1) Suppose there is a module
module Int_Sig_1 =
struct
let x = 1
end
;;
2) and an explicit signature of this module
module type INT_SIG =
sig
val x:int
end
;;
3) and i create a first-class module based on the above module and module type
let int_sig_1 = (module Int_Sig_1:INT_SIG)
4) Now i create another module with no explicit signature but same inferred signature as above
module Int_Sig_2 =
struct
let x =2
end
;;
5) As written in Chapter 10 of the book Real World OCAML "The module type doesn't need to be part of the construction of a first-class module if it can be inferred", i try to create a second first class module using the above module but with no explicit module type
let a2 = (module Int_Sig_2);;
I get the following error
Error: The signature for this packaged module couldn't be inferred.
6) then i try to do the same as above in 5, but this time I put the first class module with no module type being create as an element of a list, where the head of the list is a first class module that was created out of an explicit signature in 3 above.
let int_sig= [int_sig_1;(module Int_Sig_2)];;
val int_sig : (module INT_SIG) list = [<module>; <module>] ;;
My question is why 5 above gives me an error and 6 does not fail ?
The problem with (5) is that in general, there are multiple module types that could be infered. In your example, there is at least two valid module types which could be used to pack Int_Sig_2:
module type empty = sig end
module type with_x = sig val x:int end
In other words, both
let a2 = (module Int_Sig_2: empty)
let a2_bis = (module Int_Sig_2:with_x)
are valid. Consequently, the type checker will not try to infer a module type in this situation.
Contrarily, in your example (6), the type of the list is determined by its first element, whose type is (module INT_SIG_2), therefore the type-checker can use this information to infer that the expected type for the second element of the list is (module INT_SIG_2). Nevertheless, reversing the two elements yields a type error. In other words, this is fine:
[(module struct let x = 2 end: with_x); (module struct let x = 1 end)]
however, the reverse yields
[(module struct let x=2 end); (module struct let x = 3 end:with_x)];;
Error: The signature for this packaged module couldn't be inferred.
This is due to the fact that the type-checker has a left-to-right bias and types first the first element of the list first.
Related
Excuse my potential misuse of terminology, I'm still not very comfortable with OCaml.
We have a functor with the following (abridged) signature:
module type FUNCTORA = sig
type input
type output
type key
type inter
val my_function : input list -> (key * output) list Deferred.t
end
Next, we implement it as such. MYAPP has the same types as above.
module MyFunctor (App : MyAPP) : FUNCTORA = struct
type input = App.input
type output = App.output
type key = App.key
type inter App.value
let my_function lst = ...
end
When trying to compile the implementation, we get this error:
Error: Signature mismatch:
...
Values do not match:
val my_function :
App.input list ->
(App.key * App.output) list Async_kernel.Deferred.t
is not included in
val my_function :
input list -> (key * output) list Async.Std.Deferred.t
It doesn't consider input to include App.input etc, even though we set them to be the same type. How can we get this to type check?
If I make the following changes:
MyAPP => FUNCTORA (* Since you say they are the same *)
type inter App.value => type inter = App.inter (* Syntax/name error *)
Deferred.t => option (* To limit dependence on other modules *)
Then your code compiles for me.
Possibly the problem is with Deferred.t. There are two distinct looking types in the error message.
In a module, I have defined a type that represents a graph node, which has a polymorphic data field i.e.
type 'a t = {data: 'a; adj: 'a t list}
How can I go about creating a Set of this data? I have tried the following (as per one of the suggestions here.
let cmp (g1:int Graph.t) (g2: int Graph.t) : int=
if phys_equal g1 g2 then
0
else
Int.compare g1.data g2.data
let make_set () =
let module Ord=struct
type t=int Graph.t
let compare=cmp
end
in (module Set.Make(Ord): Set.S with type elt=Ord.t)
But when I do, I get "The signature constrained by `with' has no component named elt"
I'm not sure exactly what you're trying to do, but if you just want to make sets of nodes whose data type is int, you don't need to use anything fancier than the usual OCaml module operations.
The following code works for me:
module Graph =
struct
type 'a t = {data: 'a; adj: 'a t list}
end
let cmp (g1:int Graph.t) (g2: int Graph.t) : int =
if g1 == g2 then
0
else
compare g1.Graph.data g2.Graph.data
module GSet = Set.Make(struct type t = int Graph.t let compare = cmp end)
Here's a session with the code:
$ ocaml
OCaml version 4.01.0
# #use "g.ml";;
module Graph : sig type 'a t = { data : 'a; adj : 'a t list; } end
val cmp : int Graph.t -> int Graph.t -> int = <fun>
module GSet :
sig
type elt = int Graph.t
. . .
end
# let myset = GSet.add { Graph.data = 14; adj = [] } GSet.empty;;
val myset : GSet.t = <abstr>
# GSet.is_empty myset;;
- : bool = false
I don't see a reason to constrain the module type, as Set.S is already the module type returned by Set.Make. But I am not a sophisticated user of OCaml module types.
That code works fine in the interpreter for me. Perhaps you've opened a different Set module without the elt type defined in S?
If I define the following Set in the interpreter:
# module Set = struct
module type S = sig end
end;;
module Set : sig module type S = sig end end
And then simply redefine make_set as you wrote it, I actually get the same error message. When trying out code with the interpreter, always keep in mind that you may be working with definitions you wrote previously.
As a rule of thumb, try to avoid binding values to names already in use in the libraries you wish to employ (I know it's tempting to shorten your names there, but at least add them a small distinctive prefix - for instance use ISet instead of Set in your case).
you can always run your code as a script, i.e. by launching it from the command line as follow:
$ ocaml my_script.ml
Or by the #use directive at the interpreter prompt. This let you write code snippets before testing them out with a fresh ocaml environment.
