(*term_types.ml*) type ('a, 'b) term = | Term of 'a * ('a, 'b) term list | Var of 'b

### Analyzing syntax

To proceed, let us restrict our attention to the type`(string, string) term`

. The concrete syntax we use for such expressions is parenthesized.
*a (t*is the syntax of_{1}, ..., t_{n})`Term (a, [t1; ...; tn])`

*a (b (), c)*is the syntax of`Term ("a", [Term ("b", []); Var "c"])`

(*lexer.mll*) { open Parser (*The type token is defined in parser.mli.*) } let alpha=['a'-'z' 'A'-'Z'] rule token = parse [' ' '\t' '\n'] (* skip blanks *) { token lexbuf } | '(' { LPAREN } | ')' { RPAREN } | ',' { COMMA } | ((alpha)(alpha)*) as s { STRING s} | eof { EOI }Now the parser.

/*parser.mly*/ %tokenWith these pieces in place we can write the analyzerSTRING %token LPAREN RPAREN COMMA EOI %start main %type <(string, string) Term_types.term> main %% main: | exp EOI { $1 } ; exp: | tag exp_more { match $2 with | None -> Term_types.Var $1 | Some l -> Term_types.Term ($1, l) } ; exp_more: | /*nothing*/ { None } | LPAREN exp_list RPAREN { Some $2 } ; exp_list: | /*nothing*/ { [] } | exp exp_list_more { $1::$2 } ; exp_list_more: | /*nothing*/ {[]} | COMMA exp exp_list_more { $2::$3 } ; tag: | STRING { $1 } ;

`term_of_string`

like so
(*term.ml*) let term_of_string s = let parse_buf lexbuf = try Parser.main Lexer.token lexbuf with | Parsing.Parse_error -> begin let curr = lexbuf.Lexing.lex_curr_p in let line = curr.Lexing.pos_lnum in let cnum = curr.Lexing.pos_cnum - curr.Lexing.pos_bol in let tok = Lexing.lexeme lexbuf in raise (Failure (Printf.sprintf "file \"Reverse parsing is done with\", line %d, character %d\n\Error : Syntax error \"%s\"" line cnum tok)) end in parse_buf (Lexing.from_string s)

`string_of_term`

:
let rec string_of_term (t:(string, string)term) : string = let string_of_list (f:'a -> string) (l:'a list) : string = "(" ^ String.concat "," (List.map f l) ^ ")" in match t with | Term (s, tl) -> Printf.sprintf "%s" (s^(string_of_list string_of_term tl)) | Var s -> sA simple function for printing a term then goes like this:

let print_term (t:(string, string) term) : unit = Printf.printf "%s" (string_of_term t)For example, this program

```
let tt = term_of_string "a(b(), c)" in print_term tt
```

prints "a(b(), c)" recovering the input as we'd hope.
### Traversal

The function`term_trav`

is a workhorse for doing computations over terms. It generalizes all functions defined by recursion on the structure of a term.
(* [term_trav] is a generalization of all functions defined by recusion on the structure of an [('a, 'b) term] - a homomorphism. The definition of [term_trav] below gives the following signature. {[ val term_trav : ('a * 'b -> 'c) -> ('c -> 'b -> 'b) -> 'b -> ('d -> 'c) -> ('a, 'd) term -> 'c ]} Proof: Assume [term_trav] to operate on arguments of type [('a, 'd) term]. It "returns" either via [f] or [v], and so [f], [v] and [term_trav] must share the same return type, ['c] say. Since the intermediate list results from recursive calls to [term_trav] then it must have type ['c list]. This means [g] must be of type ['c -> 'b -> 'b] for some type ['b] which fixes [x] to ['b] whilst completing [f] as ['a * 'b -> 'c]. Lastly [v] takes arguments of type ['d] meaning it types to ['d -> 'c]. *) let rec term_trav f g x v = function | Term (a, tl) -> let l = List.map (term_trav f g x v) tl in let res = List.fold_right g l x in f (a, res) | Var b -> v bFor a trivial example, we can express a function to compute the "size" of a term this way:

let term_size (t:('a, 'b) term) : int = (term_trav (fun (_, s) -> s + 1) (fun x y -> x + y) 0 (fun _ -> 1)) tI appreciate there's an efficiency gain to be had by eliminating the intermediate list in

`term_trav`

but the resulting function definition is more challenging to reason about (well, at least for me).
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