Do I have to write a function to locally bind a type ?
Using an auxiliary function works, but trying to do so as part of a type decoration of a term during destructuring does not seem to work.
type ('a, 'b) eq
type 'a repr = TpFun : ('a, 'x -> 'y) eq -> 'a repr
let rewrite1 (type x y b f) (fxy : (f, x -> y) eq) : (f, x -> b) eq =
failwith ""
let eqp : type a b. a repr -> (a, b) eq = function
| TpFun (fxy) ->
let x = rewrite1 fxy in
_
let eqpFAIL : type a b. a repr -> (a, b) eq = function
| TpFun ((fxy : type x y. (a, x -> y) eq) ) ->
let x = failwith "" in
_
As described in https://ocaml.org/manual/gadts-tutorial.html#s%3Aexplicit-existential-name, the correct syntax for binding existential type in pattern is
let eqp : type a b. a repr -> (a, b) eq = function
| TpFun (type x y) (fxy :(a, x -> y) eq) ->
assert false
Beware also that you probably don't want to use an abstract type for eq since the type checker cannot infer any property on an abstract types that might have been defined as an abbreviation!
type ('a, 'b) eq = int
I am generally advising to use a type definition with an existing name:
type ('a,'b) eq = Eq
because that construction guarantees injectivity with respect to the type constructor parameters and allow the type checker to infer some type inequalities.
Related
I noticed that I cannot do the following in OCaml:
# let foo (f : 'a -> unit) = f 1; f "s";;
Error: This expression has type string but an expression was expected of type
int
In Haskell, this could be resolved by universally quantifying the input function f using Rank2Types:
{-# LANGUAGE Rank2Types #-}
foo :: (forall a. a -> ()) -> ()
foo f = let a = f 1 in f "2"
How can I get similar exposure to this in OCaml?
OCaml only supports semi-explicit higher-rank polymorphism: polymorphic functions arguments must be boxed either inside a record with polymorphic field:
type id = { id: 'a. 'a -> 'a }
let id = { id=(fun x -> x) }
let f {id} = id 1, id "one"
or inside an object
let id' = object method id: 'a. 'a -> 'a = fun x -> x end
let f (o: <id:'a. 'a -> 'a>) = o#id 1, o#id "one"
Beyond the syntactic heaviness, this explicit boxing of polymorphic functions has the advantage that it works well with type inference while still requiring only annotations at the definition of record types or the methods.
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
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
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.
First the code:
module Boolean = struct
exception SizeMismatch
type boolean = T | F | Vec of boolean array
let to_bool v = match v with
T -> true
| F -> false
| _ -> raise SizeMismatch
end
module Logic = struct
type 'a var_t = { name: string; mutable value: 'a }
type 'a bexp = Const of 'a
| Var of 'a var_t
let eval exp = match exp with
Const x -> x
| Var x -> x.value
let make_var s v = { name = s; value = v }
let set v n = v.value <- n
let get_var_name v = v.name
let get_var_val v = v.value
end
module type EXP =
sig
type 'a var_t
type 'a bexp
val eval_exp : 'a bexp -> bool
val get_var_name : 'a var_t -> string
val get_var_val : 'a var_t -> 'a
end
module LogicExp =
struct
include Logic
let eval_exp exp = Boolean.to_bool (Logic.eval exp)
end
module FSM ( Exp : EXP ) =
struct
let print_var v = Printf.printf "%s = %d\n" (Exp.get_var_name v)
(Exp.get_var_val v)
end
module MyFSM = FSM(LogicExp)
let myvar = Logic.make_var "foo" 1;;
MyFSM.print_var myvar ;;
When I compile it I get the following error:
File "test.ml", line 57, characters 19-27:
Error: Signature mismatch:
Modules do not match:
sig
type 'a var_t =
'a Logic.var_t = {
name : string;
mutable value : 'a;
}
type 'a bexp = 'a Logic.bexp = Const of 'a | Var of 'a var_t
val eval : 'a bexp -> 'a
val make_var : string -> 'a -> 'a var_t
val set : 'a var_t -> 'a -> unit
val get_var_name : 'a var_t -> string
val get_var_val : 'a var_t -> 'a
val eval_exp : Boolean.boolean Logic.bexp -> bool
end
is not included in
EXP
Values do not match:
val eval_exp : Boolean.boolean Logic.bexp -> bool
is not included in
val eval_exp : 'a bexp -> bool
What I don't understand is how the more specific type isn't included in the more general type?
The error message is actually quite accurate:
Values do not match:
val eval_exp : Boolean.boolean Logic.bexp -> bool
is not included in
val eval_exp : 'a bexp -> bool
The MyFSM functor expects a module argument that, amongst other things, should contain a function eval_exp of type 'a bexp -> bool. That means that given a value of type 'a bexp for whatever choice of 'a the function should produce a value of type bool. You are, however, supplying a module that contains a function that only does this for one particular choice of 'a, i.e., the one where 'a is the type boolean from the module Boolean.
This quickest fix is to define your signature EXP as
module type EXP =
sig
type b (* added *)
type 'a var_t
type 'a bexp
val eval_exp : b bexp -> bool (* changed *)
val get_var_name : 'a var_t -> string
val get_var_val : 'a var_t -> 'a
end
so that eval_exp now operates on Boolean expressions over a fixed type b and then define LogicExp as
module LogicExp =
struct
type b = Boolean.boolean (* added *)
include Logic
let eval_exp exp = Boolean.to_bool (Logic.eval exp)
end
so that it fixes b to Boolean.boolean.
Implementing these changes will make your code compile.
Now, let us look at your question on "how the more specific type isn't included in the more general type?". This assumes that 'a bexp -> bool is indeed more general than boolean bexp -> bool, but in fact it isn't. A function type A -> B is considered more general than a function type C -> D if C is more general than A and B is more general than D:
A <: C D <: B
--------------------
C -> D <: A -> B
Note the "flipping" of C and A in the premise. We say that the function-space constructor ... -> ... is contravariant in its argument position (versus covariant in its result position).
Intuitively, a type is more general than another if it contains more values. To see why the function-space constructor is contravariant in its argument position, consider a function f of type A -> C for some types A and C. Now, consider a type B that is strictly more general than A, that is, all values in A are also in B, but B contains some values that are not in A. Thus, there is at least one value b to which we can assign type B, but not type A. Its type tells us that f knows how to operate on values of type A. However, if we were to (incorrectly!) conclude from A <: B that A -> C <: B -> C, then we could use f as if it had type B -> C and, hence, then we could pass the value b as an argument to f. But b is not of type A and f only knows how to operate on values of type A!
Clearly, covariance of ... -> ... in argument positions wouldn't work. To see that contravariance does work, consider the same types A, B, and C and now also consider a function g of type B -> C. That is, g knows how to operate on all values of type B. Contravariance of the function-space constructor in its argument position allows us to conclude that g can also be safely assigned the type A -> C. As we know that all values in A are also in B and g knows how to deal with all of B this poses no problems and we can safely pass values in A to g.