Why does the order of operands affect scope? - module

I'm trying to understand why a certain constructor is accepted in one expression but not another. I would have expected it to be out of scope in both. I'm a rank beginner to OCaml (I mostly use Haskell), so I could be missing something totally obvious to someone experienced.
type zero = Zero
type 'n succ = Succ
type 'n snat =
| SZero : zero snat
| SSucc : 'm snat -> 'm succ snat
module SimpleInduction (Pred : sig type 'n pred end) = struct
open Pred
type hyps =
{ base : zero pred
; step : 'm. 'm pred -> 'm succ pred}
let rec induct : type n. hyps -> n snat -> n pred =
fun h sn -> match sn with
| SZero -> h.base
| SSucc p -> h.step (induct h p)
end;;
let module Snot = struct type 'n pred = Top end in
let module Goop = SimpleInduction(Snot) in
Goop.induct {base = Top; step = fun _ -> Top} SZero = Top;;
(*
let module Snot = struct type 'n pred = Top end in
let module Goop = SimpleInduction(Snot) in
Top = Goop.induct {base = Top; step = fun _ -> Top} SZero;;
*)
This compiles just fine, for some reason. With the second definition of Snot uncommented, I get an error:
19 | Top = Goop.induct {base = Top; step = fun _ -> Top} SZero;;
^^^
Error: Unbound constructor Top
What brings Top into scope in the first definition of Snot? Using regular modules rather than first-classlocal ones makes no difference.
If I use Snot.Top on the left-hand side, I get no complaints on the right-hand side. Why is that?

In short, type-directed disambiguation is indeed not restricted to scope.
With an explicit type annotation, the type checker can select the constructor from the type without bringing the constructor in scope.
For instance,
module M = struct type 'a t = A of 'a end
let ok: _ M.t = A ()
let wrong: _ M.t = A (A ())
the first example is valid because the type annotation is enough to know that the A in A () is an _ A.t. However, the second example does not work because the constructor has not been brought into the scope.
Moreover, type-directed disambiguation only requires the expected type of the constructor or record to be known. Typically, in this example
let expected =
let f (M.A x) = x in
f (A ())
we know that the type of the argument of f is an _ M.t, thus we know that the A in f (A ()) come from _ M.t and we can use type-directed disambiguation like in the case with the explicit annotation.
If you find this behavior exotic, the warning 42 [name-out-of-scope] can be used to warn in such situation. Compiling your example with this warning yields (among many other instances of this warning)
23 | Goop.induct {base = Top; step = fun _ -> Top} SZero = Top
^^^
Warning 40 [name-out-of-scope]: Top was selected from type Snot.pred.
It is not visible in the current scope, and will not
be selected if the type becomes unknown.
(the warning names are new in 4.12)
Concerning your second point, the order of expression may matter in the absence of explicit annotations. Indeed, without explicit annotation, type-directed disambiguation will be only be able to select the right constructor when the expected type is already known. And type checking goes from left to right in OCaml. Thus in
... = Top
the type of the left-hand side has already been inferred and thus the expected type of Top is _ Snot.pred.
When the order is reversed
Top = ...
the typechecker is trying to find a constructor Top without any type information and there are no constructor Top in scope. Thus it fails with an Unbound constructor error. If you want to avoid depending on the order, you can either
write the full name of the constructor:
Snot.Top = ...
use an explicit type annotation
(Top: _ Snot.pred) = ...
open the Snot module.
Snot.( Top ) = ...
(* or *)
let open Snot in Top = ...
I would advise to use one of those solutions since there are more robust.
After all, relying on the specific implementation of the type checking is brittle.
In fact, there is a compiler flag -principal and a warning (18) [not-principal] that takes care to emit a warning in presence of such potentially brittle inference:
23 | Goop.induct {base = Top; step = fun _ -> Top} SZero = Top
^^^
Warning 18 [not-principal]: this type-based constructor disambiguation is not principal.
Here "not principal" means that the result of the type-based disambiguation depended on the order of the type-checking.

Related

How to create a set of elements without knowing the type of the element?

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".
#

The signature for this packaged module couldn't be inferred in recursive function

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

expose a private type for module extension in OCaml

I'd like to extend a module but I need access to its private components. Here's an example:
nat.mli:
type t
val zero : t
val succ : t -> t
nat.ml:
type t = int
let zero = 0
let succ x = x + 1
I'd like to define a new module Ext_nat that defines a double function. I was trying to do something like this.
ext_nat.mli:
include (module type of Nat)
val double : t -> t
ext_nat.ml:
include Nat
let double x = 2 * x
It's not working as I don't have access to the representation of x in the last line.
Now that I'm thinking about this, it may not be such a good idea anyway because this would break the encapsulation of nat. So what is the best way to do this? I could define a new module nat_public where type t = int in the signature, and define nat and ext_nat with a private type t. What do you think?
You need to use with type statement. It is possible to write the code below in many different ways, but the idea is always the same.
module type NatSig =
sig
type t
val zero : t
val succ : t -> t
end
module type ExtNatSig =
sig
include NatSig
val double : t -> t
end
module ExtNat : ExtNatSig =
struct
type t = int
let zero = 0
let succ = fun x -> x + 1
let double = fun x -> x * 2
end
module Nat = (ExtNat : NatSig with type t = ExtNat.t)
let z = Nat.zero
let _ = ExtNat.double z
The problem is that as far as I remember it's impossible to achieve this behavior with your file structure: you define your module implicitly with signature in .mli file and structure itself in .ml, so you don't have enough control over you module, that's why I suggest you to reorganize your code a little bit (if it's not a problem).

