Naming expressions with let: Interpreter Edition

Now that we know how to access memory, we can add support for names. Our language will support OCaml-style variable binding: variables are immutable, and are bound in expressions using let. Some examples:

(let ((x 1)) x) ; => 1
(let ((x 3)) (let ((y (+ x 2)) (+ x y))) ; => 8
(let ((x 3)) (let ((x (+ x 2)) (+ x x))) ; => 10

As usual, we’ll start with our interpreter and then work on the compiler.

Names in the interpreter

How should the interpreter evaluate this expression?


It depends! If the expression is nested inside a let, e.g.,

(let ((x 1)) x)

then the interpreter should evaluate it to its bound value. Otherwise, x is meaningless. So it seems like the interpreter needs to do different things depending on context. We can’t just do interp_expr (Sym "x") and expect it to do the right thing!

We’ll supply this context via another argument to the interpreter. This argument stores a mapping between names (i.e., strings) and values. In, we’ve defined a type called symtab (short for “symbol table”) that maps strings to some other value. We can use symbol tables like this:

$ open Cs164.Util;;
$ let st : int symtab = Symtab.empty;;
$ let st' = Symtab.add "x" 2 st;;
$ Symtab.find "x" st';;
$ Symtab.find_opt "x" st;;
Exception: Not_found

Symbol tables are sort of like dictionaries in Python or Java, except that they are immutable: notice that adding a key to st above returned a new symbol table instead of changing st. (For the curious, our symbol tables use OCaml’s Map type, which is implemented using balanced binary trees.) We’ll be using only a few functions to manipulate our tables:

  • Symtab.empty is the empty symbol table, containing no mappings
  • Symtab.add var value st returns a symbol table with all of st’s mappings plus a mapping between var and value; any previous mapping for var is overwritten
  • Symtab.mem var st returns true if st has a mapping for var and false otherwise
  • Symtab.find var st returns the value var is mapped to in st, raising an exception if no such mapping is present

OK, so we’re going to need to modify our interpreter to take a value symtab as an argument:

let rec interp_exp (env: value symtab) (exp : s_exp) : value = ...

env is short for “environment”. None of our existing expressions modify the environment, so they’ll all just pass it through in their recursive calls.

Now that we have our environment, we can use it to interpret names:

let rec interp_exp (env: value symtab) (exp : s_exp) : value =
  match exp with
  (* some cases elided ... *)
  | Sym var when Symtab.mem var env ->
    Symtab.find var env

That “when” restricts when this pattern fires: we’ll only evaluate the right-hand side of the pattern if the “when” condition is true.

We’ll also need to update interp:

let interp (program : string) : string =
  parse program |> interp_exp Symtab.empty |> string_of_value

Let’s see our interpreter work:

$ Interp.interp_exp Symtab.empty (parse "x");;
<BadExpression error>
$ Interp.interp_exp (Symtab.add "x" (Number 1) Symtab.empty) (parse "x");;
Number 1
$ Interp.interp_exp (Symtab.add "x" (Number 1) Symtab.empty) (parse "(+ x x)");;
Number 2

Cool! Now we just need to implement let, which updates this environment. It’s pretty straightforward:

let rec interp_exp (env : value symtab) (exp : s_exp) : value =
  match exp with
  (* some cases elided... *)
  | Lst [Sym "let"; Lst [Lst [Sym var; e]]; body] ->
      let e_value = interp_exp env e in
      interp_exp (Symtab.add var e_value env) body

You may notice an “extra” Lst constructor; that’s because our let expression looks like (let ((var val)) body). This is how Scheme’s let works, so we’ve done it that way as well. In Homework 3, you’ll add support for multiple variables:

(let ((x 1) (y 2)) (+ x y))