Why does OCaml think that this function takes an int parameter when nothing suggests that it should be the case? - ocaml

Question

I was working on chapter 1 of Modern Compiler Implementation in ML by Andrew Appel and I decided to implement it in OCaml instead of SML. I'm new to OCaml and I came across a very frustrating problem. OCaml seems to think that the below function has the signature int * (int * 'a) -> 'a option.
let rec lookupTable = function
| name, (i, v) :: _ when name = i -> Some v
| name, (_, _) :: rest -> lookupTable (name, rest)
| _, [] -> None
But as far as I can tell, there should be nothing that suggests that the first element in the tuple is an int. This is a problem because when the lookupTable function down the line, the compiler complains that I am not passing it an integer. Perhaps I am missing something incredibly obvious, but it has been pretty mind-boggling. Here is the rest of the program
open Base
type id = string
type binop = Plus | Minus | Times | Div
type stm =
| CompoundStm of stm * stm
| AssignStm of id * exp
| PrintStm of exp list
and exp =
| IdExp of id
| NumExp of int
| OpExp of exp * binop * exp
| EseqExp of stm * exp
(* Returns the maximum number of arguments of any print
statement within any subexpression of a given statement *)
let rec maxargs s =
match s with
| CompoundStm (stm1, stm2) -> Int.max (maxargs stm1) (maxargs stm2)
| AssignStm (_, exp) -> maxargs_exp exp
(* Might be more nested expressions *)
| PrintStm exps -> Int.max (List.length exps) (maxargs_explist exps)
and maxargs_exp e = match e with EseqExp (stm, _) -> maxargs stm | _ -> 0
and maxargs_explist exps =
match exps with
| exp :: rest -> Int.max (maxargs_exp exp) (maxargs_explist rest)
| [] -> 0
type table = (id * int) list
let updateTable name value t : table = (name, value) :: t
let rec lookupTable = function
| name, (i, v) :: _ when name = i -> Some v
| name, (_, _) :: rest -> lookupTable (name, rest)
| _, [] -> None
exception UndefinedVariable of string
let rec interp s =
let t = [] in
interpStm s t
and interpStm s t =
match s with
| CompoundStm (stm1, stm2) -> interpStm stm2 (interpStm stm1 t)
| AssignStm (id, exp) ->
let v, t' = interpExp exp t in
updateTable id v t'
(* Might be more nested expressions *)
| PrintStm exps ->
let interpretAndPrint t e =
let v, t' = interpExp e t in
Stdio.print_endline (Int.to_string v);
t'
in
List.fold_left exps ~init:t ~f:interpretAndPrint
and interpExp e t =
match e with
| IdExp i -> (
match lookupTable (i, t) with
| Some v -> (v, t)
| None -> raise (UndefinedVariable i))
| NumExp i -> (i, t)
| OpExp (exp1, binop, exp2) ->
let exp1_val, t' = interpExp exp1 t in
let exp2_val, _ = interpExp exp2 t' in
let res =
match binop with
| Plus -> exp1_val + exp2_val
| Minus -> exp1_val - exp2_val
| Times -> exp1_val * exp2_val
| Div -> exp1_val / exp2_val
in
(res, t')
| EseqExp (s, e) -> interpExp e (interpStm s t)

Answer 1

Base defines = as int -> int -> bool, so when you have the expression name = i the compiler will infer them as ints.
You can access the polymorphic functions and operators through the Poly module, or use a type-specific operator by locally opening the relevant module, e.g. String.(name = i).
The reason Base does not expose polymorphic operators by default is briefly explained in the documentation's introduction:
The comparison operators exposed by the OCaml standard library are polymorphic:
What they implement is structural comparison of the runtime representation of values. Since these are often error-prone, i.e., they don't correspond to what the user expects, they are not exposed directly by Base.
There's also a performance-argument to be made, because the polymorphic/structural operators need to also inspect what kind of value it is at runtime in order to compare them correctly.