I'm having trouble with lists in OCaml. I've read conflicting statements saying whether or not the lists can be modified at runtime. Can the cons operator be used at runtime?
Additionally, why is a doberman (see below) allowed to be in a list of chihuahuas? How would one go about adding another chihuahua onto the list (as attempted with the last line)?
class virtual dog =
object
method virtual bark : unit
end;;
class chihuahua =
object
inherit dog
method bark = Printf.printf "Yip!"
end;;
class doberman =
object
inherit dog
method bark = Printf.printf "Roar!"
end;;
let c1 = new chihuahua;;
let c2 = new chihuahua;;
let c3 = new chihuahua;;
let d1 = new doberman;;
let arrayOfDogs = [c1;c2;d1];;
arrayOfDogs :: c3;;
1) You can use the cons operator at runtime, it just returns a new list rather than mutating the input list.
2) Class types in OCaml use "structural" subtyping, rather than Java-style "nominal" subtyping. The inferred type of arrayOfDogs will be "an object with a method named bark of type unit -> unit (not necessarily a dog)". For example:
# class cat = object
method bark = print_endline "meow"
end ;;
class cat : object method bark : unit end
# let c = new cat ;;
val c : cat = <obj>
# c :: arrayOfDogs ;;
- : cat list = [<obj>; <obj>; <obj>; <obj>]
3) The problem with arrayOfDogs :: c3 is you've got it the wrong way around. The type of :: is 'a -> 'a list -> 'a list. To add c3 at the beginning, use
c3 :: arrayOfDogs
To add it at the end, use the "append" operator #
arrayOfDogs # [c3]
you need to have your list on the right hand side, not the left. Ie:
c3 :: arrayOfDogs;;
This is why the last line fails.
As far as the list construction goes, given that OCaml is type-inferred, the interpreter probably figured out that you're constructing a list of dogs given you added the doberman on construction. Therefore it's not a list of Chihuahuas.
What does OCaml report as the type of arrayOfDogs?
Perhaps you mean: c3 :: arrayOfDogs;;
Related
I'm running into problems around recursive/mutually referential module definitions trying to use Caml's Map/Set stuff. I really want ones that just work on types, not modules. I feel like it should be possible to do this with first-class modules, but I'm failing to make the syntax work.
The signature I want is:
module type NonFunctorSet = sig
type 'a t
val create : ('a -> 'a -> int) -> 'a t
val add : 'a t -> 'a -> 'a t
val remove : 'a t -> 'a -> 'a t
val elements : 'a t -> 'a list
end
Possibly with other Caml.Set functions included. My idea for how this would work is something like:
type 'a t = {
m : (module Caml.Set.S with type elt = 'a);
set : m.t
}
let create (compare : 'a -> 'a -> t) =
module m = Caml.Set.Make(struct type t = 'a let compare = compare end) in
let set = m.empty in
{m = m; set = set;}
end
But that doesn't work for a number of reasons; 'a isn't exposed in the right places, I can't reference m.t in the same record where m was defined, etc.
Is there a version of this that works?
Adding more context about my use case:
I have two modules, Region and Tribe. Tribe needs access to a lot of the interface of Region, so I am currently creating Tribe as a functor, MakeTribe(Region : RegionT). Region mostly doesn't need to know about Tribe, but it does need to be able to store a mutable collection of Tribe.t that represent the tribes living in that region.
So, somehow or other, I need a RegionT like
module type RegionT = sig
type <region>
val get_local_tribes : <region> -> <tribes>
val add_tribe : <region> -> <tribe> -> unit
...
end
I don't really care about the specific syntax of <tribe>, <tribes> and <region> in this, so long as the fully built Tribe module can know that Region.get_local_tribes, etc, will yield an actual Tribe.t
The circular dependency problem is that the type <tribe> does not exist until the module Tribe is created. My idea so far has been to have RegionT.t actually be 'a RegionT.t, and then Tribe could simply refer to Tribe.t Region.t. This is all fine if I'm satisfied with keeping a <tribe> list inside Region, but I want it to be a set.
I feel this should be possible based on the following example code :
module Example : sig
type t
val compare : t -> t -> int
end = struct
type t = int
let compare = Int.compare
end
module ExampleSet = Caml.Set.Make(struct type t = Example.t let compare = Example.compare end)
All that Example exposes in its interface is a type and a function from two instances of that type to an int; why is that more than having a 'a -> 'a -> int, which has the same things?
