphic variant type dynamically using modules?

I'm trying to organzine and reuse my reasonML code. I have model module types that look like the following:
module Diet = {
type schemaType = [`DietSchema];
type idType = [`DietId(UUID.t)];
let schema = `DietSchema;
type idAsType('a) = [> | idType] as 'a;
};
module Ingredient = {
type schemaType = [`IngredientSchema];
type idType = [`IngredientId(UUID.t)];
let schema = `IngredientSchema;
type idAsType('a) = [> | idType] as 'a;
};
module Restriction = {
type schemaType = [`RestrictionSchema];
type idType = [`RestrictionId(UUID.t)];
let schema = `RestrictionSchema;
type idAsType('a) = [> | idType] as 'a;
};
And I would like to generate a types and functions from the idTypes and schemaTypes.
examples are:
type modelIdType = [
| Diet.idType
| Restriction.idType
| Ingredient.idType
];
type schemaType = [
| Diet.schemaType
| Restriction.schemaType
| Ingredient.schemaType
];
let modelIdToIdFunction = (recordIdType): (schemaType, UUID.t) =>
switch (recordIdType) {
| `DietId(uuid) => (Diet.schema, uuid)
| `RestrictionId(uuid) => (Restriction.schema, uuid)
| `IngredientId(uuid) => (Ingredient.schema, uuid)
};
So I'm attempting to construct a module using a functor passing each of the schemas through
module Diet : SchemaType = {
/* ... */
};
module type SchemaType {
type schemaType;
type idType;
let schema: [> schemaType];
type idAsType('a) = [> | idType] as 'a;
};
module ProcessSchema = (
Schema : SchemaType,
PrevFullSchema : FullSchema
) : (FullSchema) => {
type id = [> Schema.idType' | PrevFullSchema.id'('a)] as 'a;
/* type id = [PrevFullSchema.openId(PrevFullSchema.id) | Schema.idType]; */
/* type schema = [PrevFullSchema.schema | Schema.schema]; */
/* type openSchema = [PrevFullSchema.schema | Schema.schema]; */
};
The code above didn't work. I'm having trouble adding module types to the model modules at the top. I also attempted through a SchemaType module type but kept hitting The type idType is not a polymorphic variant type, When I wanted each model to have distinct polymorphic variable types.
So overall, I want to know if it is possible to create a polymorphic variant type that can be created or extended using modules and functors?
If not is it possible to construct polymorphic variant types using a "list of modules"?
Thanks
Someone asked a similar question back in 2002. According to one of the OCaml language developers, it's not possible to dynamically extend polymorphic variant types like that: https://caml-list.inria.narkive.com/VVwLM96e/module-types-and-polymorphic-variants . The relevant bit:
The functor definition is refused because
"The type M.t is not a polymorphic variant type"
Is there a workaround?
Not that I know. Polymorphic variant extension only works for known
closed variant types, otherwise it would not be sound.
The rest of the post has a suggestion which boils down to capturing the new variant types inside different tags, but again that wouldn't work for your use case of dynamically 'adding together' types using a functor.
For the types you can use an extensible variant type. But for the modelIdToIdFunction function given a list of modules I think you can only search through the list, which won't scale.
You should extend the uuid with an ID for each module so you can create a lookup table from module_id to the module from the list for fast access.

Explicit polymorphic type in record

In OCaml, it is possible to define explicit polymorphic type in a record
type foo = { f : 'a. unit -> 'a };;
It seems we can assign only general values to f like
{ f = fun () -> failwith ""; }
or
{ f = fun () -> exit 1; }
How to use this language feature in real world? Is there any good practical example?
This isn't really connected with records. If you declare any function to have type 'a. unit -> 'a (takes nothing and returns whatever the caller wanted) then you can only use it for functions that don't return.
Here's a slightly more useful example: a record containing a function for finding the length of lists (of any type).
# type foo = { f : 'a. 'a list -> int };;
type foo = { f : 'a. 'a list -> int; }
# let foo = { f = List.length };;
val foo : foo = {f = <fun>}
# foo.f [1;2;3];;
- : int = 3
It can be useful if you wanted to pass a function like List.length as an argument to another function, and have it use it on multiple types:
Say we want to pass List.length to test. We can't do it directly:
# let test fn = fn [1;2;3] + fn ["a";"b";"c"];;
Error: This expression has type string but an expression was expected of type
int
But we can use a record:
# let test foo = foo.f [1;2;3] + foo.f ["a";"b";"c"];;
val test : foo -> int = <fun>
# test foo;;
- : int = 6

Ocaml - Forward Declaration (Classes)

I need to have two classes refering to each other. Is there any way in Ocaml to make Forward Declaration of one of them?
(I don't think it's possible as with easier stuff with word and).
Or maybe it is possible, but different way than how i tried?
Ocaml doesn't have anything like forward declarations (i.e. a promise that something will be defined eventually), but it has recursive definitions (i.e. a block of things that are declared and then immediately defined in terms of each other). Recursive definitions are possible between expressions, types, classes, and modules (and more); mutually recursive modules allow mixed sets of objects to be defined recursively.
You can solve your problem using a recursive definition with the keyword and:
class foo(x : bar) = object
method f () = x#h ()
method g () = 0
end
and bar(x : foo) = object
method h () = x#g()
end
Or you could use parameterized classes. Following the previous example you have:
class ['bar] foo (x : 'bar) =
object
method f () = x#h ()
method g () = 0
end
class ['foo] bar (x : 'foo) =
object
method h () = x#g()
end
The inferred interface is:
class ['a] foo : 'a ->
object
constraint 'a = < h : unit -> 'b; .. >
method f : unit -> 'b
method g : unit -> int
end
class ['a] bar : 'a ->
object
constraint 'a = < g : unit -> 'b; .. >
method h : unit -> 'b
end