> \[[expression]]<sub>g s</sub> = (value, s')
-With that kind of framework, we can interpret `newref`, `deref`, and `setref` as follows.
+For expressions we already know how to interpret, `s'` will usually just be `s`. One exception is complex expressions like `let var = expr1 in expr2`. Part of interpreting this will be to interpret the sub-expression `expr1`, and we have to allow that in doing that, the store may have already been updated. We want to use that possibly updated store when interpreting `expr2`. Like this:
+
+ let rec eval expression g s =
+ match expression with
+ ...
+ | Let (c, expr1, expr2) ->
+ let (value, s') = eval expr1 g s
+ (* s' may be different from s *)
+ (* now we evaluate expr2 in a new environment where c has been associated
+ with the result of evaluating expr1 in the current environment *)
+ eval expr2 ((c, value) :: g) s'
+ ...
+
+Let's consider how to interpet our new syntactic forms `newref`, `deref`, and `setref`:
+
1. \[[newref starting_val]] should allocate a new reference cell in the store and insert `starting_val` into that cell. It should return some "key" or "index" or "pointer" to the newly created reference cell, so that we can do things like:
let rec eval expression g s =
match expression with
...
- | Newref expr ->
+ | Newref (expr) ->
let (starting_val, s') = eval expr g s
(* note that s' may be different from s, if expr itself contained any mutation operations *)
(* now we want to retrieve the next free index in s' *)
let rec eval expression g s =
match expression with
...
- | Deref expr ->
+ | Deref (expr) ->
let (Index n, s') = eval expr g s
(* note that s' may be different from s, if expr itself contained any mutation operations *)
in (List.nth s' n, s')
let rec eval expression g s =
match expression with
...
- | Setref expr1 expr2
+ | Setref (expr1, expr2) ->
let (Index n, s') = eval expr1 g s
(* note that s' may be different from s, if expr itself contained any mutation operations *)
in let (new_value, s'') = eval expr2 g s'
...
+
+
+
##How to implement implicit-style mutable variables##
With implicit-style mutation, we don't have new syntactic forms like `newref` and `deref`. Instead, we just treat ordinary variables as being mutable. You could if you wanted to have some variables be mutable and others not; perhaps the first sort are written in Greek and the second in Latin. But we will suppose all variables in our language are mutable.
To handle implicit-style mutation, we'll need to re-implement the way we interpret expressions like `x` and `let x = expr1 in expr2`. We will also have just one new syntactic form, `change x to expr1 then expr2`.
-Here's how to implement these. We'll suppose that our assignment function is list of pairs, as in [week6](/reader_monad_for_variable_binding).
+Here's how to implement these. We'll suppose that our assignment function is list of pairs, as in [week7](/reader_monad_for_variable_binding).
let rec eval expression g s =
match expression with
in let value = List.nth s index
in (value, s)
- | Let (c : char) expr1 expr2 ->
+ | Let ((c : char), expr1, expr2) ->
let (starting_val, s') = eval expr1 g s
(* get next free index in s' *)
in let new_index = List.length s'
(* evaluate expr2 using a new assignment function and store *)
in eval expr2 ((c, new_index) :: g) s''
- | Change (c : char) expr1 expr2 ->
+ | Change ((c : char), expr1, expr2) ->
let (new_value, s') = eval expr1 g s
(* lookup which index is associated with Var c *)
in let index = List.assoc c g