Using Polymoprhic Sets and Maps from the Base Library
In Base and Core libraries, from Jane Street, ordered data structures, such as maps, sets, hash tables, and hash sets, are all implemented as polymorphic data structures, instead of functorized versions as in the vanilla OCaml standard library.
You can read about them more in the Real World OCaml Maps and Hashtbales chapter. But here are quick recipes. When you see a comparator in the function interface, e.g., in Map.empty what it actually wants you is to give you a module that implements the comparator interface. The good news is that most of the modules in Base/Core are implementing it, so you don't have to worry or know anything about this to use it, e.g.,
# open Base;;
# let empty = Map.empty (module Int);;
val empty : (Base.Int.t, 'a, Base.Int.comparator_witness) Base.Map.t =
<abstr>
# Map.add empty 1 "one";;
- : (Base.Int.t, string, Base.Int.comparator_witness) Base.Map.t
Base.Map.Or_duplicate.t
= `Ok <abstr>
So the simple rule, if you want a set,map,hashtable,hashset where the key element has type foo, just pass (module Foo) as a comparator.
Now, what if you want to make a mapping from your custom type? E.g., a pair of ints that you would like to compare in lexicographical order.
First of all, we need to define sexp_of and compare functions. For our type. We will use ppx derivers for it, but it is easy to make it manually if you need.
module Pair = struct
type t = int * int [##deriving compare, sexp_of]
end
Now, to create a comparator, we just need to use the Base.Comparator.Make functor, e.g.,
module Lexicographical_order = struct
include Pair
include Base.Comparator.Make(Pair)
end
So now we can do,
# let empty = Set.empty (module Lexicographical_order);;
val empty :
(Lexicographical_order.t, Lexicographical_order.comparator_witness)
Base.Set.t = <abstr>
# Set.add empty (1,2);;
- : (Lexicographical_order.t, Lexicographical_order.comparator_witness)
Base.Set.t
= <abstr>
Despite that Base's data structures are polymorphic they strictly require that the module that provides the comparator is instantiated and known. You can just use the compare function to create a polymorphic data structure because Base will instantiate a witness type for each defined compare function and capture it in the data structure type to enable binary methods. Anyway, it is a complex issue, read on for easier (and harder) solutions.
Instantiating Sets on mutually dependent modules
In fact, OCaml supports mutually recursive funtors and although I would suggest you to break the recursion by introducing a common abstraction on which both Region and Tribe depend, you can still encode your problem in OCaml, e.g.,
module rec Tribe : sig
type t
val create : string -> t
val compare : t -> t -> int
val regions : t -> Region.t list
end = struct
type t = string * Region.t list
let create name = name,[]
let compare (x,_) (y,_) = String.compare x y
let regions (_,r) = r
end
and Region : sig
type t
val empty : t
val add_tribe : Tribe.t -> t -> t
val tribes : t -> Tribe.t list
end = struct
module Tribes = Set.Make(Tribe)
type t = Tribes.t
let empty = Tribes.empty
let add_tribe = Tribes.add
let tribes = Tribes.elements
end
Breaking the Dependency Loop
A much better solution would be to redesign your modules and break the dependency loop. The simplest approach would be just to choose some identifier that will be used to compare tribes, e.g., by their unique names,
module Region : sig
type 'a t
val empty : 'a t
val add_tribe : string -> 'a -> 'a t -> 'a t
val tribes : 'a t -> 'a list
end = struct
module Tribes = Map.Make(String)
type 'a t = 'a Tribes.t
let empty = Tribes.empty
let add_tribe = Tribes.add
let tribes r = Tribes.bindings r |> List.map snd
end
module Tribe : sig
type t
val create : string -> t
val name : t -> string
val regions : t -> t Region.t list
val conquer : t Region.t -> t -> t Region.t
end = struct
type t = Tribe of string * t Region.t list
let create name = Tribe (name,[])
let name (Tribe (name,_)) = name
let regions (Tribe (_,r)) = r
let conquer region tribe =
Region.add_tribe (name tribe) tribe region
end
There are also tons of other options and in general, when you have mutual dependencies it is actually an indicator of a problem in your design. So, I would still revisit the design stage and eschew the circular dependencies.
Creating Polymorphic Sets using the Vanilla OCaml Standard Library
It is not an easy task, especially if you need to handle operations that involve several sets, e.g., Set.union. The problem is that Set.Make is generating a new type for the set per each compare function so when we need to union two sets it is hard for us to prove to the OCaml compiler that they were created from the same type. It is possible but really painful, I am showing how to do this only to discourage you from doing this (and to showcase OCaml's dynamic typing capabilities).
