(C has explicit-style mutable variables, too, which it calls *pointers*. But simple variables in C are already mutable, in the implicit style.)
-When dealing with explicit-style mutation, there's a difference between the types and values of `ycell` and `!ycell` (or `(unbox ycell)`). The former has the type `int ref`: the variable `ycell` is assigned a reference cell that contains an `int`. The latter has the type `int`, and has whatever value is now stored in the relevant reference cell. In an implicit-style framework though, we only have the resources to refer to the contents of the relevant reference cell. `y` in fragment [G] or the C snippet above has the type `int`, and only ever evaluates to `int` values.
+When dealing with explicit-style mutation, there's a difference between the types and values of `ycell` and `!ycell` (or in Scheme, `(unbox ycell)`). The former has the type `int ref`: the variable `ycell` is assigned a reference cell that contains an `int`. The latter has the type `int`, and has whatever value is now stored in the relevant reference cell. In an implicit-style framework though, we only have the resources to refer to the contents of the relevant reference cell. `y` in fragment [G] or the C snippet above has the type `int`, and only ever evaluates to `int` values.
##Controlling order##
> \[[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'
+ ...
+
+Similarly:
+
+ ...
+ | Addition (expr1, expr2) ->
+ let (value1, s') = eval expr1 g s
+ in let (value2, s'') = eval expr2 g s'
+ in (value1 + value2, 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' *)
in (Index new_index, s'')
...
-2. When `expr` evaluates to a `store_index`, then `deref expr` should evaluate to whatever value is at that index in the current store. (If `expr` evaluates to a value of another type, `deref expr` is undefined.) In this operation, we don't change the store at all; we're just reading from it. So we'll return the same store back unchanged.
+2. When `expr` evaluates to a `store_index`, then `deref expr` should evaluate to whatever value is at that index in the current store. (If `expr` evaluates to a value of another type, `deref expr` is undefined.) In this operation, we don't change the store at all; we're just reading from it. So we'll return the same store back unchanged (assuming it wasn't changed during the evaluation of `expr`).
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 *)
+ (* note that s' may be different from s, if expr1 itself contained any mutation operations *)
in let (new_value, s'') = eval expr2 g s'
(* now we create a list which is just like s'' except it has new_value in index n *)
in let rec replace_nth lst m =
...
+
+
+
##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