Modules and record fields

I have stumbled across a rather simple OCaml problem, but I can't seem to find an elegant solution. I'm working with functors that are applied to relatively simple modules (they usually define a type and a few functions on that type) and extend those simple modules by adding additional more complex functions, types and modules. A simplified version would be:
module type SIMPLE = sig
type t
val to_string : t -> string
val of_string : string -> t
end
module Complex = functor (S:SIMPLE) -> struct
include S
let write db id t = db # write id (S.to_string t)
let read db id = db # read id |> BatOption.map S.of_string
end
There is no need to give the simple module a name because all its functionality is present in the extended module, and the functions in the simple module are generated by camlp4 based on the type. The idiomatic use of these functors is:
module Int = Complex(struct
type t = int
end)
The problem appears when I'm working with records:
module Point2D = Complex(struct
type t = { x : int ; y : int }
end)
let (Some location) = Point2D.read db "location"
There seems to be no simple way of accessing the x and y fields defined above from outside the Point2D module, such as location.x or location.Point2D.x. How can I achieve this?
EDIT: as requested, here's a complete minimal example that displays the issue:
module type TYPE = sig
type t
val default : t
end
module Make = functor(Arg : TYPE) -> struct
include Arg
let get = function None -> default | Some x -> (x : t)
end
module Made = Make(struct
type t = {a : int}
let default = { a = 0 } (* <-- Generated by camlp4 based on type t above *)
end)
let _ = (Made.get None).a (* <-- ERROR *)
Let's look at the signature of some of the modules involved. These are the signatures generated by Ocaml, and they're principal signatures, i.e. they are the most general signatures allowed by the theory.
module Make : functor (Arg : TYPE) -> sig
type t = Arg.t
val default : t
val get : t option -> t
end
module Made : sig
type t
val default : t
val get : t option -> t
end
Notice how the equation Make(A).t = A.t is retained (so Make(A).t is a transparent type abbreviation), yet Made.t is abstract. This is because Made is the result of applying the functor to an anonymous structure, so there is no canonical name for the argument type in this case.
Record types are generative. At the level of the underlying type theory, all generative types behave like abstract types with some syntactic sugar for constructors and destructors. The only way to designate a generative type is to give its name, either the original name or one that expands to the original name via a series of type equations.
Consider what happens if you duplicate the definition of Made:
module Made1 = Make(struct
type t = {a : int}
let default = { a = 0 } (* <-- Generated by camlp4 based on type t above *)
end)
module Made2 = Make(struct
type t = {a : int}
let default = { a = 0 } (* <-- Generated by camlp4 based on type t above *)
end)
You get two different types Made1.t and Made2.t, even though the right-hand sides of the definitions are the same. That's what generativity is all about.
Since Made.t is abstract, it's not a record type. It doesn't have any constructor. The constructors were lost when the structure argument was closed, for a lack of a name.
It so happens that with records, one often wants the syntactic sugar but not the generativity. But Ocaml doesn't have any structural record types. It has generative record types, and it has objects, which from a type theoretical view subsume records but in practice can be a little more work to use and have a small performance penalty.
module Made_object = Make(struct
type t = <a : int>
let default = object method a = 0 end
end)
Or, if you want to keep the same type definition, you need to provide a name for the type and its constructors, which means naming the structure.
module A = struct
type t = {a : int}
let default = { a = 0 } (* <-- Generated by camlp4 based on type t above *)
end
module MadeA = Make(A)
Note that if you build Make(A) twice, you get the same types all around.
module MadeA1 = Make(A)
module MadeA2 = Make(A)
(Ok, this isn't remarkable here, but you'd still get the same abstract types in MadeA1 and MakeA2, unlike the Made1 and Made2 case above. That's because now there's a name for these types: MadeA1.t = Make(A).t.)
First of all, in your last code sample, last line, you probably mean .a rather than .x.
The problem with your code is that, with the way you define your Make functor, the type t is abstract in Made: indeed, the functors use the TYPE signature which seals {a : int} as an abstract type.
The following design circumvent the issue, but, well, its a different design.
module type TYPE = sig
type t
val default : t
end
module Extend = functor(Arg : TYPE) -> struct
open Arg
let get = function None -> default | Some x -> (x : t)
end
module T = struct
type t = {a : int}
let default = { a = 0 }
end
module Made = struct
include T
include Extend(T)
end
let _ = Made.((get None).a)
The problem is that OCaml doesn't have a name to refer to the qualified components of the type t (in this case a record, but the same problem would be present with normal variants) outside Made. Naming the unnamed solves the problem:
module F = struct
type t = {a : int}
let default = { a = 0 }
end
module Made = Make(F)
let _ = (Made.get None).F.a (* <-- WORKS *)
You can also declare explicitly the type outside the functorial application:
type rcd = {a : int}
module Made = Make(struct
type t = rcd
let default = { a = 0 }
end)
let _ = (Made.get None).a (* <-- WORKS *)

Can I avoid committing to particular types in a module type and still get pattern matching?

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.