First of all we need a witness type that will reify an OCaml type for the set into a concrete value.
type _ witness = ..
module type Witness = sig
type t
type _ witness += Id : t witness
end
Now we can define our polymorphic set as an existential that holds the set itself and the module with operations. It also holds the tid (for type identifier) that we will later use to recover the type 's of the set.
type 'a set = Set : {
set : 's;
ops : (module Set.S with type elt = 'a and type t = 's);
tid : (module Witness with type t = 's);
} -> 'a set
Now we can write the create function that will take the compare function and turn it into a set,
let create : type a s. (a -> a -> int) -> a set =
fun compare ->
let module S = Set.Make(struct
type t = a
let compare = compare
end) in
let module W = struct
type t = S.t
type _ witness += Id : t witness
end in
Set {
set = S.empty;
ops = (module S);
tid = (module W);
}
The caveat here is that each call to create will generate a new instance of the set type 's so we can compare/union/etc two sets that were created with the same create function. In other words, all sets in our implementation shall share the same ancestor. But before that lets take a pain and implement at least two operations, add and union,
let add : type a. a -> a set -> a set =
fun elt (Set {set; tid; ops=(module Set)}) -> Set {
set = Set.add elt set;
ops = (module Set);
tid;
}
let union : type a. a set -> a set -> a set =
fun (Set {set=s1; tid=(module W1); ops=(module Set)})
(Set {set=s2; tid=(module W2)}) ->
match W1.Id with
| W2.Id -> Set {
set = Set.union s1 s2;
tid = (module W1);
ops = (module Set);
}
| _ -> failwith "sets are potentially using different types"
Now, we can play with it a bit,
# let empty = create compare;;
val empty : '_weak1 set = Set {set = <poly>; ops = <module>; tid = <module>}
# let x1 = add 1 empty;;
val x1 : int set = Set {set = <poly>; ops = <module>; tid = <module>}
# let x2 = add 2 empty;;
val x2 : int set = Set {set = <poly>; ops = <module>; tid = <module>}
# let x3 = union x1 x2;;
val x3 : int set = Set {set = <poly>; ops = <module>; tid = <module>}
# let x4 = create compare;;
val x4 : '_weak2 set = Set {set = <poly>; ops = <module>; tid = <module>}
# union x3 x4;;
Exception: Failure "sets are potentially using different types".
#
I'm still trying to figure out how to split code when using mirage and it's myriad of first class modules.
I've put everything I need in a big ugly Context module, to avoid having to pass ten modules to all my functions, one is pain enough.
I have a function to receive commands over tcp :
let recvCmds (type a) (module Ctx : Context with type chan = a) nodeid chan = ...
After hours of trial and errors, I figured out that I needed to add (type a) and the "explicit" type chan = a to make it work. Looks ugly, but it compiles.
But if I want to make that function recursive :
let rec recvCmds (type a) (module Ctx : Context with type chan = a) nodeid chan =
Ctx.readMsg chan >>= fun res ->
... more stuff ...
|> OtherModule.getStorageForId (module Ctx)
... more stuff ...
recvCmds (module Ctx) nodeid chan
I pass the module twice, the first time no problem but
I get an error on the recursion line :
The signature for this packaged module couldn't be inferred.
and if I try to specify the signature I get
This expression has type a but an expression was expected of type 'a
The type constructor a would escape its scope
And it seems like I can't use the whole (type chan = a) thing.
If someone could explain what is going on, and ideally a way to work around it, it'd be great.
I could just use a while of course, but I'd rather finally understand these damn modules. Thanks !
The pratical answer is that recursive functions should universally quantify their locally abstract types with let rec f: type a. .... = fun ... .
More precisely, your example can be simplified to
module type T = sig type t end
let rec f (type a) (m: (module T with type t = a)) = f m
which yield the same error as yours:
Error: This expression has type (module T with type t = a)
but an expression was expected of type 'a
The type constructor a would escape its scope
This error can be fixed with an explicit forall quantification: this can be done with
the short-hand notation (for universally quantified locally abstract type):
let rec f: type a. (module T with type t = a) -> 'never = fun m -> f m
The reason behind this behavior is that locally abstract type should not escape
the scope of the function that introduced them. For instance, this code
let ext_store = ref None
let store x = ext_store := Some x
let f (type a) (x:a) = store x
should visibly fail because it tries to store a value of type a, which is a non-sensical type outside of the body of f.
