Using OCaml Menhir, is there a way to access something before it is processed?

Question

I am writing a parser to parse and compute function derivatives in a calculator.

I got a problem with implementing product and quotient rules : for a product the derivation formula is (u*v)' = u'v+uv', thus I need the value of u and u' in the final output. And with the parser I currently have, whenever comes the need to write u, it has already been replaced by u' and I quite don't know how to save its value, nor if it's even possible...

Here's the parser :

%token <string> VAR FUNCTION CONST
%token LEFT_B RIGHT_B PLUS MINUS TIMES DIV
%token DERIV
%token EOL

%start<string> main

%%

main:
t = toDeriv; EOL {t}
;

toDeriv:
DERIV; LEFT_B; e = expr; RIGHT_B {e}
;

expr:
u = expr; PLUS v = hat_funct {u^"+"^v}
  | u = expr; MINUS; v = hat_funct {u^"-"^v}
  | u = hat_funct {u}
;

hat_funct:
u = hat_funct TIMES v = funct {Printf.sprintf "%s * %s + %s * %s" u (A way to save v) v (A way to save u)}
  | u = hat_funct DIV v = funct {Printf.sprintf "(%s * %s - %s * %s)/%s^2" u (A way to save v) (A way to save u) v (A way to save v)}
  | u = funct {u}
;

funct:
f = func; LEFT_B; c = content; RIGHT_B {Derivatives.deriv_func f c}
;

content:
e = expr {e}
  | x = VAR {x}
  | k = CONST {k}

func:
f = FUNCTION {f}
  | k = CONST {k}
;

P.S : I know it might not be the greatest grammar definition at all, it's still a work in progress

score 1 · Accepted Answer · answered May 15 '20 at 18:29

Answering directly your question, yes you can maintain the state of what is being already processed. But that is not how things are done. The idiomatic solution is to write a parser that parses the input language into the abstract syntax tree and then write a solver that will take this tree as input and computes it. You shouldn't do anything in the parser, this is a simple automaton which shall not have any side-effects.

To keep it less abstract, what you want from the parser is the function string -> expr, where expr type is defined something like

type expr = 
  | Var of string
  | Const of string 
  | Binop of binop * expr * expr
and binop = Add | Mul | Sub | Div

Thank you for the quick answer, I spent a part of the night reading things about it because I'm still new to OCaml and didn't really know ho wall of that worked, I've implemented it and figured out how to do things, thank you very much. But just by curiosity, how is it possible to maintain the state of what is being already processed ? — Zeti, May 16 '20 at 15:34

Using OCaml Menhir, is there a way to access something before it is processed?

1 Answers1