Finally, as in #Jeffrey's provided answer, an unpacked module is good enough for most purpose; your code was about instantiating a first class module, which is only interesting if you intend to pass that module around without the use of functors. See the documentation on modules (and related extensions of the language) for further explanation.
module type Arity =
sig
val arity : nat (* in my real code it has another type *)
end
module S =
functor (A : Arity) -> struct
let check = ...
end
I would like to use the function check inside the functor S without implement signature Arity. I read the first-class module but still not understand how to write it (in practice). Here is my draft code:
let A = has type of (module Arity)
then
let M = S (A)
then I can call check function by
M.check
I tried:
let f arity = (module (val arity : Arity) : Arity)
it returns : val f : (module Arity) -> (module Arity)
Could you please help me to write this first-class module? Am I able to write it in Ocaml?
Also in (http://caml.inria.fr/pub/docs/manual-ocaml-4.00/manual021.html#toc81) section 7.14 it says :
"The module expression (val expr : package-type) cannot be used in the body of a functor,..."
I am not understand it. Could you please help me understand by giving an example?
Thank you for your help.
I do not understand clearly what you want to know here. Apparently you get confused with some words of the normal OCaml modules and functors and, rather newer "first class modules" of OCaml. Anyway, I give you a short working example with OCaml 4.00.1 (do not try with 3.12.1 since things are improved in 4), probably it would help you:
module type Arity = sig
val arity :int
end
module S = functor (A : Arity) -> struct
let check = A.arity = 2 (* or whatever *)
end
The above is what you gave us with some trivial fixes to get compiled. Normally to use check, you give an implementation of signature Arity and give it to the functor S:
module AR = struct
let arity = 3
end
module SAR = S(AR)
let () = Printf.printf "%b\n" SAR.check
Let's use first class modules:
let a = (module AR : Arity)
This translates the module AR to a value and bind it to the variable a. Note that the parens are mandatory for syntax. You also need to give the siganture Arity. You can also write as follows:
let a' : (module Arity) = (module AR)
So the type of a and a' are (module Arity) and you need to give it to the compiler somehow. Unfortunately the type inference does not help us here.
You can make the value back to a module as follows:
module A' = (val a)
Now you can also make a first class module value of the functor S:
module type RESULT = sig
val check : bool
end
let s (a : (module Arity)) =
let module A = (val a) in
let module SA = S(A) in
(module SA : RESULT)
What s does is: take a value, make it back to a module, apply the functor S to it, then make another value from the result of functor application. The singature RESULT is necessary for the conversion. You cannot write (module SA : sig val check bool end). I am not good at things around here, but the typing of first class module values are not structural but nominal, I heard. You need to give a name to the signature at (module M : X).
The s's type is (module Arity) -> (module RESULT). Let's apply s to a:
let m = s a
To access check inside m, you need to make it back a module:
let m_check =
let module M = (val m) in
M.check
You might be disappointed to see that value<->module conversions are explicit, but this is how it works...
I have a module StringMap built by the functor Map.Make given a type String:
module StringMap = Map.Make(String)
Besides the ordinary operations provided by Map, I would like to add more definitions in this module, for instance, my_own_function, such that I could call StringMap.my_own_function. Does anyone know where I should define this kind of functions and their signature?
You can use the include keyword inside a new module to add all the same functions. This was also extended to the signature in OCaml 3.12.
module StringMap =
struct
include Map.Make(String)
end
If you want to access the structure of the map, you'll have to add some Obj.magic or the %identity special external function. The redefinition of the type must be exact since no type checking is happening,
module Make (Ord : Map.OrderedType) =
struct
include Map.Make(Ord)
type 'a impl = Empty
| Node of 'a impl * key * 'a * 'a impl * int
external impl_of_t : 'a t -> 'a impl = "%identity"
external t_of_impl : 'a impl -> 'a t = "%identity"
let cardinal map =
let rec cardinal = function
| Empty -> 0
| Node(l,_,_,r,_) -> cardinal l + 1 + cardinal r
in
cardinal (impl_of_t map)
end
module type ELEMENT =
sig
type element_i
end
module Element:ELEMENT =
struct
type element_i = N of int | CNN of cnn
end
module type SEMIRING =
functor (E:ELEMENT)->
sig
type elements
end
module Semiring:SEMIRING =
functor(Element:ELEMENT) ->
struct
let zero = (Element.element_i 0) (*ERROR: Unbounded Value; Same with N 0*)
end
How can I create objects of the type element_i inside Semiring module here?
You can allow the programmer to create values of type element_i inside Semiring by not hiding the constructors of this type, as you are currently doing.
Instead, define the signature ELEMENT as:
module type ELEMENT =
sig
type element_i = N of int | CNN of cnn
end
This makes your functor Semiring expect more of its Element argument: instead of any type Element.element_i, it now only accepts a type with exactly these constructors. But on the plus side it can now apply the constructors to build values of this type, for instance Element.N 12
There's actually two problems with your example. The first is pointed out by Pascal (i.e., the constructors of element_i are hidden by the signature). The second is that the module Element in the functor is not the same as module Element you declared above. The Element argument to the functor is a "hiding" the definition of Element the same way a function parameter would "hide" a variable:
let x = 0
let f = fun x -> (* x here is a different x... *)
module type T = sig (* ... *) end
module M : T = struct (* ... *) end
module F = functor (M : T) -> (* M here is a different M... *)