By consequence, values with a locally abstract type can only be used by polymorphic function. For instance, this example
let id x = x
let f (x:a) : a = id x
is fine because id x works for any x.
The problem with a function like
let rec f (type a) (m: (module T with type t = a)) = f m
is then that the type of f is not yet generalized inside its body, because type generalization in ML happens at let definition. The fix is therefore to explicitly tell to the compiler that f is polymorphic in its argument:
let rec f: 'a. (module T with type t = 'a) -> 'never =
fun (type a) (m:(module T with type t = a)) -> f m
Here, 'a. ... is an universal quantification that should read forall 'a. ....
This first line tells to the compiler that the function f is polymorphic in its first argument, whereas the second line explicitly introduces the locally abstract type a to refine the packed module type. Splitting these two declarations is quite verbose, thus the shorthand notation combines both:
let rec f: type a. (module T with type t = a) -> 'never = fun m -> f m
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 two module types:
module type ORDERED =
sig
type t
val eq : t * t -> bool
val lt : t * t -> bool
val leq : t * t -> bool
end
module type STACK =
sig
exception Empty
type 'a t
val empty : 'a t
val isEmpty : 'a t -> bool
val cons : 'a * 'a t -> 'a t
val head : 'a t -> 'a
val tail : 'a t -> 'a t
val length : 'a t -> int
end
I want to write a functor which "lifts" the order relation from the basic ORDERED type to STACKs of that type. That can be done by saying that, for example, two stacks of elements will be equal if all its individual elements are equal. And that stacks s1 and s2 are s.t. s1 < s2 if the first of each of their elements, e1 and e2, are also s.t. e1 < e2, etc.
Now, if don't commit to explicitly defining the type in the module type, I will have to write something like this (or won't I?):
module StackLift (O : ORDERED) (S : STACK) : ORDERED =
struct
type t = O.t S.t
let rec eq (x,y) =
if S.isEmpty x
then if S.isEmpty y
then true
else false
else if S.isEmpty y
then false
else if O.eq (S.head x,S.head y)
then eq (S.tail x, S.tail y)
else false
(* etc for lt and leq *)
end
which is a very clumsy way of doing what pattern matching serves so well. An alternative would be to impose the definition of type STACK.t using explicit constructors, but that would tie my general module somewhat to a particular implementation, which I don't want to do.
Question: can I define something different above so that I can still use pattern matching while at the same time keeping the generality of the module types?
As an alternative or supplement to the other access functions, the module can provide a view function that returns a variant type to use in pattern matching.
type ('a, 's) stack_view = Nil | Cons of 'a * 's
module type STACK =
sig
val view : 'a t -> ('a , 'a t) stack_view
...
end
module StackLift (O : ORDERED) (S : STACK) : ORDERED =
struct
let rec eq (x, y) =
match S.view x, S.view y with
Cons (x, xs), Cons (y, ys) -> O.eq (x, y) && eq (xs, ys)
| Nil, Nil -> true
| _ -> false
...
end
Any stack with a head and tail function can have a view function too, regardless of the underlying data structure.
I believe you've answered your own question. A module type in ocaml is an interface which you cannot look behind. Otherwise, there's no point. You cannot keep the generality of the interface while exposing details of the implementation. The only thing you can use is what's been exposed through the interface.
My answer to your question is yes, there might be something you can do to your definition of stack, that would make the type of a stack a little more complex, thereby making it match a different pattern than just a single value, like (val,val) for instance. However, you've got a fine definition of a stack to work with, and adding more type-fluff is probably a bad idea.
Some suggestions with regards to your definitions:
Rename the following functions: cons => push, head => peek, tail => pop_. I would also add a function val pop : 'a t -> 'a * 'a t, in order to combine head and tail into one function, as well as to mirror cons. Your current naming scheme seems to imply that a list is backing your stack, which is a mental leak of the implementation :D.
Why do eq, lt, and leq take a pair as the first parameter? In constraining the type of eq to be val eq : 'a t * 'a t -> 'a t, you're forcing the programmer that uses your interface to keep around one side of the equality predicate until they've got the other side, before finally applying the function. Unless you have a very good reason, I would use the default curried form of the function, since it provides a little more freedom to the user (val eq : 'a t -> 'a t -> 'a t). The freedom comes in that they can partially apply eq and pass the function around instead of the value and function together.